Springer onographs in Mathematics - Université Lorrainemath.univ-metz.fr/~gnc/bibliographie/Ceccherini... · 2013-03-22 · models for self-reproducing machines. About twenty years

Springer Monographs in Mathematics

For further volumes:www.springer.com/series/3733

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Tullio Ceccherini-Silberstein � Michel Coornaert

Cellular Automataand Groups

Tullio Ceccherini-SilbersteinDipartimento di IngegneriaUniversità del SannioC.so Garibaldi 10782100 [email protected]

Michel CoornaertInstitut de Recherche Mathématique AvancéeUniversité de Strasbourg7 rue René-Descartes67084 Strasbourg [email protected]

ISSN 1439-7382ISBN 978-3-642-14033-4 e-ISBN 978-3-642-14034-1DOI 10.1007/978-3-642-14034-1Springer Heidelberg Dordrecht London New York

Library of Congress Control Number: 2010934641

Mathematics Subject Classification (2010): 37B15, 68Q80, 20F65, 43A07, 16S34, 20C07

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To Katiuscia, Giacomo, and Tommaso

To Martine and Nathalie

Preface

Two seemingly unrelated mathematical notions, namely that of an amenablegroup and that of a cellular automaton, were both introduced by John vonNeumann in the first half of the last century. Amenability, which originatedfrom the study of the Banach-Tarski paradox, is a property of groups gener-alizing both commutativity and finiteness. Nowadays, it plays an importantrole in many areas of mathematics such as representation theory, harmonicanalysis, ergodic theory, geometric group theory, probability theory, and dy-namical systems. Von Neumann used cellular automata to serve as theoreticalmodels for self-reproducing machines. About twenty years later, the famouscellular automaton associated with the Game of Life was invented by JohnHorton Conway and popularized by Martin Gardner. The theory of cellularautomata flourished as one of the main branches of computer science. Deepconnections with complexity theory and logic emerged from the discoverythat some cellular automata are universal Turing machines.

A group G is said to be amenable (as a discrete group) if the set of allsubsets of G admits a right-invariant finitely additive probability measure.All finite groups, all solvable groups (and therefore all abelian groups), andall finitely generated groups of subexponential growth are amenable. VonNeumann observed that the class of amenable groups is closed under theoperation of taking subgroups and that the free group of rank two F2 is non-amenable. It follows that a group which contains a subgroup isomorphic toF2 is non-amenable. However, there are examples of groups which are non-amenable and contain no subgroups isomorphic to F2 (the first examplesof such groups were discovered by Alexander Y. Ol’shanskii and by SergeiI. Adyan).

Loosely speaking, a general cellular automaton can be described as fol-lows. A configuration is a map from a set called the universe into anotherset called the alphabet. The elements of the universe are called cells and theelements of the alphabet are called states. A cellular automaton is then amap from the set of all configurations into itself satisfying the following localproperty: the state of the image configuration at a given cell only depends on

vii

viii Preface

the states of the initial configuration on a finite neighborhood of the givencell. In the classical setting, for instance in the cellular automata constructedby von Neumann and the one associated with Conway’s Game of Life, thealphabet is finite, the universe is the two dimensional infinite square lattice,and the neighborhood of a cell consists of the cell itself and its eight adjacentcells. By iterating a cellular automaton one gets a discrete dynamical system.Such dynamical systems have proved very useful to model complex systemsarising from natural sciences, in particular physics, biology, chemistry, andpopulation dynamics.

∗∗ ∗

In this book, the universe will always be a group G (in the classical settingthe corresponding group was G = Z

2) and the alphabet may be finite orinfinite. The left multiplication in G induces a natural action of G on the setof configurations which is called the G-shift and all cellular automata will berequired to commute with the shift.

It was soon realized that the question whether a given cellular automatonis surjective or not needs a special attention. From the dynamical viewpoint,surjectivity means that each configuration may be reached at any time. Thefirst important result in this direction is the celebrated theorem of Moore andMyhill which gives a necessary and sufficient condition for the surjectivity ofa cellular automaton with finite alphabet over the group G = Z

2. Edward F.Moore and John R. Myhill proved that such a cellular automaton is surjectiveif and only if it is pre-injective. As the term suggests it, pre-injectivity is aweaker notion than injectivity. More precisely, a cellular automaton is said tobe pre-injective if two configurations are equal whenever they have the sameimage and coincide outside a finite subset of the group. Moore proved the“surjective ⇒ pre-injective” part and Myhill proved the converse implicationshortly after. One often refers to this result as to the Garden of Eden theorem.This biblical terminology is motivated by the fact that, regarding a cellularautomaton as a dynamical system with discrete time, a configuration whichis not in the image of the cellular automaton may only appear as an initialconfiguration, that is, at time t = 0.

The surprising connection between amenability and cellular automata wasestablished in 1997 when Antonio Machı, Fabio Scarabotti and the first au-thor proved the Garden of Eden theorem for cellular automata with finitealphabets over amenable groups. At the same time, and completely inde-pendently, Misha Gromov, using a notion of spacial entropy, presented amore general form of the Garden of Eden theorem where the universe isan amenable graph with a dense holonomy and cellular automata are calledmaps of bounded propagation. Machı, Scarabotti and the first author alsoshowed that both implications in the Garden of Eden theorem become falseif the underlying group contains a subgroup isomorphic to F2. The questionwhether the Garden of Eden theorem could be extended beyond the class of

Preface ix

amenable groups remained open until 2008 when Laurent Bartholdi provedthat the Moore implication fails to hold for non-amenable groups. As a conse-quence, the whole Garden of Eden theorem only holds for amenable groups.This gives a new characterization of amenable groups in terms of cellular au-tomata. Let us mention that, up to now, the validity of the Myhill implicationfor non-amenable groups is still an open problem.

Following Walter H. Gottschalk, a group G is said to be surjunctive if everyinjective cellular automaton with finite alphabet over G is surjective. WayneLawton proved that all residually finite groups are surjunctive and that ev-ery subgroup of a surjunctive group is surjunctive. Since injectivity impliespre-injectivity, an immediate consequence of the Garden of Eden theorem foramenable groups is that every amenable group is surjunctive. Gromov andBenjamin Weiss introduced a class of groups, called sofic groups, which in-cludes all residually finite groups and all amenable groups, and proved thatevery sofic group is surjunctive. Sofic groups can be defined in three equivalentways: in terms of local approximation by finite symmetric groups equippedwith their Hamming distance, in terms of local approximation of their Cayleygraphs by finite labelled graphs, and, finally, as being the groups that canbe embedded into ultraproducts of finite symmetric groups (this last charac-terization is due to Gabor Elek and Endre Szabo). The class of sofic groupsis the largest known class of surjunctive groups. It is not known, up to now,whether all groups are surjunctive (resp. sofic) or not.

Stimulated by Gromov ideas, we considered cellular automata whose al-phabets are vector spaces. In this framework, the space of configurations hasa natural structure of a vector space and cellular automata are required tobe linear. An analogue of the Garden of Eden theorem was proved for linearcellular automata with finite dimensional alphabets over amenable groups. Inthe proof, the role of entropy, used in the finite alphabet case, is now playedby the mean dimension, a notion introduced by Gromov. Also, examples oflinear cellular automata with finite dimensional alphabets over groups con-taining F2 showing that the linear version of the Garden of Eden theoremmay fail to hold in this case, were provided. It is not known, up to now, ifthe Garden of Eden theorem for linear cellular automata with finite dimen-sional alphabet only holds for amenable groups or not. We also introducedthe notion of linear surjunctivity: a group G is said to be L-surjunctive if ev-ery injective linear cellular automaton with finite dimensional alphabet overG is surjective. We proved that every sofic group is L-surjunctive. Linearcellular automata over a group G with alphabet of finite dimension d overa field K may be represented by d × d matrices with entries in the groupring K[G]. This leads to the following characterization of L-surjunctivity: agroup is L-surjunctive if and only if it satisfies Kaplansky’s conjecture on thestable finiteness of group rings (a ring is said to be stably finite if one-sidedinvertible finite dimensional square matrices with coefficients in that ring arein fact two-sided invertible). As a corollary, one has that group rings of soficgroups are stably finite, a result previously established by Elek and Szabo

x Preface

using different methods. Moreover, given a group G and a field K, the pre-injectivity of all nonzero linear cellular automata with alphabet K over G isequivalent to the absence of zero-divisors in K[G]. As a consequence, anotherimportant problem on the structure of group rings also formulated by IrvingKaplansky may be expressed in terms of cellular automata. Is every nonzerolinear cellular automaton with one-dimensional alphabet over a torsion-freegroup always pre-injective?

∗∗ ∗

The material presented in this book is entirely self-contained. In fact, itsreading only requires some acquaintance with undergraduate general topol-ogy and abstract algebra. Each chapter begins with a brief overview of itscontents and ends with some historical notes and a list of exercises at variousdifficulty levels. Some additional topics, such as subshifts and cellular au-tomata over subshifts, are treated in these exercises. Hints are provided eachtime help may be needed. In order to improve accessibility, a few appendicesare included to quickly introduce the reader to facts he might be not toofamiliar with.

In the first chapter, we give the definition of a cellular automaton. Wepresent some basic examples and discuss general methods for constructingcellular automata. We equip the set of configurations with its prodiscreteuniform structure and prove the generalized Curtis-Hedlund theorem: a nec-essary and sufficient condition for a self-mapping of the configuration spaceto be a cellular automaton is that it is uniformly continuous and commuteswith the shift.

Chapter 2 is devoted to residually finite groups. We give several equiv-alent characterizations of residual finiteness and prove that the class ofresidually finite groups is closed under taking subgroups and projective lim-its. We establish in particular the theorems, respectively due to Anatoly I.Mal’cev and Gilbert Baumslag, which assert that finitely generated residuallyfinite groups are Hopfian and that their automorphism group is residually fi-nite.

Surjunctive groups are introduced in Chap. 3. We show that every sub-group of a surjunctive group is surjunctive and that locally residually finitegroups are surjunctive. We also prove a theorem of Gromov which says thatlimits of surjunctive marked groups are surjunctive.

The theory of amenable groups is developed in Chap. 4. The class ofamenable groups is closed under taking subgroups, quotients, extensions, andinductive limits. We prove the theorems due to Erling Følner and AlfredTarski which state the equivalence between amenability, the existence of aFølner net, and the non-existence of a paradoxical decomposition.

The Garden of Eden theorem is established in Chap. 5. It is proved byshowing that both surjectivity and pre-injectivity of the cellular automatonare equivalent to the fact that the image of the configuration space has maxi-

Preface xi

mal entropy. We give an example of a cellular automaton with finite alphabetover F2 which is pre-injective but not surjective. Following Bartholdi’s con-struction, we also prove the existence of a surjective but not pre-injectivecellular automaton with finite alphabet over any non-amenable group.

In Chap. 6 we present the basic elementary notions and results on growthof finitely generated groups. We prove that finitely generated nilpotent groupshave polynomial growth. We then introduce the Grigorchuk group and showthat it is an infinite finitely generated periodic group of intermediate growth.We show that every finitely generated group of subexponential growth isamenable. We also establish the Kesten-Day characterization of amenabilitywhich asserts that a group with a finite (not necessarily symmetric) generat-ing subset is amenable if and only if 0 is in the �2-spectrum of the associatedLaplacian. Finally, we consider the notion of quasi-isometry for not neces-sarily countable groups and we show that amenability is a quasi-isometryinvariant.

In Chap. 7 we consider the notion of local embeddability of groups intoa class of groups. For the class of finite groups, this gives the class of LEFgroups introduced by Anatoly M. Vershik and Edward I. Gordon. We dis-cuss several stability properties of local embeddability and show that locallyembeddable groups are closed in marked groups spaces. The remaining ofthe chapter is devoted to the class of sofic groups. We show that the threedefinitions, namely analytic, geometric, and algebraic, we alluded to before,are equivalent. We then prove the Gromov-Weiss theorem which states thatevery sofic group is surjunctive.

The last chapter is devoted to linear cellular automata. We prove the linearversion of the Garden of Eden theorem and show that every sofic group isL-surjunctive. We end the chapter with a discussion on the stable finitenessand the zero-divisors conjectures of Kaplansky and their reformulation interms of linear cellular automata.

Appendix A gives a quick overview of a few fundamental notions and re-sults of topology (nets, compactness, product topology, and the Tychonoffproduct theorem). Appendix B is devoted to Andre Weil’s theory of uniformspaces. It includes also a detailed exposition of the Hausdorff-Bourbaki uni-form structure on subsets of a uniform space. In Appendix C, we establishsome basic properties of symmetric groups and prove the simplicity of thealternating groups. The definition and the construction of free groups aregiven in Appendix D. The proof of Klein’s ping-pong lemma is also includedthere. In Appendix E we shortly describe the constructions of inductive andprojective limits of groups. Appendix F treats topological vector spaces, theweak-∗ topology, and the Banach-Alaoglu theorem. The proof of the Markov-Kakutani fixed point theorem is presented in Appendix G. In the subsequentappendix, of a pure graph-theoretical and combinatorial flavour, we considerbipartite graphs and their matchings. We prove Hall’s marriage theorem andits harem version which plays a key role in the proof of Tarski’s theorem onamenability. The Baire theorem, the open mapping theorem, as well as other

xii Preface

complements of functional analysis including uniform convexity are treatedin Appendix I. The last appendix deals with the notions of filters and ultra-filters.

We would like to express our deep gratitude to Dr. Catriona Byrne, Dr.Marina Reizakis and Annika Eling from Springer Verlag and to DonatasAkmanavicius for their constant and kindest help at all stages of the editorialprocess.

Rome and Strasbourg Tullio Ceccherini-SilbersteinMichel Coornaert

Contents

1 Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The Configuration Set and the Shift Action . . . . . . . . . . . . . . . . 11.2 The Prodiscrete Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Periodic Configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Minimal Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.6 Cellular Automata over Quotient Groups . . . . . . . . . . . . . . . . . . 151.7 Induction and Restriction of Cellular Automata . . . . . . . . . . . . 161.8 Cellular Automata with Finite Alphabets . . . . . . . . . . . . . . . . . . 201.9 The Prodiscrete Uniform Structure . . . . . . . . . . . . . . . . . . . . . . . 221.10 Invertible Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2 Residually Finite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.1 Definition and First Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.2 Stability Properties of Residually Finite Groups . . . . . . . . . . . . 402.3 Residual Finiteness of Free Groups . . . . . . . . . . . . . . . . . . . . . . . 422.4 Hopfian Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.5 Automorphism Groups of Residually Finite Groups . . . . . . . . . 452.6 Examples of Finitely Generated Groups Which Are Not

Residually Finite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.7 Dynamical Characterization of Residual Finiteness . . . . . . . . . . 50Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3 Surjunctive Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.2 Stability Properties of Surjunctive Groups . . . . . . . . . . . . . . . . . 583.3 Surjunctivity of Locally Residually Finite Groups . . . . . . . . . . . 59

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3.4 Marked Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.5 Expansive Actions on Uniform Spaces . . . . . . . . . . . . . . . . . . . . . 643.6 Gromov’s Injectivity Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.7 Closedness of Marked Surjunctive Groups . . . . . . . . . . . . . . . . . 67Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

4 Amenable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.1 Measures and Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.2 Properties of the Set of Means . . . . . . . . . . . . . . . . . . . . . . . . . . . 824.3 Measures and Means on Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 834.4 Definition of Amenability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.5 Stability Properties of Amenable Groups . . . . . . . . . . . . . . . . . . 884.6 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924.7 The Følner Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 944.8 Paradoxical Decompositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 984.9 The Theorems of Tarski and Følner . . . . . . . . . . . . . . . . . . . . . . . 994.10 The Fixed Point Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

5 The Garden of Eden Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1115.1 Garden of Eden Configurations and Garden of Eden Patterns 1115.2 Pre-injective Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1125.3 Statement of the Garden of Eden Theorem . . . . . . . . . . . . . . . . 1145.4 Interiors, Closures, and Boundaries . . . . . . . . . . . . . . . . . . . . . . . 1155.5 Mutually Erasable Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1215.6 Tilings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1225.7 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1255.8 Proof of the Garden of Eden Theorem . . . . . . . . . . . . . . . . . . . . 1285.9 Surjunctivity of Locally Residually Amenable Groups . . . . . . . 1315.10 A Surjective but Not Pre-injective Cellular Automaton

over F2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1335.11 A Pre-injective but Not Surjective Cellular Automaton

over F2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1335.12 A Characterization of Amenability in Terms of Cellular

Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1355.13 Garden of Eden Patterns for Life . . . . . . . . . . . . . . . . . . . . . . . . . 136Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

6 Finitely Generated Amenable Groups . . . . . . . . . . . . . . . . . . . . 1516.1 The Word Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1516.2 Labeled Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1536.3 Cayley Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1566.4 Growth Functions and Growth Types . . . . . . . . . . . . . . . . . . . . . 160

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6.5 The Growth Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1686.6 Growth of Subgroups and Quotients . . . . . . . . . . . . . . . . . . . . . . 1706.7 A Finitely Generated Metabelian Group with Exponential

Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1736.8 Growth of Finitely Generated Nilpotent Groups . . . . . . . . . . . . 1756.9 The Grigorchuk Group and Its Growth . . . . . . . . . . . . . . . . . . . . 1786.10 The Følner Condition for Finitely Generated Groups . . . . . . . . 1916.11 Amenability of Groups of Subexponential Growth . . . . . . . . . . 1926.12 The Theorems of Kesten and Day . . . . . . . . . . . . . . . . . . . . . . . . 1936.13 Quasi-Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

7 Local Embeddability and Sofic Groups . . . . . . . . . . . . . . . . . . . . 2337.1 Local Embeddability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2347.2 Local Embeddability and Ultraproducts . . . . . . . . . . . . . . . . . . . 2437.3 LEF-Groups and LEA-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 2467.4 The Hamming Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2517.5 Sofic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2547.6 Sofic Groups and Metric Ultraproducts of Finite Symmetric

Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2607.7 A Characterization of Finitely Generated Sofic Groups . . . . . . 2657.8 Surjunctivity of Sofic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

8 Linear Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2838.1 The Algebra of Linear Cellular Automata . . . . . . . . . . . . . . . . . 2848.2 Configurations with Finite Support . . . . . . . . . . . . . . . . . . . . . . . 2888.3 Restriction and Induction of Linear Cellular Automata . . . . . . 2898.4 Group Rings and Group Algebras . . . . . . . . . . . . . . . . . . . . . . . . 2918.5 Group Ring Representation of Linear Cellular Automata . . . . 2948.6 Modules over a Group Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2998.7 Matrix Representation of Linear Cellular Automata . . . . . . . . . 3018.8 The Closed Image Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3058.9 The Garden of Eden Theorem for Linear Cellular Automata . 3088.10 Pre-injective but not Surjective Linear Cellular Automata . . . 3148.11 Surjective but not Pre-injective Linear Cellular Automata . . . 3158.12 Invertible Linear Cellular Automata . . . . . . . . . . . . . . . . . . . . . . 3178.13 Pre-injectivity and Surjectivity of the Discrete Laplacian . . . . 3218.14 Linear Surjunctivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3248.15 Stable Finiteness of Group Algebras . . . . . . . . . . . . . . . . . . . . . . 3278.16 Zero-Divisors in Group Algebras and Pre-injectivity

of One-Dimensional Linear Cellular Automata . . . . . . . . . . . . . 330Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338

xvi Contents

A Nets and the Tychonoff Product Theorem . . . . . . . . . . . . . . . . 343A.1 Directed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343A.2 Nets in Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343A.3 Initial Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346A.4 Product Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346A.5 The Tychonoff Product Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 347Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349

B Uniform Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351B.1 Uniform Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351B.2 Uniformly Continuous Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353B.3 Product of Uniform Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355B.4 The Hausdorff-Bourbaki Uniform Structure on Subsets . . . . . . 356Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358

C Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359C.1 The Symmetric Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359C.2 Permutations with Finite Support . . . . . . . . . . . . . . . . . . . . . . . . 360C.3 Conjugacy Classes in Sym0(X) . . . . . . . . . . . . . . . . . . . . . . . . . . . 362C.4 The Alternating Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363

D Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367D.1 Concatenation of Words . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367D.2 Definition and Construction of Free Groups . . . . . . . . . . . . . . . . 367D.3 Reduced Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373D.4 Presentations of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375D.5 The Klein Ping-Pong Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 376

E Inductive Limits and Projective Limits of Groups . . . . . . . . 379E.1 Inductive Limits of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379E.2 Projective Limits of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

F The Banach-Alaoglu Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383F.1 Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383F.2 The Weak-∗ Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384F.3 The Banach-Alaoglu Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

G The Markov-Kakutani Fixed Point Theorem . . . . . . . . . . . . . . 387G.1 Statement of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387G.2 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389

H The Hall Harem Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391H.1 Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391H.2 Matchings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393

Contents xvii

H.3 The Hall Marriage Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394H.4 The Hall Harem Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401

I Complements of Functional Analysis . . . . . . . . . . . . . . . . . . . . . . 403I.1 The Baire Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403I.2 The Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404I.3 Spectra of Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406I.4 Uniform Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407

J Ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409J.1 Filters and Ultrafilters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409J.2 Limits Along Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421

List of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433

Notation

Throughout this book, the following conventions are used:

• N is the set of nonnegative integers so that 0 ∈ N;• the notation A ⊂ B means that each element in the set A is also in the

set B so that A and B may coincide;• a countable set is a set which admits a bijection onto a subset of N so that

finite sets are countable;• all group actions are left actions;• all rings are assumed to be associative (but not necessarily commutative)

with a unity element;• a field is a nonzero commutative ring in which each nonzero element is

invertible.

xix

Chapter 1

Cellular Automata

In this chapter we introduce the notion of a cellular automaton. We fix agroup and an arbitrary set which will be called the alphabet. A configura-tion is defined as being a map from the group into the alphabet. Thus, aconfiguration is a way of attaching an element of the alphabet to each el-ement of the group. There is a natural action of the group on the set ofconfigurations which is called the shift action (see Sect. 1.1). A cellular au-tomaton is a self-mapping of the set of configurations defined from a systemof local rules commuting with the shift (see Definition 1.4.1). We equip theconfiguration set with the prodiscrete topology, that is, the topology of point-wise convergence associated with the discrete topology on the alphabet (seeSect. 1.2). It turns out that every cellular automaton is continuous with re-spect to the prodiscrete topology (Proposition 1.4.8) and commutes with theshift (Proposition 1.4.4). Conversely, when the alphabet is finite, every con-tinuous self-mapping of the configuration space which commutes with theshift is a cellular automaton (Theorem 1.8.1). Another important fact in thefinite alphabet case is that every bijective cellular automaton is invertible, inthe sense that its inverse map is also a cellular automaton (Theorem 1.10.2).We give examples showing that, when the alphabet is infinite, a continuousself-mapping of the configuration space which commutes with the shift mayfail to be a cellular automaton and a bijective cellular automaton may failto be invertible. In Sect. 1.9, we introduce the prodiscrete uniform structureon the configuration space. We show that a self-mapping of the configurationspace is a cellular automaton if and only if it is uniformly continuous andcommutes with the shift (Theorem 1.9.1).

1.1 The Configuration Set and the Shift Action

Let G be a group. For g ∈ G, denote by Lg the left multiplication by g in G,that is, the map Lg : G → G given by

T. Ceccherini-Silberstein, M. Coornaert, Cellular Automata and Groups,Springer Monographs in Mathematics,DOI 10.1007/978-3-642-14034-1 1, © Springer-Verlag Berlin Heidelberg 2010

1

http://dx.doi.org/10.1007/978-3-642-14034-1_1

2 1 Cellular Automata

Lg(g′) = gg′ for all g′ ∈ G.

Observe that for all g1, g2, g′ ∈ G one has

(Lg1 ◦ Lg2) (g′) = Lg1(Lg2(g′)) = Lg1(g2g

′) = g1g2g′ = Lg1g2(g

′)

which shows thatLg1 ◦ Lg2 = Lg1g2 . (1.1)

Let A be a set. Consider the set AG consisting of all maps from G to A:

AG =∏

g∈G

A = {x : G → A}.

The set A is called the alphabet . The elements of A are called the letters, orthe states, or the symbols, or the colors. The group G is called the universe.The set AG is called the set of configurations.

Given an element g ∈ G and a configuration x ∈ AG, we define the con-figuration gx ∈ AG by

gx = x ◦ Lg−1 . (1.2)

Thus one hasgx(g′) = x(g−1g′) for all g′ ∈ G.

The map

G × AG → AG

(g, x) �→ gx

is a left action of G on AG. Indeed, for all g1, g2 ∈ G and x ∈ AG, one has

g1(g2x) = g1(x ◦ Lg−12

) = x ◦ Lg−12

◦ Lg−11

= x ◦ Lg−12 g−1

1= x ◦ L(g1g2)−1

= (g1g2)x,

where the third equality follows from (1.1). Also, denoting by 1G the identityelement of G and by IdG : G → G the identity map, one has

1Gx = x ◦ L1G= x ◦ IdG = x.

This left action of G on AG is called the G-shift on AG.A pattern over the group G and the alphabet A is a map p : Ω → G defined

on some finite subset Ω of G. The set Ω is then called the support of p.

1.3 Periodic Configurations 3

1.2 The Prodiscrete Topology

Let G be a group and let A be a set.We equip each factor A of AG with the discrete topology (all subsets of

A are open) and AG with the associated product topology (see Sect. A.4).This topology is called the prodiscrete topology on AG. This is the small-est topology on AG for which the projection map πg : AG → A, given byπg(x) = x(g), is continuous for every g ∈ G (cf. Sect. A.4). The elementarycylinders

C(g, a) = π−1g ({a}) = {x ∈ AG : x(g) = a} (g ∈ G, a ∈ A)

are both open and closed in AG. A subset U ⊂ AG is open if and only ifU can be expressed as a (finite or infinite) union of finite intersections ofelementary cylinders.

For a subset Ω ⊂ G and a configuration x ∈ AG let x|Ω ∈ AΩ denote therestriction of x to Ω, that is, the map x|Ω : Ω → A defined by x|Ω(g) = x(g)for all g ∈ Ω.

If x ∈ AG, a neighborhood base of x is given by the sets

V (x,Ω) = {y ∈ AG : x|Ω = y|Ω} =⋂

g∈Ω

C(g, x(g)), (1.3)

where Ω runs over all finite subsets of G.

Proposition 1.2.1. The space AG is Hausdorff and totally disconnected.

Proof. The discrete topology on A is Hausdorff and totally disconnected,and, by Proposition A.4.1 and Proposition A.4.2, a product of Hausdorff(resp. totally disconnected) topological spaces is Hausdorff (resp. totally dis-connected). ��

Recall that an action of a group G on a topological space X is said to becontinuous if the map ϕg : X → X given by ϕg(x) = gx is continuous on Xfor each g ∈ G.

Proposition 1.2.2. The action of G on AG is continuous.

Proof. Let g ∈ G and consider the map ϕg : AG → AG defined by ϕg(x) = gx.The map πh◦ϕg is equal to πg−1h and is therefore continuous on AG for everyh ∈ G. Consequently, ϕg is continuous (cf. Sect. A.4). ��

1.3 Periodic Configurations

Let G be a group and let A be a set. Let H be a subgroup of G. A configurationx ∈ AG is called H-periodic if x is fixed by H, that is, if one has


hx = x for all h ∈ H.

Let Fix(H) denote the subset of AG consisting of all H-periodic configura-tions.

Examples 1.3.1. (a) One has Fix({1G}) = AG.(b) The set Fix(G) consists of all constant configurations and may be

therefore identified with A.(c) For G = Z and H = nZ, n ≥ 1, the set Fix(H) is the set of sequences

x : Z → A which admit n as a (not necessarily minimal) period, that is, suchthat x(i + n) = x(i) for all i ∈ Z.

Proposition 1.3.2. Let H be a subgroup of G. Then the set Fix(H) is closedin AG for the prodiscrete topology.

Proof. We haveFix(H) =

⋂

h∈H

{x ∈ AG : hx = x}. (1.4)

The space AG is Hausdorff by Proposition 1.2.1 and the action of G on AG

is continuous by Proposition 1.2.2. Thus the set of fixed points of the mapx �→ gx is closed in AG for each g ∈ G. Therefore Fix(H) is closed in AG

by (1.4). ��Consider the set H\G = {Hg : g ∈ G} consisting of all right cosets of H

in G and the canonical surjective map

ρ : G → H\Gg �→ Hg.

Given an element y ∈ AH\G, i.e., a map y : H\G → A, we can form thecomposite map y ◦ ρ : G → A which is an element of AG. In fact, we havey ◦ ρ ∈ Fix(H) since

(h(y ◦ ρ))(g) = y ◦ ρ(h−1g) = y(ρ(h−1g)) = y(ρ(g)) = y ◦ ρ(g)

for all g ∈ G and h ∈ H.

Proposition 1.3.3. Let H be a subgroup of G and let denote by ρ : G →H\G the canonical surjection. Then the map ρ∗ : AH\G → Fix(H) defined byρ∗(y) = y ◦ ρ for all y ∈ AH\G is bijective.

Proof. If y1, y2 ∈ AH\G satisfy y1◦ρ = y2◦ρ, then y1 = y2 since ρ is surjective.Thus ρ∗ is injective.

If x ∈ Fix(H), then hx = x for all h ∈ H, that is,

x(h−1g) = x(g) for all h ∈ H, g ∈ G.

Thus, the configuration x is constant on each right coset of G modulo H,that is, x is in the image of ρ∗. This shows that ρ∗ is surjective. ��

1.3 Periodic Configurations 5

Corollary 1.3.4. If the set A is finite and H is a subgroup of finite indexof G, then the set Fix(H) is finite and one has |Fix(H)| = |A|[G:H], where[G : H] denotes the index of H in G. ��

Example 1.3.5. Let G = Z and H = nZ, where n ≥ 1. If A is finite ofcardinality k, then |Fix(H)| = kn.

Suppose now that N is a normal subgroup of G, that is, gN = Ng for allg ∈ G. Then, there is a natural group structure on G/N = N\G for whichthe canonical surjection ρ : G → G/N is a homomorphism.

Proposition 1.3.6. Let N be a normal subgroup of G. Then Fix(N) is aG-invariant subset of AG.

Proof. Let x ∈ Fix(N) and g ∈ G. Given h ∈ N , then there exists h′ ∈ Nsuch that hg = gh′, since N is normal in G. Thus, we have

h(gx) = g(h′x) = gx

which shows that gx ∈ Fix(N). ��

Since every element of Fix(N) is fixed by N , the action of G on Fix(N)induces an action of G/N on Fix(N) which satisfies ρ(g)x = gx for all g ∈ Gand x ∈ Fix(N).

Suppose that a group Γ acts on two sets X and Y . A map ϕ : X → Y iscalled Γ -equivariant if one has ϕ(γx) = γϕ(x) for all γ ∈ Γ and x ∈ X.

Proposition 1.3.7. Let N be a normal subgroup of G and let ρ : G → G/Ndenote the canonical epimorphism. Then the map ρ∗ : AG/N → Fix(N) de-fined by ρ∗(y) = y ◦ ρ for all y ∈ AG/N is a G/N -equivariant bijection.

Proof. We already know that ρ∗ is bijective (Proposition 1.3.3).Let g ∈ G and y ∈ AG/N . For all g′ ∈ G, we have

ρ(g)ρ∗(y)(g′) = gρ∗(y)(g′)

= ρ∗(y)(g−1g′)

= (y ◦ ρ)(g−1g′)

= y(ρ(g−1g′))

= y((ρ(g))−1ρ(g′))= ρ(g)y(ρ(g′))= ρ∗(ρ(g)y)(g′).

Thus ρ(g)ρ∗(y) = ρ∗(ρ(g)y). This shows that ρ∗ is G/N -equivariant. ��


1.4 Cellular Automata

Let G be a group and let A be a set.

Definition 1.4.1. A cellular automaton over the group G and the alphabetA is a map τ : AG → AG satisfying the following property: there exist a finitesubset S ⊂ G and a map μ : AS → A such that

τ(x)(g) = μ((g−1x)|S) (1.5)

for all x ∈ AG and g ∈ G, where (g−1x)|S denotes the restriction of theconfiguration g−1x to S.

Such a set S is called a memory set and μ is called a local defining mapfor τ .

Observe that formula (1.5) says that the value of the configuration τ(x) at anelement g ∈ G is the value taken by the local defining map μ at the patternobtained by restricting to the memory set S the shifted configuration g−1x.

Remark 1.4.2. (a) Equality (1.5) may also be written

τ(x)(g) = μ((x ◦ Lg)|S) (1.6)

by (1.2).(b) For g = 1G, formula (1.5) gives us

τ(x)(1G) = μ(x|S). (1.7)

As the restriction map AG → AS , x �→ x|S , is surjective, this shows that ifS is a memory set for the cellular automaton τ , then there is a unique mapμ : AS → A which satisfies (1.5). Thus one says that this unique μ is the localdefining map for τ associated with the memory set S.

Examples 1.4.3. (a) The cellular automaton associated with the Gameof Life. Consider an infinite two-dimensional orthogonal grid of square cells,each of which is in one of two possible states, live or dead. Every cell cinteracts with its eight neighboring cells, namely the North, North-East, East,South-East, South, South-West, West and North-West cells (see Fig. 1.1).

At each step in time, the following rules for the evolution of the states ofthe cells are applied (in Figs. 1.2–1.5 we label with a “•” a live cell and witha “◦” a dead cell):

• (birth): a cell that is dead at time t becomes alive at time t + 1 if andonly if three of its neighbors are alive at time t (cf. Fig. 1.2);

• (survival): a cell that is alive at time t will remain alive at time t + 1if and only if it has exactly two or three live neighbors at time t (cf.Fig. 1.3);

1.4 Cellular Automata 7

Fig. 1.1 The cell c and its eight neighboring cells

Fig. 1.2 A cell that is dead at time t becomes alive at time t + 1 if and only if three ofits neighbors are alive at time t

Fig. 1.3 A cell that is alive at time t will remain alive at time t + 1 if and only if it has

exactly two or three live neighbors at time t


Fig. 1.4 A live cell that has at most one live neighbor at time t will be dead at time t+1

• (death by loneliness): a live cell that has at most one live neighbor attime t will be dead at time t + 1 (cf. Fig. 1.4);

• (death by overcrowding): a cell that is alive at time t and has four ormore live neighbors at time t, will be dead at time t + 1 (cf. Fig. 1.5).

Fig. 1.5 A cell that is alive at time t and has four or more live neighbors at time t, willbe dead at time t + 1

Let us show that the map which transforms a configuration of cells at timet into the configuration at time t + 1 according to the above rules is indeeda cellular automaton.

Consider the group G = Z2 and the finite set S = {−1, 0, 1}2 ⊂ G. Then

there is a one-to-one correspondence between the cells in the grid and theelements in G in such a way that the following holds. If c is a given cell, thenc+(0, 1) is the neighboring North cell, c+(1, 1) is the neighboring North-Eastcell, and so on; in other words, c and its eight neighboring cells correspondto the group elements c + s with s ∈ S (see Fig. 1.6).

Consider the alphabet A = {0, 1}. The state 0 (resp. 1) corresponds toabsence (resp. presence) of life. With each configuration of the states of thecells in the grid we associate a map x ∈ AG defined as follows. Given a cell cwe set x(c) = 1 (resp. 0) if the cell c is alive (resp. dead).


Fig. 1.6 The cell c and its eight neighboring cells c + s, s ∈ S = {−1, 0, 1}2

Consider the map μ : AS → A given by

μ(y) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

1 if

⎧⎪⎨

⎪⎩

∑s∈S y(s) = 3

or∑s∈S y(s) = 4 and y((0, 0)) = 1,

0 otherwise

(1.8)

for all y ∈ AS .A moment of thought tells us that μ just expresses the rules for the Game

of Life.The cellular automaton τ : AG → AG with memory set S and local defining

map μ is called the cellular automaton associated with the Game of Life.(b) The Discrete Laplacian. Let G = Z and A = R. Consider the map

Δ : RZ → R

Z defined by

Δ(x)(n) = 2x(n) − x(n − 1) − x(n + 1).

Then Δ is the cellular automaton over Z with memory set S = {−1, 0, 1}and local defining map μ : R

S → R given by

μ(y) = 2y(0) − y(−1) − y(1) for all y ∈ RS .

This may be generalized in the following way. Let G be an arbitrary groupand let S be a nonempty finite subset of G. Let K be a field. Consider themap ΔS = ΔG

S : KG → K

G defined by

ΔS(x)(g) = |S|x(g) −∑

s∈S

x(gs).


Then ΔS is a cellular automaton over G with memory set S∪{1G} and localdefining map μ : K

S∪{1G} → K given by

μ(y) = |S|y(1G) −∑

s∈S

y(s) for all y ∈ KS∪{1G}.

This cellular automaton is called the discrete Laplacian over K associatedwith G and S.

(c) The Majority action cellular automaton. Let G be a group and letS be a finite subset of G. Take A = {0, 1} and consider the map τ : AG → AG

defined by

τ(x)(g) =

⎧⎪⎨

⎪⎩

1 if∑

s∈S x(gs) > |S|2

0 if∑

s∈S x(gs) < |S|2

x(g) if∑

s∈S x(gs) = |S|2

for all x ∈ AG. Then τ is a cellular automaton over G with memory setS ∪ {1G} and local defining map μ : AS∪{1G} → A given by

μ(y) =

⎧⎪⎨

⎪⎩

1 if∑

s∈S y(s) > |S|2

0 if∑

s∈S y(s) < |S|2

y(1G) if∑

s∈S y(s) = |S|2

for all y ∈ AS∪{1G}.The cellular automaton τ is called the majority action cellular automaton

associated with G and S (see Figs. 1.7–1.8).The terminology comes from the fact that given x ∈ AG and g ∈ G, the

value τ(x)(g) is equal to a ∈ {0, 1} if there is a strict majority of elementsof gS at which the configuration x takes the value a, or to x(g) if no suchmajority exists.

(d) Let G be a group, A a set, and f : A → A a map from A into itself.Then the map τ : AG → AG defined by τ(x) = f ◦ x is a cellular automatonwith memory set S = {1G} and local defining map μ : AS → A given byμ(y) = f(y(1G)). Note that, if f is the identity map IdA on A, then τ equalsthe identity map IdAG on AG.

(e) Let G be a group, A a set, and s0 an element of G. Let Rs0 : G → Gdenote the right multiplication by s0 in G, that is, the map Rs0 : G → Gdefined by Rs0(g) = gs0. Then the map τ : AG → AG defined by τ(x) =x ◦Rs0 is a cellular automaton with memory set S = {s0} and local definingmap μ : AS → A given by μ(y) = y(s0).

Proposition 1.4.4. Let G be a group and let A be a set. Then every cellularautomaton τ : AG → AG is G-equivariant.


Fig. 1.7 The local defining map μ for the majority action on Z associated with

S = {+1,−1}

Fig. 1.8 The majority action τ on Z associated with S = {+1,−1}

Proof. Let S be a memory set for τ and let μ : AS → A be the associatedlocal defining map. For all g, h ∈ G and x ∈ AG, we have

τ(gx)(h) = μ((h−1gx)|S) = μ(((g−1h)−1x)|S) = τ(x)(g−1h) = gτ(x)(h).

Thus τ(gx) = gτ(x). ��

Corollary 1.4.5. Let τ : AG → AG be a cellular automaton and let H be asubgroup of G. Then one has τ(Fix(H)) ⊂ Fix(H).

Proof. Let x ∈ Fix(H). By the previous Proposition, we have, for everyh ∈ H,

hτ(x) = τ(hx) = τ(x).

Thus τ(x) ∈ Fix(H). ��

The following characterization of cellular automata will be useful in thesequel.


Proposition 1.4.6. Let G be a group and let A be a set. Consider a mapτ : AG → AG. Let S be a finite subset of G and let μ : AS → A. Then thefollowing conditions are equivalent:

(a) τ is a cellular automaton admitting S as a memory set and μ as theassociated local defining map;

(b) τ is G-equivariant and one has τ(x)(1G) = μ(x|S) for every x ∈ AG.

Proof. The fact that (a) implies (b) follows from Proposition 1.4.4 and for-mula (1.7)

Conversely, suppose (b). Then, by using the G-equivariance of τ , we get

τ(x)(g) = τ(g−1x)(1G) = μ((g−1x)|S)

for all x ∈ AG and g ∈ G. Consequently, τ satisfies (a). ��

An important feature of cellular automata is their continuity (with respectto the prodiscrete topology). In the proof of this property, we shall use thefollowing.

Lemma 1.4.7. Let G be a group and let A be a set. Let τ : AG → AG be acellular automaton with memory set S and let g ∈ G. Then τ(x)(g) dependsonly on the restriction of x to gS.

Proof. This is an immediate consequence of (1.5) since (g−1x)(s) = x(gs) forall s ∈ S. ��

Proposition 1.4.8. Let G be a group and let A be a set. Then every cellularautomaton τ : AG → AG is continuous.

Proof. Let S be a memory set for τ . Let x ∈ AG and let W be a neigh-borhood of τ(x) in AG. Then we can find a finite subset Ω ⊂ G such that(cf. equation (1.3))

V (τ(x), Ω) ⊂ W.

Consider the finite set ΩS = {gs : g ∈ Ω, s ∈ S}. If y ∈ AG coincide with xon ΩS, then τ(y) and τ(x) coincide on Ω by Lemma 1.4.7. Thus, we have

τ(V (x,ΩS)) ⊂ V (τ(x), Ω) ⊂ W.

This shows that τ is continuous. ��

Proposition 1.4.9. Let G be a group and let A be a set. Let σ : AG → AG

and τ : AG → AG be cellular automata. Then the composite map σ◦τ : AG →AG is a cellular automaton. Moreover, if S (resp. T ) is a memory set for σ(resp. τ), then ST = {st : s ∈ S, t ∈ T} is a memory set for σ ◦ τ .

Proof. It is clear that the map σ ◦ τ is G-equivariant since σ and τ areG-equivariant (by Proposition 1.4.4). Let S (resp. T ) be a memory set for


σ (resp. τ). For every x ∈ AG, we have σ ◦ τ(x)(1G) = σ(τ(x))(1G). ByLemma 1.4.7, σ(τ(x))(1G) depends only on the restriction of τ(x) to S. Byusing Lemma 1.4.7 again, we deduce that, for every s ∈ S, the element τ(x)(s)depends only on the restriction of x to sT . Therefore, σ ◦ τ(x)(1G) dependsonly on the restriction of x to ST . By applying Proposition 1.4.6, we concludethat σ ◦ τ is a cellular automaton admitting ST as a memory set. ��

Remark 1.4.10. With the hypotheses and notation of the previous proposi-tion, denote by μ : AS → A and ν : AT → A the local defining maps for σand τ , respectively. Then, the local defining map κ : AST → A for σ ◦ τ maybe described in the following way.

For y ∈ AST and s ∈ S define ys ∈ AT by setting ys(t) = y(st) for allt ∈ T . Also, denote by y ∈ AS the map defined by y(s) = ν(ys) for all s ∈ S.We finally define the map κ : AST → A by setting

κ(y) = μ(y) (1.9)

for all y ∈ AST .Let x ∈ AG, g ∈ G, s ∈ S, and t ∈ T . We then have

(s−1g−1x)|T (t) = s−1g−1x(t)

= g−1x(st)

= (g−1x)|ST (st)

=((g−1x)|ST

)s(t).

This shows that(s−1g−1x)|T =

((g−1x)|ST

)s

and therefore

τ(g−1x)(s) = ν((s−1g−1x)|T

)= ν

(((g−1x)|ST

)s

)= (g−1x)|ST (s).

As a consequence,τ(g−1x)|S = (g−1x)|ST . (1.10)

Finally, one has

(σ ◦ τ)(x)(g) = σ(τ(x))(g)

= μ((g−1τ(x))|S

)

= μ(τ(g−1x)|S)

(by (1.10)) = μ((g−1x)|ST )

(by (1.9)) = κ((g−1x)|ST

).

(1.11)

Recall that a monoid is a set equipped with an associative binary operationadmitting an identity element. Denote by CA(G;A) the set consisting of all


cellular automata τ : AG → AG. In Example 1.4.3(d) we have seen that theidentity map IdAG : AG → AG is a cellular automaton. Thus we have:

Corollary 1.4.11. The set CA(G;A) is a monoid for the composition ofmaps. ��

1.5 Minimal Memory

Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton.Let S be a memory set for τ and let μ : AS → A be the associated definingmap. If S′ is a finite subset of G such that S ⊂ S′, then S′ is also a memory setfor τ and the local defining map associated with S′ is the map μ′ : AS′ → Agiven by μ′ = μ◦p, where p : AS′ → AS is the canonical projection (restrictionmap). This shows that the memory set of a cellular automaton is not uniquein general. However, we shall see that every cellular automaton admits aunique memory set of minimal cardinality. Let us first establish the followingresult.

Lemma 1.5.1. Let τ : AG → AG be a cellular automaton. Let S1 and S2 bememory sets for τ . Then S1 ∩ S2 is also a memory set for τ .

Proof. Let x ∈ AG. Let us show that τ(x)(1G) depends only on the restrictionof x to S1 ∩ S2. To see this, consider an element y ∈ AG such that x|S1∩S2 =y|S1∩S2 . Let us choose an element z ∈ AG such that z|S1 = x|S1 and z|S2 =y|S2 (we may take for instance the configuration z ∈ AG which coincides withx on S1 and with y on G \ S1). We have τ(x)(1G) = τ(z)(1G) since x andz coincide on S1, which is a memory set for τ . On the other hand, we haveτ(y)(1G) = τ(z)(1G) since y and z coincide on S2, which is also a memoryset for τ . It follows that τ(x)(1G) = τ(y)(1G).

Thus there exists a map μ : AS1∩S2 → A such that

τ(x)(1G) = μ(x|S1∩S2) for all x ∈ AG.

As τ is G-equivariant (Proposition 1.4.4), we deduce that S1∩S2 is a memoryset for τ by using Proposition 1.4.6. ��

Proposition 1.5.2. Let τ : AG → AG be a cellular automaton. Then thereexists a unique memory set S0 ⊂ G for τ of minimal cardinality. Moreover,if S is a finite subset of G, then S is a memory set for τ if and only ifS0 ⊂ S.

Proof. Let S0 be a memory set for τ of minimal cardinality. As we have seenat the beginning of this section, every finite subset of G containing S0 is alsoa memory set for τ . Conversely, let S be a memory set for τ . As S ∩ S0 isa memory set for τ by Lemma 1.5.1, we have |S ∩ S0| ≥ |S0|. This implies

1.6 Cellular Automata over Quotient Groups 15

S ∩ S0 = S0, that is, S0 ⊂ S. In particular, S0 is the unique memory set ofminimal cardinality. ��

The memory set of minimal cardinality of a cellular automaton is calledits minimal memory set.

Remark 1.5.3. A map F : AG → AG is constant if there exists a configurationx0 ∈ AG such that F (x) = x0 for all x ∈ AG. By G-equivariance, a cellularautomaton τ : AG → AG is constant if and only if there exists a ∈ A suchthat τ(x)(g) = a for all x ∈ AG and g ∈ G. Observe that a cellular automatonτ : AG → AG is constant if and only if its minimal memory set is the emptyset.

1.6 Cellular Automata over Quotient Groups

Let G be a group and let A be a set. Let τ : AG → AG be a cellular au-tomaton. Suppose that N is a normal subgroup of G and let ρ : G → G/Ndenote the canonical epimorphism. It follows from Proposition 1.3.7 that themap ρ∗ : AG/N → Fix(N), defined by ρ∗(y) = y ◦ ρ for all y ∈ AG/N , is abijection from the set AG/N of configurations over the group G/N onto theset Fix(N) ⊂ AG of N -periodic configurations over G. On the other hand,the set Fix(N) satisfies τ(Fix(N)) ⊂ Fix(N) by Corollary 1.4.5. Thus, wecan define a map τ : AG/N → AG/N by setting

τ = (ρ∗)−1 ◦ τ |Fix(N) ◦ ρ∗. (1.12)

In other words, the map τ is obtained by conjugating by ρ∗ the restriction ofτ to Fix(N), so that the diagram

AG/N ρ∗

−−−−→ Fix(N) ⊂ AG

τ

⏐⏐ ⏐⏐ τ |Fix(N)

AG/N −−−−→ρ∗

Fix(N)

is commutative.Suppose that S ⊂ G is a memory set for τ and that μ : AS → A is the

associated local defining map. Consider the finite subset S = ρ(S) ⊂ G/N

and the map μ : AS → A defined by μ = μ ◦ π, where π : AS → AS is theinjective map induced by ρ.

Proposition 1.6.1. The map τ : AG/N → AG/N is a cellular automaton overthe group G/N admitting S as a memory set and μ : AS → A as the associatedlocal defining map.


Proof. Let y ∈ AG/N , g ∈ G, and g = ρ(g). We have

τ(y)(g) = τ(y ◦ ρ)(g)

= μ((g−1(y ◦ ρ))|S)

= μ((g−1y)|S).

Thus τ is a cellular automaton with memory set S and local defining mapμ : AS → A. ��

Consider now the map Φ : CA(G;A) → CA(G/N ; A) given by Φ(τ) = τ ,where τ is defined by (1.12). We have the following:

Proposition 1.6.2. The map Φ : CA(G;A) → CA(G/N ; A) is a monoid epi-morphism.

Proof. Let σ : AG/N → AG/N be a cellular automaton over G/N with mem-ory set T ⊂ G/N and local defining map ν : AT → A. Let S ⊂ G be a finiteset such that ρ induces a bijection φ : S → T . Consider the map μ : AS → Adefined by μ(y) = ν(y ◦ φ−1) for all y ∈ AS . Let τ : AG → AG be the cellularautomaton over G with memory set S and local defining map μ. We have

μ(z) = (μ ◦ π)(z) = ν(π(z) ◦ φ−1) = ν(z)

for all z ∈ AS . It follows that μ = ν and τ = σ. This shows that Φ issurjective.

The fact that Φ is a monoid morphism immediately follows from (1.12).��

Examples 1.6.3. Let G be a group, S ⊂ G a finite subset, and N a normalsubgroup of G. Denote by ρ : G → G/N the canonical epimorphism andsuppose that ρ induces a bijection between S and S = ρ(S) ⊂ G/N .

(a) Consider the discrete laplacian ΔS : RG → R

G associated with G and S(cf. Example 1.4.3(b)). Then Φ(ΔS) : R

G/N → RG/N is the discrete laplacian

associated with G/N and S.(b) Consider the majority action cellular automaton τ : {0, 1}G → {0, 1}G

associated with G and S (cf. Example 1.4.3(c)). Then Φ(τ) : {0, 1}G/N →{0, 1}G/N is the majority action cellular automaton associated with G/Nand S.

1.7 Induction and Restriction of Cellular Automata

Let G be a group and let A be a set. Let H be a subgroup of G.Let CA(G, H; A) denote the set consisting of all cellular automata τ : AG →

AG admitting a memory set S such that S ⊂ H. Thus, CA(G, H; A) is the

1.7 Induction and Restriction of Cellular Automata 17

subset of CA(G;A) consisting of the cellular automata whose minimal mem-ory set is contained in H.

Recall that a subset N of a monoid M is called a submonoid if the identityelement 1M is in N and N is stable under the monoid operation (that is,xy ∈ N for all x, y ∈ N). If N is a submonoid of a monoid M , then themonoid operation induces by restriction a monoid structure on N .

Proposition 1.7.1. The set CA(G, H; A) is a submonoid of CA(G;A).

Proof. The identity element of CA(G;A) is the identity map IdAG . We haveIdAG ∈ CA(G, H; A) since {1G} is a memory set for IdAG and {1G} ⊂ H.Let σ, τ ∈ CA(G, H; A). Let S (resp. T ) be a memory set for σ (resp. τ) suchthat S ⊂ H (resp. T ⊂ H). It follows from Proposition 1.4.9 that ST is amemory set for σ ◦ τ . Since ST ⊂ H, this implies that σ ◦ τ ∈ CA(G, H; A).This shows that CA(G, H; A) is a submonoid of CA(G;A). ��

Let τ ∈ CA(G, H; A). Let S be a memory set for τ such that S ⊂ Hand let μ : AS → A denote the associated local defining map. Then, the mapτH : AH → AH defined by

τH(x)(h) = μ((h−1x)|S) for all x ∈ AH , h ∈ H,

is a cellular automaton over the group H with memory set S and local definingmap μ. Observe that if x ∈ AG is such that x|H = x, then

τH(x)(h) = τ(x)(h) for all h ∈ H. (1.13)

This shows in particular that τH does not depend on the choice of the memoryset S ⊂ H. One says that τH is the restriction of the cellular automaton τto H.

Conversely, let σ : AH → AH be a cellular automaton with memory set Sand local defining map μ : AS → A. Then the map σG : AG → AG defined by

σG(x)(g) = μ((g−1x)|S) for all x ∈ AG, g ∈ G,

is a cellular automaton over G with memory set S and local defining mapμ. If S0 is the minimal memory set of σ and μ0 : AS0 → A is the associatedlocal defining map then μ = μ0 ◦ π, where π : AS → AS0 is the restrictionmap (see Sect. 1.5). Thus, one has

σG(x)(g) = μ((g−1x)|S) = μ0 ◦ π((g−1x)|S) = μ0((g−1x)|S0)

for all x ∈ AG and g ∈ G. This shows in particular that σG does not dependon the choice of the memory set S ⊂ H. One says that σG ∈ CA(G, H; A) isthe cellular automaton induced by σ ∈ CA(H; A).

Proposition 1.7.2. The map τ �→ τH is a monoid isomorphism fromCA(G, H; A) onto CA(H; A) whose inverse is the map σ �→ σG.


Proof. To simplify notation, denote by α : CA(G, H; A) → CA(H; A) andβ : CA(H; A) → CA(G, H; A) the maps defined by α(τ) = τH and β(σ) = σG

respectively. It is clear from the definitions given above that β ◦ α and α ◦ βare the identity maps. Therefore, α is bijective with inverse β.

It remains to show that α is a monoid homomorphism.Let x ∈ AH and let x ∈ AG extending x. By applying (1.13), we get

α(IdAG)(x)(h) = IdAG(x)(h) = x(h) = x(h)

for all h ∈ H. This shows that α(IdAG)(x) = x for all x ∈ AH , that is,α(IdAG) = IdAH .

Let σ, τ ∈ CA(G, H; A). Let x ∈ AH and let x ∈ AG extending x. Byapplying (1.13) again, we have

α(σ ◦ τ)(x)(h) = (σ ◦ τ)(x)(h) = σ(τ(x))(h) (1.14)

for all h ∈ H. On the other hand, since τ(x) extends α(τ)(x), we have

α(σ)(α(τ)(x))(h) = σ(τ(x))(h)

that is,(α(σ) ◦ α(τ))(x)(h) = σ(τ(x))(h) (1.15)

for all h ∈ H. From (1.14) and (1.15), we deduce that α(σ ◦ τ)(x) = (α(σ) ◦α(τ))(x) for all x ∈ AH , that is, α(σ ◦ τ) = α(σ) ◦ α(τ). ��

Let τ ∈ CA(G, H; A). In order to analyze the way τ transforms a configu-ration x ∈ AG, we now introduce the set G/H = {gH : g ∈ G} consisting ofall left cosets of H in G. Since the cosets c ∈ G/H form a partition of G, wehave a natural identification AG =

∏c∈G/H Ac. With this identification, we

havex = (x|c)c∈G/H

for each x ∈ AG, where x|c ∈ Ac denotes the restriction of x to c. Observe nowthat if c ∈ G/H and g ∈ c, then τ(x)(g) depends only on x|c (this directlyfollows from Lemma 1.4.7 since if S is a memory set for τ with S ⊂ H, thengS ⊂ c). This implies that τ may be written as a product

τ =∏

c∈G/H

τc, (1.16)

where τc : Ac → Ac is the unique map which satisfies τc(x|c) = (τ(x))|c forall x ∈ AG. Note that the notation is coherent when c = H, since, in thiscase, τc = τH : AH → AH is the cellular automaton obtained by restrictionof τ to H.

Given a coset c ∈ G/H and an element g ∈ c, denote by φg : H → cthe bijective map defined by φg(h) = gh for all h ∈ H. Then φg induces abijective map φ∗

g : Ac → AH given by

1.7 Induction and Restriction of Cellular Automata 19

φ∗g(x) = x ◦ φg (1.17)

for all x ∈ Ac. It turns out that the maps τc and τH are conjugate by φ∗g:

Proposition 1.7.3. With the above notation, we have,

τc = (φ∗g)

−1 ◦ τH ◦ φ∗g. (1.18)

In other words, the following diagram

Ac τc−−−−→ Ac

φ∗g

⏐⏐ ⏐⏐ φ∗

g

AH −−−−→τH

AH

is commutative.

Proof. Let x ∈ Ac and let x ∈ AG extending x. For all h ∈ H, we have

(φ∗g ◦ τc)(x)(h) = φ∗

g(τc(x))(h)

= (τc(x) ◦ φg)(h)= τc(x)(gh)= τ(x)(gh)

= g−1τ(x)(h)

= τ(g−1x)(h),

where the last equality follows from the G-equivariance of τ (Proposition1.4.4). Now observe that the configuration g−1x ∈ AG extends x ◦ φg ∈ AH .Thus, we have

(φ∗g ◦ τc)(x)(h) = τH(x ◦ φg)(h) = τH(φ∗

g(x))(h) = (τH ◦ φ∗g)(x)(h).

This shows that φ∗g ◦ τc = τH ◦φ∗

g, which gives (1.18) since φ∗g is bijective. ��

The following statement will be used in the proof of Proposition 3.2.1:

Proposition 1.7.4. Let G be a group and let A be a set. Let H be a sub-group of G and let τ ∈ CA(G, H;A). Let τH : AH → AH denote the cellularautomaton obtained by restriction of τ to H. Then the following hold:

(i) τ is injective if and only if τH is injective;(ii) τ is surjective if and only if τH is surjective;(iii) τ is bijective if and only if τH is bijective.


Proof. It immediately follows from (1.16) that τ is injective (resp. surjective,resp. bijective) if and only if τc is injective (resp. surjective, resp. bijective)for all c ∈ G/H.

Now, (1.18) says that, given c ∈ G/H and g ∈ G, the map τc and τH areconjugate by the bijection φg. We deduce that τc is injective (resp. surjective,resp. bijective) if and only if τH is injective (resp. surjective, resp. bijective).

Thus, τ is injective (resp. surjective, resp. bijective) if and only if τH isinjective (resp. surjective, resp. bijective). ��

1.8 Cellular Automata with Finite Alphabets

Let G be a group and let A be a finite alphabet. As a product of finite spacesis compact by Tychonoff theorem (see Corollary A.5.3), it follows that AG

is compact. This topological property is very useful in the study of cellularautomata over finite alphabets. In particular, it may be used to prove thefollowing:

Theorem 1.8.1 (Curtis-Hedlund theorem). Let G be a group and let Abe a finite set. Let τ : AG → AG be a map and equip AG with its prodiscretetopology. Then the following conditions are equivalent:

(a) the map τ is a cellular automaton;(b) the map τ is continuous and G-equivariant.

Proof. The fact that (a) implies (b) directly follows from Proposition 1.4.4and Proposition 1.4.8 (this implication does not require the finiteness as-sumption on the alphabet A).

Conversely, suppose (b). Let us show that τ is a cellular automaton. Asthe map ϕ : AG → A defined by ϕ(x) = τ(x)(1G) is continuous, we can find,for each x ∈ AG, a finite subset Ωx ⊂ G such that if y ∈ AG coincide with xon Ωx, that is, if y ∈ V (x,Ωx), then τ(y)(1G) = τ(x)(1G). The sets V (x,Ωx)form an open cover of AG. As AG is compact, there is a finite subset F ⊂ AG

such that the sets V (x,Ωx), x ∈ F , cover AG. Let us set S = ∪x∈F Ωx andsuppose that two configurations y, z ∈ AG coincide on S. Let x0 ∈ F besuch that y ∈ V (x0, Ωx0), that is, y|Ωx0

= x0|Ωx0. As S ⊃ Ωx0 we have

y|Ωx0= z|Ωx0

and therefore τ(y)(1G) = τ(x0)(1G) = τ(z)(1G). Thus thereis a map μ : AS → A such that τ(x)(1G) = μ(x|S) for all x ∈ AG. As τ isG-equivariant, it follows from Proposition 1.4.6 that τ is a cellular automatonwith memory set S and local defining map μ. ��

When the alphabet A is infinite, a continuous and G-equivariant mapτ : AG → AG may fail to be a cellular automaton. In other words, the impli-cation (b) ⇒ (a) in Theorem 1.8.1 becomes false if we suppress the finitenesshypothesis on A. This is shown by the following example.

1.8 Cellular Automata with Finite Alphabets 21

Example 1.8.2. Let G be an arbitrary infinite group and take A = G as thealphabet set. To avoid confusion, we denote by g · h the product of twoelements g and h in G. Consider the map τ : AG → AG defined by

τ(x)(g) = x(g · x(g))

for all x ∈ AG and g ∈ G.Given x ∈ AG and g, h ∈ G we have

g(τ(x))(h) = τ(x)(g−1 · h)

= x(g−1 · h · x(g−1 · h))

= x(g−1 · h · [gx](h))= gx(h · [gx](h))= τ(gx)(h).

This shows that g(τ(x)) = τ(gx) for all x ∈ AG and g ∈ G. Therefore, τ isG-equivariant.

Moreover, τ is continuous. Indeed, given x ∈ AG and a finite set K ⊂ G,let us show that there exists a finite set F ⊂ G such that, if y ∈ AG andy ∈ V (x, F ), then τ(y) ∈ V (τ(x), K). Set F = K ∪ {k · x(k) : k ∈ K}. Then,if y ∈ V (x, F ), then, for all k ∈ K one has

τ(x)(k) = x(k · x(k)) = y(k · x(k)) = y(k · y(k)) = τ(y)(k).

This shows that τ(y) ∈ V (τ(x), K). Thus, τ is continuous.However, τ is not a cellular automaton. Indeed, fix g0 ∈ G \ {1G} and, for

all g ∈ G, consider the configurations xg and yg in AG defined by

xg(h) =

⎧⎪⎨

⎪⎩

g if h = 1G

g0 if h = g

1G otherwise

and

yg(h) =

{g if h = 1G

1G otherwise

for all h ∈ G.Note that xg|G\{g} = yg|G\{g}. Let F ⊂ G be a finite set and choose

g ∈ G \ F (this is possible because G is infinite). Then one has xg|F = yg|Fwhile

τ(xg)(1G) = xg(xg(1G)) = xg(g) = g0

andτ(yg)(1G) = yg(yg(1G)) = yg(g) = 1G,


so that τ(xg)(1G) �= τ(yg)(1G). It follows that there is no finite set F ⊂ Gsuch that, for all x ∈ AG, the value of τ(x) at 1G only depends on the valuesof x|F . This shows that τ is not a cellular automaton (cf. Remark 1.4.2(b)).

1.9 The Prodiscrete Uniform Structure

Let G be a group and let A be a set. The prodiscrete uniform structure onAG is the product uniform structure obtained by taking the discrete uniformstructure on each factor A of AG =

∏g∈G A (see Appendix B for definition

and basic facts about uniform structures).A base of entourages for the prodiscrete uniform structure on AG is given

by the sets WΩ ⊂ AG × AG, where

WΩ = {(x, y) ∈ AG × AG : x|Ω = y|Ω} (1.19)

and Ω runs over all finite subsets of G.Observe that, using the notation introduced in (1.3), we have

V (x,Ω) = {y ∈ AG : (x, y) ∈ WΩ}

for all x ∈ AG.The following statement gives a global characterization of cellular au-

tomata in terms of the prodiscrete uniform structure and the G-shift onAG.

Theorem 1.9.1. Let A be a set and let G be a group. Let τ : AG → AG be amap and equip AG with its prodiscrete uniform structure. Then the followingconditions are equivalent:

(a) τ is a cellular automaton;(b) τ is uniformly continuous and G-equivariant.

Proof. Suppose that τ : AG → AG is a cellular automaton. We already knowthat τ is G-equivariant by Proposition 1.4.4. Let us show that τ is uniformlycontinuous. Let S be a memory set for τ . It follows from Lemma 1.4.7 that iftwo configurations x, y ∈ AG coincide on gS for some g ∈ G, then τ(x)(g) =τ(y)(g). Consequently, if the configurations x and y coincide on ΩS = {gs :g ∈ Ω, s ∈ S} for some subset Ω ⊂ G, then τ(x) and τ(y) coincide on Ω.Observe that ΩS is finite whenever Ω is finite. Using the notation introducedin (1.19), we deduce that

(τ × τ)(WΩS) ⊂ WΩ

for every finite subset Ω of G. As the sets WΩ, where Ω runs over all finitesubsets of G, form a base of entourages for the prodiscrete uniform structure

1.9 The Prodiscrete Uniform Structure 23

on AG, it follows that τ is uniformly continuous. This shows that (a) implies(b).

Conversely, suppose that τ is uniformly continuous and G-equivariant. Letus show that τ is a cellular automaton. Consider the subset Ω = {1G} ⊂ G.Since τ is uniformly continuous, there exists a finite subset S ⊂ G such that(τ×τ)(WS) ⊂ WΩ. This means that τ(x)(1G) only depends on the restrictionof x to S. Thus, there is a map μ : AS → A such that

τ(x)(1G) = μ(x|S)

for all x ∈ AG. Using the G-equivariance of τ , we get

τ(x)(g) = [g−1τ(x)](1G) = τ(g−1x)(1G) = μ((g−1x)|S)

for all x ∈ AG and g ∈ G. This shows that τ is a cellular automaton withmemory set S and local defining map μ. Consequently, (b) implies (a). ��

Every uniformly continuous map between uniform spaces is continuouswith respect to the associated topologies, and the converse is true when thesource space is compact (Theorem B.2.3). The topology defined by the prodis-crete uniform structure on AG is the prodiscrete topology (see Example (1)in Sect. B.3). In the case when A is finite, the prodiscrete topology on AG

is compact by Tychonoff theorem (Theorem A.5.2). Thus Theorem 1.9.1 re-duces to the Curtis-Hedlund theorem (Theorem 1.8.1) in this case.

Remark 1.9.2. Suppose that G is countable and A is an arbitrary set. Thenthe prodiscrete uniform structure (and hence the prodiscrete topology) onAG is metrizable. To see this, choose an increasing sequence

∅ = E0 ⊂ E1 ⊂ · · · ⊂ En ⊂ · · ·

of finite subsets of G such that⋃

n≥0 En = G. Then the sets WEn , n ≥ 0, forma base of entourages for the prodiscrete uniform structure on AG. Considernow the metric d on AG defined by

d(x, y) =

{0 if x = y,

2−max{n≥0: x|En=y|En} if x �= y.

for all x, y ∈ AG. Then we have

WEn = {(x, y) ∈ AG × AG : d(x, y) < 2−n+1}

for every n ≥ 0. Consequently, d defines the prodiscrete uniform structure onAG.

Let G be a group and let A be a set. Let H be a subgroup of G. Let usequip AH\G with its prodiscrete uniform structure and Fix(H) ⊂ AG with the


uniform structure induced by the prodiscrete uniform structure on AG. Recallfrom Proposition 1.3.7 that there is a natural bijection ρ∗ : AH\G → AG

defined by ρ∗(y) = y ◦ ρ, where ρ : G → H\G is the canonical surjection.

Proposition 1.9.3. The map ρ∗ : AH\G → Fix(H) is a uniform isomor-phism.

Proof. For g ∈ G, let πg : AG → A and π′g : AH\G → A denote the projec-

tion maps given by x �→ x(g) and y �→ y(ρ(g)) respectively. Observe thatπg ◦ ρ∗ = π′

g is uniformly continuous for all g ∈ G. This shows that ρ∗ is uni-formly continuous. Similarly, the uniform continuity of (ρ∗)−1 follows fromthe fact that π′

g ◦ (ρ∗)−1 = πg|Fix(H) is uniformly continuous for each g ∈ G.Consequently, ρ∗ is a uniform isomorphism. ��

1.10 Invertible Cellular Automata

Let G be a group and let A be a set. One says that a cellular automatonτ : AG → AG is invertible (or reversible) if τ is bijective and the inversemap τ−1 : AG → AG is also a cellular automaton. This is equivalent to theexistence of a cellular automaton σ : AG → AG such that τ ◦ σ = σ ◦ τ =IdAG . Thus, the set of invertible cellular automata over the group G andthe alphabet A is exactly the group ICA(G;A) consisting of all invertibleelements of the monoid CA(G;A).

Theorem 1.10.1. Let A be a set and let G be a group. Let τ : AG → AG be amap and equip AG with its prodiscrete uniform structure. Then the followingconditions are equivalent:

(a) τ is an invertible cellular automaton;(b) τ is a G-equivariant uniform automorphism of AG.

Proof. It is clear that the inverse map of a bijective G-equivariant map fromAG onto itself is also G-equivariant. Therefore, the equivalence of conditions(a) and (b) follows from the characterization of cellular automata given inTheorem 1.9.1. ��

Bijective cellular automata over finite alphabets are always invertible:

Theorem 1.10.2. Let G be a group and let A be a finite set. Then everybijective cellular automaton τ : AG → AG is invertible.

Proof. Let τ : AG → AG be a bijective cellular automaton. The map τ−1 isG-equivariant since τ is G-equivariant. On the other hand, τ−1 is continuouswith respect to the prodiscrete topology by compactness of AG. Consequently,τ−1 is a cellular automaton by Theorem 1.8.1. ��

1.10 Invertible Cellular Automata 25

The following example shows that Theorem 1.10.2 becomes false if we omitthe finiteness hypothesis on the alphabet set A.

Example 1.10.3. Let K be a field. Let us take as the alphabet set the ringA = K[[t]] of all formal power series in one indeterminate t with coefficientsin K. Thus, an element of A is just a sequence a = (ki)i∈N of elements of K

written in the form

a = k0 + k1t + k2t2 + k3t

3 + · · · =∑

i∈N

kiti,

and the addition and multiplication of two elements a =∑

i∈Nkit

i and b =∑i∈N

k′it

i are respectively given by a+b =∑

i∈N(ki+k′

i)ti and ab =

∑i∈N

k′′i ti

with k′′i =

∑i1+i2=i ki1k

′i2

for all i ∈ N. We take G = Z. Thus, a configurationx ∈ AG is a map x : Z → K[[t]]. Consider the map τ : AG → AG defined by

τ(x)(n) = x(n) − tx(n + 1)

for all x ∈ AG and n ∈ Z. Clearly τ is a cellular automaton admittingS = {0, 1} as a memory set (the local defining map associated with S is themap μ : AS → A defined by μ(x0, x1) = x0 − tx1 for all x0, x1 ∈ A).

Let us show that τ is bijective. Consider the map σ : AG → AG given by

σ(x)(n) = x(n) + tx(n + 1) + t2x(n + 2) + t3x(n + 3) + · · ·

for all x ∈ AG and n ∈ Z. Observe that σ(x)(n) ∈ K[[t]] is well defined bythe preceding formula. In fact, if we develop x(n) ∈ K[[t]] in the form

x(n) =∑

i∈N

xn,iti (n ∈ Z, xn,i ∈ K),

then

σ(x)(n) =∑

i∈N

⎛

⎝i∑

j=0

xn+j,i−j

⎞

⎠ ti.

One immediately checks that σ ◦ τ = τ ◦ σ = IdAG . Therefore, τ is bijectivewith inverse map τ−1 = σ.

Let us show that the map σ : AG → AG is not a cellular automaton.Let F be a finite subset of Z and choose an integer M ≥ 0 such that

F ⊂ (−∞, M ]. Consider the configuration y defined by y(n) = 0 if n ≤ Mand y(n) = 1 if n ≥ M + 1, and the configuration z defined by z(n) = 0 forall n ∈ Z. Then y and z coincide on F . However, the value at 0 of σ(y) is

σ(y)(0) = tM+1 + tM+2 + tM+3 + · · ·

while the value of σ(z) at 0 is σ(z)(0) = 0. It follows that there is no finitesubset F ⊂ Z such that σ(x)(0) only depends on the restriction of x ∈ AG


to F . This shows that σ is not a cellular automaton. Consequently, τ is abijective cellular automaton which is not invertible.

In the next proposition we show that invertibility is preserved under theoperations of induction and restriction.

Proposition 1.10.4. Let G be a group and let A be a set. Let H be a sub-group of G and let τ ∈ CA(G, H; A). Let τH ∈ CA(H; A) denote the cellularautomaton obtained by restriction of τ to H. Then the following conditionsare equivalent:

(a) τ is invertible;(b) τH is invertible.

Moreover, if τ is invertible, then τ−1 ∈ CA(G, H; A) and one has

(τ−1)H = (τH)−1. (1.20)

Proof. First recall from (1.16) the factorizations

AG =∏

c∈G/H

Ac and τ =∏

c∈G/H

τc, (1.21)

where τc : Ac → Ac satisfies τc(x|c) = (τ(x))|c for all x ∈ AG.Suppose that τ is invertible. Denote by σ ∈ CA(G;A) the inverse cellular

automaton τ−1. It follows from (1.21) that the map τc : Ac → Ac is bijectivefor each c ∈ G/H and that

σ =∏

c∈G/H

(τc)−1, (1.22)

where (τc)−1 : Ac → Ac is the inverse map of τc. Let us show that σ ∈CA(G, H; A). Let S ⊂ G be a memory set for σ. Let x ∈ AG. It follows from(1.22) that (σ(x))|H = (τH)−1(x|H). Thus, we have

σ(x)(1G) = (σ(x))|H(1G) = (τH)−1(x|H)(1G).

This shows that σ(x)(1G) only depends on x|H . Arguing as in the proof ofLemma 1.5.1, we deduce that S ∩ H is a memory set for σ. Indeed, sup-pose that two configurations x, y ∈ AG coincide on S ∩ H. Consider theconfiguration z ∈ AG which coincide with x on S and with y on G \ S. Wehave σ(x)(1G) = σ(z)(1G) since x and z coincide on S. On the other hand,we have σ(y)(1G) = σ(z)(1G) since y and z coincide on H. This impliesσ(x)(1G) = σ(y)(1G). Thus, there is a map μ : AS∩H → A such that

σ(x)(1G) = μ(x|S∩H)

for all x ∈ AG. By applying Proposition 1.4.6, it follows that S ∩ H is amemory set for σ. Since S∩H ⊂ H, this shows that τ−1 = σ ∈ CA(G, H; A).

Notes 27

Moreover, it follows from (1.22) that

(τ−1)H = σH = (τH)−1

which gives us (1.20).The equivalence (a) ⇔ (b) is then an immediate consequence of Proposi-

tion 1.7.2 which tells us that the restriction map CA(G, H; A) → CA(H; A)is a monoid isomorphism. ��

Notes

Cellular automata were introduced by J. von Neumann (see [vNeu2]) whoused them to describe theoretical models of self-reproducing machines. Hefirst attempted to get such models by means of partial differential equationsin R

3. Later he changed the perspective and tried to use ideas and methodscoming from robotics and electrical engineering. Eventually, in 1952, followinga suggestion of S. Ulam, his former colleague at the Los Alamos Laboratories,he constructed a cellular automaton over the group Z

2 with an alphabet con-sisting of 29 states. He then outlined the construction of a pattern, containingapproximatively 200,000 cells, which would reproduce itself. The details werelater filled in by A.W. Burks in the 1960s [Bur].

The branch of mathematics which is concerned with the study of the dy-namical properties of the shift action is known as symbolic dynamics. Manyauthors trace the birth of symbolic dynamics back to a paper published in1898 by J. Hadamard [Had] in which words on two letters were used to codegeodesics on certain surfaces with negative curvature. However, as it waspointed out by E.M. Coven and Z.W. Nitecki [CovN], Hadamard’s symbolicdescription of geodesics is purely static and involves only finite words. Ac-cording to the authors of [CovN], the beginning of symbolic dynamics shouldbe placed in a paper by G.A. Hedlund [Hed-1] published in 1944. Symbolicdynamics has important applications in dynamical systems, especially in thestudy of hyperbolic dynamical systems for which symbolic codings may beobtained from Markov partitions. One of the first examples of such an appli-cation was the use of the properties of the Thue-Morse sequence (see Exer-cise 3.41) by M. Morse [Mors] in 1921 to prove the existence of non-periodicrecurrent geodesics on surfaces with negative curvature. A detailed exposi-tion of symbolic dynamics over Z may be found for example in the books byB. Kitchens [Kit], by P. Kurka [Kur], and by D. Lind and B. Marcus [LiM].

In the mid-1950s, Hedlund studied the so-called shift-commuting blockmaps which turn out to be exactly cellular automata over the group Z.The Curtis-Hedlund theorem (Theorem 1.8.1), also called Curtis-Hedlund-Lyndon’s theorem or Hedlund’s theorem, is named after Hedlund [Hed-3]


who proved it in 1969. Its generalization to infinite alphabets, as stated as inTheorem 1.9.1, was proved by the authors in [CeC7].

Cellular automata were intensely studied from the 1960s, both by pure andapplied mathematicians, under different names such as tessellation automata,parallel maps, cellular spaces, iterative automata, homogeneous structures,universal spaces, and sliding block codes (cf. [LiM, Section 1.1]). In mostcases, these researches focused on cellular automata with finite alphabet overthe groups Z or Z

2 (see the surveys [BanMS], [BKM], [Kar3], [Wolfr3]).The Game of Life was invented by the British mathematician J.H. Con-

way. This cellular automaton was described for the first time by M. Gard-ner [Gar-1] in the October 1970 issue of the Scientific American. From atheoretical computer science point of view, it is important because it has thepower of a universal Turing machine, that is, anything that can be computedalgorithmically can be computed by using the Game of Life.

In the 1980s, S. Wolfram [Wolfr1], [Wolfr2] started a systematic study andempirical classification of elementary cellular automata, that is, of cellularautomata over Z with alphabet A = {0, 1} and memory set S = {−1, 0, 1}.There are 2(23) = 256 such elementary cellular automata. Wolfram introduceda naming scheme for them which is nowadays widely used. Each elementarycellular automaton τ : AZ → AZ is uniquely determined by the eight bitsequence

μ(111)μ(110)μ(101)μ(100)μ(011)μ(010)μ(001)μ(000) ∈ A8,

where μ : AS → A is the associated local defining map. This bitsequence is the binary expansion of an integer in the interval [0, 255],called the Wolfram number of τ . For example, the majority actionτ : AZ → AZ associated with S = {−1, 0, 1} (cf. Example 1.4.3(c)) is anelementary cellular automaton. Its local defining map μ givesμ(111)μ(110)μ(101)μ(100)μ(011)μ(010)μ(001)μ(000) = 11101000 (cf. Fig.1.7). It follows that the Wolfram number of τ is 232. Let us also mentionthat the elementary cellular automaton with Wolfram number 110 was re-cently proved computationally universal by M. Cook.

Wolfram introduced an empirical classification of elementary cellular au-tomata into four classes according to the behavior of random initial configu-rations under iterations. These are known as Wolfram classes and are definedas follows:(W1) Almost all initial configurations lead to the same uniform fixed-pointconfiguration,(W2) Almost all initial configurations lead to a periodic configuration,(W3) Almost all initial configurations lead to chaos,(W4) Localized structures with complex interactions emerge.

The survey paper [BKM] contains other dynamical classifications of cellu-lar automata over Z with finite alphabet due to R. Gilman and to M. Hurleyand P. Kurka.

Exercises 29

Invertible cellular automata are used to model time-reversible processesoccurring in physics and biology. A group G is called periodic if every el-ement g ∈ G has finite order. In [CeC11] it is shown that if G is a non-periodic group, then for every infinite set A there exists a bijective cellularautomaton τ : AG → AG which is not invertible (cf. Theorem 1.10.2 andExample 1.10.3). It was shown by Amoroso and Patt in 1972 [Amo] that itis decidable whether a given cellular automaton with finite alphabet over Z

is invertible. This means that there exists an algorithm which establishes,after a finite number of steps, whether the cellular automaton correspond-ing to a given local defining map is invertible or not. On the other hand,J. Kari [Kar1], [Kar2], [Kar3] proved that the similar problem for cellular au-tomata with finite alphabet over Z

d, d ≥ 2, is undecidable. Its proof is basedon R. Berger’s undecidability result for the tiling problem of Wang tiles.

Exercises

1.1. An action of a group Γ on a topological space X is said to be topologicallymixing if for each pair of nonempty subsets U and V of X there exists afinite subset F ⊂ Γ such that U ∩ γV �= ∅ for all γ ∈ Γ \ F . Show that if Gis a group and A is a set then the G-shift on AG is topologically mixing forthe prodiscrete topology on AG.

1.2. Let G be a group and let A be a set. Let x ∈ AG and let Ω1 and Ω2

be two subsets of G. Show that V (x,Ω1 ∪ Ω2) = V (x,Ω1) ∩ V (x,Ω2) andWΩ1∪Ω2 = WΩ1 ∩WΩ2 (see (1.3) and (1.19) for the definition of V (x,Ω) andWΩ).

1.3. Let G be a countable group and let A be a set. Show that the metric don AG introduced in Remark 1.9.2 is complete.

1.4. Let G be an uncountable group and let A be a set having at least twoelements. Prove that the prodiscrete topology on AG is not metrizable. Hint:Prove that this topology does not satisfy the first axiom of countability.

1.5. Let G be a group. Let A and B be two sets. Let τA : AG → AG andτB : BG → BG be cellular automata. For x ∈ (A × B)G, let xA ∈ AG andxB ∈ BG be the configurations defined by x(g) = (xA(g), xB(g)) for allg ∈ G. Show that the map τ : (A × B)G → (A × B)G given by τ(x)(g) =(τA(xA)(g), τB(xB)(g)) for all g ∈ G is a cellular automaton.

1.6. Let G be a group and let S be a finite subset of G of cardinality k. LetA be a finite set of cardinality n. Show that there are exactly nnk

cellularautomata τ : AG → AG admitting S as a memory set.


1.7. Let G = Z2 and A = {0, 1}. Let τ : AG → AG denote the cellular

automaton associated with the Game of Life. Let y ∈ AG be the constantconfiguration defined by y(g) = 1 for all g ∈ G (all cells are alive). Find aconfiguration x ∈ AG such that y = τ(x).

1.8. Let A be a set and suppose that G is a trivial group. Show that themonoid CA(G;A) is canonically isomorphic to the monoid consisting of allmaps from A to A (with composition of maps as the monoid operation). Alsoshow that the group ICA(G;A) is canonically isomorphic to the symmetricgroup of A.

1.9. Let G be a group and let A be a set. Let τ : AG → AG be a cellularautomaton. Show that τ admits a memory set which is reduced to a singleelement if and only if there exist an element s ∈ G and a map f : A → Asuch that one has τ(x)(g) = f(x(gs)) for all x ∈ AG and g ∈ G.

1.10. Prove that there are exactly 218 cellular automata τ : {0, 1}Z → {0, 1}Z

whose minimal memory set is {−1, 0, 1}.

1.11. Let τ ∈ CA(Z2; {0, 1}) denote the cellular automaton associated withthe Game of Life. Show that the minimal memory set of τ is the set{−1, 0, 1}2.

1.12. Let G be a group and let A be a set. Let σ, τ ∈ CA(G;A). Let S0 (resp.T0, resp. C0) denote the minimal memory set of σ (resp. τ , resp. σ ◦τ). Provethat C0 ⊂ S0T0. Give an example showing that this inclusion may be strict.

1.13. Let G be a group and let A be a set. Let H be a subgroup of G and letτ ∈ CA(G, H; A). Show that τ and τH have the same minimal memory set.

1.14. Prove Proposition 1.4.9 by applying Theorem 1.9.1. Hint: Observe thatthe composite of two uniformly continuous maps is a uniformly continuousmap.

1.15. Let G be a group and let A be a set. Let F be a nonempty finitesubset of G and set B = AF . The sets AG and BG are equipped with theirprodiscrete uniform structures and with the G-shift action. Show that themap ΦF : AG → BG defined by ΦF (x)(g) = (g−1x)|F for all x ∈ AG andg ∈ G is a G-equivariant uniform embedding.

1.16. Let G be a group, H ⊂ G a subgroup of G, and let A be a set. LetH\G = {Hg : g ∈ G} be the set of all right cosets of H in G and setB = AH\G. The set AG (resp. BH) is equipped with its prodiscrete uniformstructure and with the G-shift (resp. H-shift) action. Let T ⊂ G be a completeset of representatives for the right cosets of H in G so that G =

∐t∈T Ht.

Show that the map Ψ = Ψ(H,T ) : AG → BH defined by Ψ(x)(h)(Ht) = x(ht)for all x ∈ AG, h ∈ H and t ∈ T is an H-equivariant uniform isomorphism.

Exercises 31

1.17. Let G be a group and let A be a set. For each s ∈ G, let τs : AG → AG

be the cellular automaton defined by τs(x)(g) = x(gs) for all x ∈ AG, g ∈ G(cf. Example 1.4.3(e)).

(a) Show that τs ∈ ICA(G;A) for every s ∈ G.(b) Prove that the map Φ : G → ICA(G;A) defined by φ(s) = τs for all

s ∈ G is a group homomorphism.(c) Prove that if A has at least two elements, then Φ is injective but not

surjective.

1.18. Let G be a group and let A be a set. Prove that the set consisting ofall invertible cellular automata τ : AG → AG admitting a memory set whichis reduced to a single element is a subgroup of ICA(G;A) isomorphic to thedirect product G × Sym(A).

1.19. (cf. [Amo]) Let G = Z and A = {0, 1}. Fix an integer n ≥ 3 andlet S = {−1, 0, 1, . . . , n}. Consider the element α ∈ AS (resp. β ∈ AS)defined by α(−1) = α(n) = 0 and α(k) = 1 for 0 ≤ k ≤ n − 1 (resp.β(−1) = β(0) = β(n) = 0 and β(k) = 1 for 1 ≤ k ≤ n − 1) and the mapμ : AS → A defined by μ(α) = 0, μ(β) = 1 and μ(y) = y(0) for y ∈ AS\{α, β}.Let τ : AG → AG be the cellular automaton with memory set S and localdefining map μ.

(a) Show that S is the minimal memory set of τ .(b) Show that τ is an invertible cellular automaton and that τ−1 = τ .

1.20. Show that the inverse map of the bijective cellular automaton τ : AZ →AZ studied in Example 1.10.3 is discontinuous, with respect to the prodiscretetopology on AZ, at every configuration x ∈ AZ.

1.21. Let G be a group and let A be a set. A subshift of the configurationspace AG is a subset X ⊂ AG which is G-invariant (i.e., such that gx ∈ Xfor all x ∈ X and g ∈ G) and closed in AG with respect to the prodiscretetopology.

(a) Show that ∅ and AG are subshifts of AG.(b) Show that if x ∈ AG then its orbit closure Gx ⊂ AG is a subshift.(c) Show that if (Xi)i∈I is a family of subshifts of AG then

⋂i∈I Xi is a

subshift of AG.(d) Show that if (Xi)i∈I is a finite family of subshifts of AG then

⋃i∈I Xi

is a subshift of AG.(e) Suppose that A is finite. Show that if X ⊂ AG is a subshift and

τ : AG → AG is a cellular automaton then τ(X) is a subshift of AG. Note:This last statement becomes false when A is infinite (see Example 3.3.3).

1.22. Let G be a group and let A and B be two sets. Let f : A → B be amap and consider the map f∗ : AG → BG defined by f∗(x) = f ◦ x for allx ∈ AG.

(a) Show that if A is finite and X is a subshift of AG then f∗(X) is asubshift of BG. Hint: Use the compactness of the configuration space AG.


(b) Let G = Z, A = Z, B = {0, 1} and let f : A → B be defined byf(n) = 0 if n < 0 and f(n) = 1 otherwise. Let X = {xn : n ∈ Z} ⊂ AZ,where xn(m) = n + m for all n, m ∈ Z. In other words, X is the Z-orbit Zx0

of the configurations x0. Show that X is a subshift of AZ but f∗(X) is not asubshift of BZ.

1.23. Let G be a group and let A be a set. Given a set of patterns P ⊂⋃

AΩ ,where the union runs over all finite subsets Ω of G, we set

XP = {x ∈ AG : (gx)|Ω /∈ P for all g ∈ G and all finite subsets Ω ⊂ G},

where (gx)|Ω denotes the restriction of the configuration gx to Ω.(a) Let P ⊂

⋃AΩ be a set of patterns. Show that XP is a subshift of AG.

(b) Conversely, show that if X ⊂ AG is a subshift, then there exists asubset P ⊂

⋃AΩ such that X = XP . Such a set P is called a defining set of

forbidden patterns for X.

1.24. Let G be a group and let A be a set. Given a finite subset Ω ⊂ G anda finite subset A ⊂ AΩ we set

X(Ω,A) = {x ∈ AG : (gx)|Ω ∈ A for all g ∈ G}.

(a) Let Ω ⊂ G and A ⊂ AΩ be finite subsets. Show that X(Ω,A) is asubshift of AG. A subshift X ⊂ AG is said to be of finite type if there existsa finite subset Ω ⊂ G and a finite subset A ⊂ AΩ such that X = X(Ω,A).Such a set A is then called a defining set of admissible patterns for X andthe subset Ω is called a memory set for X.

(b) Suppose that A is finite. Show that a subshift X ⊂ AG is of finite typeif and only if it admits a finite defining set of forbidden patterns.

1.25. Let G be a countable group and let A be a finite set. Show that thereare at most countably many distinct subshifts X ⊂ AG of finite type.

1.26. Let G be a group and let A be a finite set.(a) Let τ1, τ2 : AG → AG be two cellular automata. Show that the set

{x ∈ AG : τ1(x) = τ2(x)} ⊂ AG is a subshift of finite type.(b) Deduce from (a) that if τ : AG → AG is a cellular automaton then the

set Fix(τ) = {x ∈ AG : τ(x) = x} ⊂ AG is a subshift of finite type.(c) Conversely, show that if X ⊂ AG is a subshift of finite type then there

exists a cellular automaton τ : AG → AG such that X = Fix(τ).

1.27. Suppose that a group Γ acts continuously on a topological space Z.One says that the action of Γ on Z is topologically transitive if for any pairof nonempty open subsets U and V of Z there exists an element γ ∈ Γ suchthat U ∩ γV �= ∅.

Let G be a group and let A be a set. A subshift X ⊂ AG is said to beirreducible if for any finite subset Ω of G and any two elements x1, x2 ∈ X,

Exercises 33

there exist a configuration x ∈ X and an element g ∈ G such that x|Ω = x1|Ωand (gx)|Ω = x2|Ω.

Suppose that X ⊂ AG is a subshift. Show that the action of G on Xinduced by the G-shift is topologically transitive if and only if X is irreducible.

1.28. Let G be a group and let A be a set. Let B ⊂ A and consider thesubsets X,Y ⊂ AG defined by X = {b : b ∈ B} ⊂ AG, where b denotes theconstant configuration given by b(g) = b for all g ∈ G, and Y = {y ∈ AG :y(g) ∈ B for all g ∈ G}.

(a) Show that X and Y are subshifts of AG.(b) Show that if G is infinite then Y is irreducible.(c) Suppose that B has at least two distinct elements. Show that X is not

irreducible.

1.29. Let G be a group acting continuously on a nonempty complete metricspace X whose topology satisfies the second axiom of countability (i.e., ad-mitting a countable base of open subsets). Show that the following conditionsare equivalent:

(i) the action of G on X is topologically transitive;(ii) there is a point x ∈ X whose G-orbit is dense in X;(iii) there is a dense subset D ⊂ X such that the G-orbit of each point

x ∈ D is dense in X.Hint: The implications (iii) ⇒ (ii) and (ii) ⇒ (i) are straightforward. To

prove (i) ⇒ (iii), consider a sequence (Un)n∈N of nonempty open subsetsof X which form a base of the topology and denote by Ωn the set of pointsx ∈ X whose G-orbit meets Un. Then observe that each Ωn is and open densesubset of X if (i) is satisfied and apply Baire’s theorem (Theorem I.1.1, alsocf. Remark I.1.2(ii)).

1.30. Let G be a countable group and let A be a countable (e.g. finite) set.Let X ⊂ AG be a nonempty subshift. Show that the following conditions areequivalent:

(i) the subshift X is irreducible;(ii) there is a configuration x ∈ X whose G-orbit is dense in X;(iii) there is a dense subset D ⊂ X such that the G-orbit of each configu-

ration x ∈ D is dense in X.Hint: Use the results of Exercises 1.3 and 1.29.

1.31. Let G be a group and let A be a set. One says that a subshift X ⊂ AG

is topologically mixing if the action of G on X induced by the G-shift istopologically mixing (cf. Exercise 1.1).

(a) Let X ⊂ AG be a subshift. Show that X is topologically mixing if andonly if for any finite subset Ω of G and any two configurations x1, x2 ∈ X,there exists a finite subset F ⊂ G such that, for all g ∈ G \ F , there exists aconfiguration x ∈ X satisfying x|Ω = x1|Ω and (gx)|Ω = x2|Ω.

(b) Show that if G is infinite then every topologically mixing subshiftX ⊂ AG is irreducible.


1.32. Let G be a group and let A be a set. Let Δ ⊂ G be a finite subset.A subshift X ⊂ AG is said to be Δ-irreducible if it satisfies the followingcondition: if Ω1 and Ω2 are two finite subsets of G such that Ω1 and Ω2δ aredisjoint for all δ ∈ Δ, then, given any two configurations x1, x2 ∈ X, thereexists a configuration x ∈ X which satisfies x|Ω1 = x1|Ω1 and x|Ω2 = x2|Ω2 .A subshift X ⊂ AG is said to be strongly irreducible if there exists a finitesubset Δ ⊂ G such that X is Δ-irreducible.

(a) Show that the subshift Y ⊂ AG described in Exercise 1.28 is {1G}-irreducible and therefore strongly irreducible.

(b) Show that every strongly irreducible subshift is topologically mixing.

1.33. Let G be a group and let A be a set. Let H be a subgroup of G.For x ∈ AG and g ∈ G, denote by xH

g ∈ AH the configuration defined byxH

g = (gx)|H .(a) Check that hxH

g = xHhg for all x ∈ AG, h ∈ H and g ∈ G.

(b) Let X ⊂ AH be a subshift. Show that the set X(G) defined by X(G) ={x ∈ AG : xH

g ∈ X for all g ∈ G} is a subshift of AG.(c) Show that if H �= G then the subshift X(G) ⊂ AG is irreducible for

any subshift X ⊂ AH .(d) Let X ⊂ AH be a subshift. Show that X(G) ⊂ AG is of finite type

(resp. topologically mixing, resp. strongly irreducible) if and only if X is offinite type (resp. topologically mixing, resp. strongly irreducible).

(e) Suppose that σ : AH → AH is a cellular automaton and let σG : AG →AG be the induced cellular automaton (cf. Sect. 1.7). Check that one hasσG(x)H

g = σ(xHg ) for all x ∈ AG and g ∈ G.

(f) Show that if X ⊂ AH is a subshift such that σ(X) ⊂ X then one hasσG(X(G)) ⊂ X(G).

1.34. Let G be a group and let A be a set. Let F be a nonempty finite subsetof G and consider the map ΦF : AG → BG defined in Exercise 1.15, whereB = AF . Let X ⊂ AG be a subshift and set X [F ] = ΦF (X).

(a) Show that X [F ] is a subshift of BG.(b) Show that X is irreducible (resp. topologically mixing, resp. strongly

irreducible) if and only if X [F ] is irreducible (resp. topologically mixing, resp.strongly irreducible).

(c) Show that if X is of finite type then X [F ] is of finite type.

1.35. Let G be a group, H ⊂ G a subgroup of G, and let A be a set. Let alsoT ⊂ G be a complete set of representatives for the right cosets of H in G andconsider the map Ψ : AG → BH defined in Exercise 1.16, where B = AH\G.Let X ⊂ AG be a subshift and set X(H,T ) = Ψ(X).

(a) Show that X(H,T ) is a subshift of BH .(b) Show that if X is of finite type then X(H,T ) is of finite type.

1.36. Let A be a set. Let A∗ denote the monoid consisting of all words inthe alphabet A (cf. Sect. D.1). Recall that any word w ∈ A∗ can be uniquely

Exercises 35

written in the form w = a1a2 · · · an, where n ≥ 0 and ai ∈ A for 1 ≤ i ≤ n.The integer n is called the length of the word w and it is denoted by �(w).In the sequel, we shall identify the word w = a1a2 · · · an with the patternp : {1, 2, . . . , n} → A defined by p(i) = ai for 1 ≤ i ≤ n. Given a subshiftX ⊂ AZ and an integer n ≥ 0, we denote by Ln(X) ⊂ A∗ the set consistingof all words w ∈ A∗ for which there exists an element x ∈ X such thatw = x(1)x(2) · · ·x(n). The set L(X) = ∪n∈NLn(X) is called the languageof X. The elements w ∈ L(X) are called the admissible words of X (or,simply, the X-admissible words). The elements w ∈ A∗ \L(X) are called theforbidden words of X.

(a) Let X and Y be two subshifts of AZ. Show that one has X ⊂ Y (resp.X = Y ) if and only if L(X) ⊂ L(Y ) (resp. L(X) = L(Y )).

(b) One says that a word u ∈ A∗ is a subword of a word w ∈ A∗ if thereexist v1, v2 ∈ A∗ such that w = v1uv2. Let X ⊂ AZ be a subshift and letL = L(X). Show that L satisfies the following conditions:

(i) if w ∈ L, then u ∈ L for every subword u of w;(ii) if w ∈ L, then there exist a, a′ ∈ A such that awa′ ∈ L.(c) Conversely, show that if a subset L ⊂ A∗ satisfies conditions (i) and

(ii) in (b), then there exists a unique subshift X ⊂ AZ such that L = L(X).

1.37. Let A be a set and let X ⊂ AZ be a subshift.(a) Show that X is of finite type if and only if the following holds: there

exists an integer n0 ≥ 0 such that if the words u, v, w ∈ A∗ satisfy �(v) ≥ n0

and uv, vw ∈ L(X), then one has uvw ∈ L(X).(b) Show that X is irreducible if and only if for every pair of words u and

v in L(X), there exists a word w ∈ A∗ such that uwv ∈ L(X).(c) Show that X is topologically mixing if and only if the following holds:

for every pair of words u and v in L(X), there exists an integer n0 ≥ 0 suchthat for every integer n ≥ n0 there exists a word w ∈ A∗ of length �(w) = nsatisfying uwv ∈ L(X).

(d) Show that X is strongly irreducible if and only if the following holds:there exists an integer n0 ≥ 0 such that, for every pair of words u and v inL(X), and for every integer n ≥ n0, there exists a word w ∈ A∗ of length�(w) = n such that uwv ∈ L(X).

1.38. Let A = {0, 1} and let X ⊂ AZ be the set of all x ∈ AZ such thatthe following holds: if x(n) = 1, x(n + 1) = x(n + 2) = · · · = x(n + k) =0, x(n + k + 1) = 1, for some n ∈ Z and k ∈ N, then k is even.

(a) Show that X is a subshift (it is called the even subshift).(b) Show that X is not of finite type.(c) Show that X is strongly irreducible (and therefore topologically mixing

and irreducible).

1.39. Let A = {0, 1} and consider the subshift of finite type X ⊂ AZ definedby X = X{11} = {x ∈ AZ : (x(n), x(n+1)) �= (1, 1) for all n ∈ Z}. Show thatX is strongly irreducible (and therefore topologically mixing and irreducible).The subshift X is called the golden mean subshift.


1.40. Let A = {0, 1} and let X ⊂ AZ be the set consisting of the two config-urations x, y ∈ AZ defined by

x(n) =

{0 if n is even1 otherwise

and y(n) =

{1 if n is even0 otherwise,

for all n ∈ Z. Show that X is an irreducible subshift of finite type which isnot topologically mixing (and therefore not strongly irreducible either).

1.41. Let A be a set. Let X ⊂ AZ be a subshift of finite type. Show that Xis topologically mixing if and only if X is strongly irreducible.

1.42. Let A = {0, 1} and let X ⊂ AZ be the subshift with defining setof forbidden words {01k0h1 : 1 ≤ h ≤ k, k = 1, 2, . . .}. Show that X istopologically mixing (and therefore irreducible) but not strongly irreducible.

1.43. Cellular automata between subshifts. Let G be a group and let A be aset. The set AG is equipped with its prodiscrete uniform structure and withthe G-shift action. Let X,Y ⊂ AG be two subshifts and τ : X → Y a map.Then the following are equivalent:

(i) there exists a cellular automaton τ : AG → AG such that τ(x) = τ(x)for all x ∈ X;

(ii) τ is G-equivariant and uniformly continuous.One says that τ : X → Y is a cellular automaton if the two equivalent

conditions above are satisfied.

1.44. Let G be a group and let A be a finite set. Let τ : X → Y be amap between subshifts X,Y ⊂ AG. Show that the following conditions areequivalent:

(i) τ is a cellular automaton,(ii) τ is G-equivariant and continuous (with respect to the topologies in-

duced on X and Y by the prodiscrete topology on AG).

1.45. Let G be a group and let A be a finite set. Let τ : AG → AG be acellular automaton and let X ⊂ AG be an irreducible (resp. topologicallymixing, resp. strongly irreducible) subshift. Show that τ(X) is an irreducible(resp. topologically mixing, resp. strongly irreducible) subshift of AG.

Chapter 2

Residually Finite Groups

This chapter is devoted to the study of residually finite groups, which forma class of groups of special importance in several branches of mathematics.As their name suggests it, residually finite groups generalize finite groups.They are defined as being the groups whose elements can be distinguishedafter taking finite quotients (see Sect. 2.1). There are many other equivalentdefinitions. For example, a group is residually finite if and only if it can be em-bedded into the direct product of a family of finite groups (Corollary 2.2.6).The class of residually finite groups is closed under taking subgroups andtaking projective limits. It contains in particular all finite groups, all finitelygenerated abelian groups, and all free groups (Theorem 2.3.1). Every finitelygenerated residually finite group is Hopfian (Theorem 2.4.3) and the automor-phism group of a finitely generated residually finite group is itself residuallyfinite (Theorem 2.5.1). Examples of finitely generated groups which are notresidually finite are presented in Sect. 2.6. The following dynamical charac-terization of residually finite groups is given in Sect. 2.7: a group is residuallyfinite if and only if there is a Hausdorff topological space on which the groupacts continuously and faithfully with a dense subset of points with finiteorbit.

2.1 Definition and First Examples

Definition 2.1.1. A group G is called residually finite if for each elementg ∈ G with g �= 1G, there exist a finite group F and a homomorphismφ : G → F such that φ(g) �= 1F .

Proposition 2.1.2. Let G be a group. Then the following conditions areequivalent:

(a) G is residually finite;


37

http://dx.doi.org/10.1007/978-3-642-14034-1_2

38 2 Residually Finite Groups

(b) for all g, h ∈ G with g �= h, there exist a finite group F and a homomor-phism φ : G → F such that φ(g) �= φ(h).

Proof. The fact that (b) implies (a) is obvious, since (b) gives (a) by takingh = 1G. Conversely, suppose that G is residually finite. Let g, h ∈ G withg �= h. As gh−1 �= 1G, there exist a finite group F and a homomorphismφ : G → F such that φ(gh−1) �= 1F . Since φ(gh−1) = φ(g)(φ(h))−1, it followsthat φ(g) �= φ(h). Therefore, (a) implies (b). ��

Proposition 2.1.3. Every finite group is residually finite.

Proof. Let G be a finite group and consider g ∈ G such that g �= 1G. Ifφ = IdG : G → G is the identity map, we have φ(g) = g �= 1G. ��

Proposition 2.1.4. The additive group Z is residually finite.

Proof. Consider k ∈ Z such that k �= 0. Choose an integer m such that |k| <m. Then the canonical homomorphism φ : Z → Z/mZ (reduction modulo m)satisfies φ(k) �= 0. ��

A similar argument gives us the following:

Proposition 2.1.5. The group GLn(Z) is residually finite for every n ≥ 1.

Proof. Let A = (aij) ∈ GLn(Z) with A �= In = 1GLn(Z). Choose an integerm such that |aij | < m for all i, j. Then the homomorphism φ : GLn(Z) →GLn(Z/mZ) given by reduction modulo m satisfies φ(A) �= 1GLn(Z/mZ). ��

We shall now give examples of groups which are not residually finite.A group G is called divisible if for each g ∈ G and each integer n ≥ 1,

there is an element h ∈ G such that hn = g.

Example 2.1.6. The additive groups Q, R, and C are divisible. More generally,every Q-vector space, with its additive underlying group structure, is divisi-ble. In particular, every field of characteristic 0, with its underlying additivegroup structure, is divisible.

Lemma 2.1.7. Let G be a divisible group and let F be a finite group. Thenevery homomorphism φ : G → F is trivial.

Proof. Set n = |F |. Let g ∈ G. As G is divisible, we can find h ∈ G suchthat g = hn. If φ : G → F is a homomorphism, then we have φ(g) = φ(hn) =φ(h)n = 1F . ��

The preceding lemma immediately yields the following result.

Proposition 2.1.8. A nontrivial divisible group cannot be residually finite.��

2.1 Definition and First Examples 39

Example 2.1.9. The additive group underlying a field of characteristic 0 is notresidually finite. In particular, the additive group Q is not residually finite.

In order to give another characterization of residually finiteness, we shalluse the following:

Lemma 2.1.10. Let G be a group and let H be a subgroup of G. Let K =⋂g∈G gHg−1. Then K is a normal subgroup of G contained in H. Moreover,

if H is of finite index in G, then Kis of finite index in G.

Proof. Since gHg−1 = H for g = 1G, we have K ⊂ H. Let Sym(G/H)denote the group of permutations of G/H. Consider the action of G on G/Hgiven by left multiplication and let ρ : G → Sym(G/H) be the associatedhomomorphism. For each g ∈ G, the stabilizer of gH is gHg−1. Consequently,we have K = Ker(ρ), which shows that K is a normal subgroup of G. Thegroup G/K is isomorphic to Im(ρ) ⊂ Sym(G/H). Suppose that H is of finiteindex in G. This means that the set G/H is finite. This implies that thegroup Sym(G/H) is finite. We deduce that the group G/K is finite, that is,K is of finite index in G. ��

Given a group G, the intersection of all subgroups of finite index of G iscalled the residual subgroup (or profinite kernel) of G.

Proposition 2.1.11. Let G be a group and let N denote the residual sub-group of G. Then:

(i) N is equal to the intersection of all normal subgroups of finite indexin G;

(ii) N is a normal subgroup of G;(iii) G is residually finite if and only if N = {1G}.

Proof. Denote by N ′ the intersection of all normal subgroups of finite indexof G. The inclusion N ⊂ N ′ is trivial. If H is a subgroup of finite index of G,then K = ∩g∈GgHg−1 is a normal subgroup of finite index of G containedin H, by Lemma 2.1.10. This implies N ′ ⊂ K ⊂ H. It follows that N ′ ⊂ N .This shows (i).

Assertion (ii) follows from (i) since the intersection of a family of normalsubgroups of G is a normal subgroup of G.

Suppose that G is residually finite. Let g ∈ G such that g �= 1G. Then thereexist a finite group F and a homomorphism φ : G → F such that φ(g) �= 1F .Thus g /∈ Ker(φ). As the group G/Ker(φ) is isomorphic to F , the subgroupKer(φ) is of finite index in G. This shows that N = {1G}.

Conversely, suppose that N = {1G}. Let g ∈ G such that g �= 1G. By (i),we can find a normal subgroup of finite index K ⊂ G such that g /∈ K. Ifφ : G → G/K is the canonical homomorphism, we have φ(g) �= 1G/K . Thisshows that G is residually finite. ��


2.2 Stability Properties of Residually Finite Groups

Proposition 2.2.1. Every subgroup of a residually finite group is residuallyfinite.

Proof. Let G be a residually finite group and let H be a subgroup of G. Leth ∈ H such that h �= 1G. Since G is residually finite, there exist a finite groupF and a homomorphism φ : G → F such that φ(h) �= 1F . If φ′ : H → F isthe restriction of φ to H, we have φ′(h) = φ(h) �= 1F . Consequently, H isresidually finite. ��

Proposition 2.2.2. Let (Gi)i∈I be a family of residually finite groups. Thentheir direct product G =

∏i∈I Gi is residually finite.

Proof. Let g = (gi)i∈I ∈ G such that g �= 1G. Then there exists i0 ∈ Isuch that gi0 �= 1Gi0

. Since Gi0 is residually finite, we can find a finite groupF and a homomorphism φ : Gi0 → F such that φ(gi0) �= 1F . Consider thehomomorphism φ′ : G → F defined by φ′ = π ◦ φ, where π : G → Gi0 isthe projection onto Gi0 . We have φ′(g) = φ(gi0) �= 1F . Consequently, G isresidually finite. ��

Corollary 2.2.3. Let (Gi)i∈I be a family of residually finite groups. Thentheir direct sum G =

⊕i∈I Gi is residually finite.

Proof. This follows immediately from Proposition 2.2.1 and Proposition 2.2.2,since G is the subgroup of the direct product P =

∏i∈I Gi consisting of all

g = (gi) ∈ P for which gi = 1Gi for all but finitely many i ∈ I. ��

Corollary 2.2.4. Every finitely generated abelian group is residually finite.

Proof. If G is a finitely generated abelian group, then there exist an integerr ≥ 0 and a finite abelian group T such that G is isomorphic to Z

r × T . Byusing Proposition 2.1.3 and Proposition 2.1.4, we deduce that G is residuallyfinite. ��

Remark 2.2.5. An arbitrary abelian group need not be residually finite. Forexample, the additive group Q is not residually finite (Example 2.1.9).

Corollary 2.2.6. Let G be a group. Then the following conditions are equiv-alent:

(a) the group G is residually finite;(b) there exist a family (Fi)i∈I of finite groups such that the group G is iso-

morphic to a subgroup of the direct product group∏

i∈I Fi.

Proof. The fact that (b) implies (a) follows from Proposition 2.1.3, Propo-sition 2.2.2 and Proposition 2.2.1. Conversely, suppose that G is residuallyfinite. Then, for each g ∈ G \ {1G}, we can find a finite group Fg and ahomomorphism φg : G → Fg such that φg(g) �= 1Fg . Consider the group

2.2 Stability Properties of Residually Finite Groups 41

H =∏

g∈G\{1G}Fg.

The homomorphism ψ : G → H defined by

ψ =∏

g∈G\{1G}φg

is injective. Therefore, G is isomorphic to a subgroup of H. This shows that(a) implies (b). ��

The class of residually finite groups is closed under taking projective limits(see Sect. E.2 for the definition of the limit of a projective system of groups):

Proposition 2.2.7. If a group G is the limit of a projective system of resid-ually finite groups, then G is residually finite.

Proof. Let (Gi)i∈I be a projective system of residually finite groups such thatG = lim←−Gi. By construction of a projective limit (see Appendix E), G is asubgroup of the group

∏i∈I Gi. We deduce that G is residually finite by using

Proposition 2.2.2 and Proposition 2.2.1. ��

A group G is called profinite if G is the limit of some projective system offinite groups. An immediate consequence of Proposition 2.2.7 is the following:

Corollary 2.2.8. Every profinite group is residually finite. ��

Example 2.2.9. Let p be a prime number. Given integers n ≥ m ≥ 0, letφn,m : Z/pm

Z → Z/pnZ denote reduction modulo pn. Then (Z/pn

Z, φn,m)is a projective system of groups over N. The limit of this projective systemis called the group of p-adic integers and is denoted by Zp. Since Zp is theprojective limit of finite groups, it is profinite and hence residually finite byCorollary 2.2.8.

Remark 2.2.10. A group which is the limit of an inductive system of residuallyfinite groups need not be residually finite. For instance, we have seen inExample 2.1.9 that the additive group underlying a field of characteristic 0is not residually finite. However, such a group is the limit of the inductivesystem formed by its finitely generated subgroups, which are all residuallyfinite by Corollary 2.2.4.

If P is a property of groups, one says that a group G is virtually P if Gcontains a subgroup of finite index which satisfies P.

Lemma 2.2.11. Let G be a group. Let H be a subgroup of finite index of Gand let K be a subgroup of finite index of H. Then K is a subgroup of finiteindex of G.


Proof. Let h1, . . . , hn be a complete set of representatives of the left cosetsof G modulo H and let k1, . . . , kp be a complete set of representatives ofthe left cosets of H modulo K. Observe that the elements hikj , 1 ≤ i ≤ n,1 ≤ j ≤ p, form a complete set of representatives of the left cosets of Gmodulo K. Therefore [G : K] = np = [G : H][H : K] < ∞. ��

Proposition 2.2.12. Every virtually residually finite group is residually fi-nite.

Proof. Let G be a group and let H be a subgroup of finite index of G.By Lemma 2.2.11, the intersection of the subgroups of finite index of G iscontained in the intersection of the subgroups of finite index of H. Sincea group is residually finite if and only if the intersection of its subgroups offinite index is reduced to the identity element (Proposition 2.1.11), we deducethat G is residually finite if H is residually finite. ��

2.3 Residual Finiteness of Free Groups

The goal of this section is to establish the following result:

Theorem 2.3.1. Every free group is residually finite.

To prove this theorem, we shall use the following:

Lemma 2.3.2. The subgroup of SL2(Z) generated by the matrices

a =(

1 20 1

)and b =

(1 02 1

)

is a free group of rank 2.

Proof. The group GL2(R) naturally acts on the set of lines of R2 passing

through the origin, that is, on the projective line P1(R). In nonhomogeneouscoordinates this action is given by

gt =g11t + g12

g21t + g22

for g = (gij)1≤i,j≤2 ∈ GL2(R) and t ∈ P1(R) = R∪{∞} representing the lineof R

2 with slope 1/t passing through (0, 0) (see Fig. 2.1). Note that we have

akt = t + 2k and bkt =t

2kt + 1

for all k ∈ Z. Consider the subsets Y and Z of P1(R) defined by

Y = ] − 1, 1[ and Z = P1(R) \ [−1, 1].

2.3 Residual Finiteness of Free Groups 43

Fig. 2.1 The action of GL2(R) on P1(R). Here, t = 1 and g = a

One immediately checks that, for all k ∈ Z \ {0}, one has

akY = ]2k − 1, 2k + 1[ ⊂ Z and bkZ = ]1/(2k + 1), 1/(2k − 1)[ ⊂ Y.

By applying the Klein Ping-Pong theorem (Theorem D.5.1), we deduce thata and b generate a free group of rank 2. ��

Proof of Theorem 2.3.1. Since GL2(Z) is residually finite by Proposition2.1.5, it follows from Lemma 2.3.2, Corollary D.5.3, and Proposition 2.2.1that every free group of finite rank is residually finite.

Consider now an arbitrary set X and let F (X) denote the free groupbased on X. Let g ∈ F (X) such that g �= 1F (X). Let Y ⊂ X denote the setof elements x ∈ X such that xk appear in the reduced form of g for somek ∈ Z\{0}. The subgroup F (Y ) ⊂ F (X) generated by Y is a free group withbase Y (see Proposition D.2.4). We have g ∈ F (Y ). As the group F (Y ) isfree of finite rank, it is residually finite by the first part of the proof. Thuswe can find a finite group H and a homomorphism φ : F (Y ) → H such thatφ(g) �= 1H . Consider the unique homomorphism π : F (X) → F (Y ) such thatπ(y) = y for every y ∈ Y and π(x) = 1F (Y ) for every x ∈ X \ Y . Then thehomomorphism φ ◦ π : F (X) → H satisfies φ ◦ π(g) = φ(π(g)) = φ(g) �= 1H .This shows that F (X) is residually finite. ��


2.4 Hopfian Groups

Definition 2.4.1. A group G is called Hopfian if every surjective endomor-phism of G is injective.

Examples 2.4.2. (a) Every finite group is Hopfian.(b) The additive group Q is Hopfian. Indeed, every endomorphism of the

group Q is of the form x �→ ax for some a ∈ Q, and it is clear that such anendomorphism is surjective (resp. injective) if and only if a �= 0.

(c) Every simple group is Hopfian. (we recall that a simple group is anontrivial group G such that the only normal subgroups of G are {1G} and G).

(d) The additive group Q/Z is not Hopfian. Indeed, the endomorphismψ : Q/Z → Q/Z defined by ψ(x) = 2x is surjective but not injective.

Theorem 2.4.3. Every finitely generated residually finite group is Hopfian.

Given groups G1 and G2, we shall denote by Hom(G1, G2) the set of allhomomorphisms from G1 to G2. We begin by establishing the following result.

Lemma 2.4.4. Let G be a finitely generated group and let F be a finite group.Then the set Hom(G, F ) is finite.

Proof. Let A be a finite generating subset of G. Let us set n = |A| andp = |F |. As A generates G, any homomorphism u : G → F is completelydetermined by the elements u(a), a ∈ A. Thus the set Hom(G, F ) containsat most pn elements. ��

Proof of Theorem 2.4.3. Let G be a finitely generated residually finite group.Suppose that ψ : G → G is a surjective endomorphism of G. Let K be a nor-mal subgroup of finite index of G and let ρ : G → G/K denote the canonicalhomomorphism. Consider the map

Φ : Hom(G, G/K) → Hom(G, G/K)

defined by Φ(u) = u ◦ ψ for all u ∈ Hom(G, G/K). The map Φ is injectivesince ψ is surjective by our hypothesis. As the set Hom(G, G/K) is finite byLemma 2.4.4, we deduce that Φ is also surjective. In particular, there existsa homomorphism u0 ∈ Hom(G, G/K) such that ρ = u0 ◦ ψ. This impliesKer(ψ) ⊂ Ker(ρ) = K. It follows that Ker(ψ) is contained in the intersectionof all normal subgroups of finite index of G. As G is residually finite, wededuce that Ker(ψ) = {1G} by Proposition 2.1.11(iii). Thus ψ is injective.This shows that G is Hopfian. ��

Since every free group is residually finite by Theorem 2.3.1, an immediateconsequence of Theorem 2.4.3 is the following:

Corollary 2.4.5. Every free group of finite rank is Hopfian.

2.5 Automorphism Groups of Residually Finite Groups 45

Remarks 2.4.6. (a) Let X be an infinite set and let F (X) denote the freegroup based on X. Every surjective but non injective map f : X → X ex-tends to a surjective endomorphism ψ : F (X) → F (X) which is not injective.Consequently, the group F (X) is not Hopfian. Since F (X) is residually finiteby Theorem 2.3.1, this shows that we cannot suppress the hypothesis that Gis finitely generated in Theorem 2.4.3.

(b) An example of a finitely generated Hopfian group which is not residu-ally finite will be given in Sect. 2.6 (see Proposition 2.6.1).

2.5 Automorphism Groups of Residually Finite Groups

Let G be a group. Recall that an automorphism of G is a bijective homomor-phism α : G → G. Clearly, the set Aut(G) consisting of all automorphismsof G is a subgroup of the symmetric group of G. The group Aut(G) is calledthe automorphism group of G.

Theorem 2.5.1. Let G be a finitely generated residually finite group. Thenthe group Aut(G) is residually finite.

Let us first establish the following result.

Lemma 2.5.2. Let G be a group. Let H1 and H2 be subgroups of finite indexof G. Then the subgroup H = H1 ∩ H2 is of finite index in G.

Proof. Two elements in G are left congruent modulo H if and only if theyare both left congruent modulo H1 and left congruent modulo H2. Therefore,there is an injective map from G/H into G/H1 × G/H2 given by gH �→(gH1, gH2). As the sets G/H1 and G/H2 are finite by hypothesis, we deducethat G/H is finite, that is, H is of finite index in G. ��

Proof of Theorem 2.5.1. Let α0 ∈ Aut(G) such that α0 �= IdG. Then we canfind an element g0 ∈ G such that α0(g0) �= g0. As G is residually finite,there exist a finite group F and a homomorphism φ : G → F satisfyingφ(α0(g0)) �= φ(g0). Consider the set H defined by

H =⋂

ψ∈Hom(G,F )

Ker(ψ),

where Hom(G, F ) denotes, as above, the set of all homomorphisms from G toF . Observe that H is a normal subgroup of G since it is the intersection of afamily of normal subgroups of G. On the other hand, for every α ∈ Aut(G),one has


α(H) = α

(⋂

ψ∈Hom(G,F )

Ker(ψ)

)

=⋂

ψ∈Hom(G,F )

α(Ker(ψ))

=⋂

ψ∈Hom(G,F )

Ker(ψ ◦ α−1).

As the map from Hom(G, F ) to itself defined by ψ �→ ψ ◦ α−1 is bijective(with ψ �→ ψ ◦ α as inverse map), we get

α(H) =⋂

ψ∈Hom(G,F )

Ker(ψ) = H.

Therefore α induces an automorphism α of G/H, given by α(gH) = α(g)Hfor all g ∈ G. The map α �→ α is clearly a homomorphism from Aut(G)to Aut(G/H). Let us show that the group Aut(G/H) is finite and thatα0 �= 1Aut(G/H) = IdG/H . Observe first that the set Hom(G, F ) is finiteby Lemma 2.4.4. As Ker(ψ) is of finite index in G for every ψ ∈ Hom(G, F ),we deduce that H is of finite index in G by applying Lemma 2.5.2. Thisimplies that the group Aut(G/H) is finite. On the other hand, we haveα0(g0H) = α0(g0)H �= g0H since H is a subgroup of Ker(φ) and φ(g0) �= g0.Therefore α0 �= IdG/H . This shows that the group Aut(G) is residually finite.

��

Every free group is residually finite by Theorem 2.3.1. Therefore we deducefrom Theorem 2.5.1 the following result.

Corollary 2.5.3. The group Aut(Fn) is residually finite for every n ≥ 1. ��

Every finitely generated abelian group is residually finite by Corollary 2.2.4.Thus we have:

Corollary 2.5.4. The automorphism group of a finitely generated abeliangroup is residually finite. ��

Remark 2.5.5. The automorphism group of a free abelian group of finite rankn is isomorphic to GLn(Z). Thus, the residual finiteness of GLn(Z) (Propo-sition 2.1.5) may also be deduced from Corollary 2.5.4.

Corollary 2.5.6. Let R be a ring and let M be a left (or right) module overR. Suppose that M is finitely generated as a Z-module. Then the automor-phism group AutR(M) of the R-module M is residually finite.

Proof. The group AutR(M) is a subgroup of AutZ(M). Since AutZ(M) isresidually finite by Corollary 2.5.4, we deduce that AutR(M) is residuallyfinite by applying Proposition 2.2.1. ��

2.6 Examples of Finitely Generated Groups Which Are Not Residually Finite 47

Corollary 2.5.7. Let R be a ring. Suppose that R is finitely generated as aZ-module. Then the group GLn(R) is residually finite for every n ≥ 1.

Proof. This is an immediate consequence of the preceding corollary sincethe group GLn(R) is isomorphic to AutR(Rn), where Rn is viewed as a leftmodule over R. ��

Example 2.5.8. The ring of Gaussian integers Z[i] = {a + bi : a, b ∈ Z} isa free abelian Z-module of rank 2. Thus, the group GLn(Z[i]) is residuallyfinite for every n ≥ 1 by Corollary 2.5.7. More generally, consider a numberfield K, that is, a field extension of Q such that d = dimQ K < ∞. Let Adenote the ring of algebraic integers of K (we recall that an element x ∈ Kis called an algebraic integer of K if x is a root of a monic polynomial withintegral coefficients). It is a standard fact in algebraic number theory that Ais a free Z-module of rank d. Thus, the group GLn(A) is residually finite forevery n ≥ 1 by Corollary 2.5.7.

2.6 Examples of Finitely Generated Groups Which AreNot Residually Finite

We have seen in Example 2.1.9 that the additive group Q is not residuallyfinite. Observe that the group Q is not finitely generated. In fact, any finitelygenerated abelian group is residually finite by Corollary 2.2.4. The purposeof this section is to give two examples of finitely generated groups G1 and G2

which are not residually finite.Our first example is a subgroup of the symmetric group Sym(Z) generated

by two elements:

Proposition 2.6.1. Let G1 denote the subgroup of Sym(Z) generated by thetranslation T : n �→ n + 1 and the transposition S = (0 1). Then G1 is afinitely generated Hopfian group which is not residually finite.

Let us first establish the following lemmas:

Lemma 2.6.2. Let G be an infinite simple group. Then G is not residuallyfinite.

Proof. The only normal subgroup of finite index of G is G itself. ThereforeG is not residually finite by Proposition 2.1.11(iii). ��

Given a set X, we recall that Sym0(X) denotes the subgroup of Sym(X)consisting of all permutations of X whose support is finite (see Appendix C).

Lemma 2.6.3. Let X be an infinite set. Then the group Sym0(X) is notresidually finite.


Proof. The subgroup Sym+0 (X) ⊂ Sym(X) consisting of all permutations of

X with finite support and signature 1 is an infinite simple group by Theo-rem C.4.3. Therefore Sym+

0 (X) is not residually finite by Lemma 2.6.2. Wededuce that Sym0(X) is not residually finite by using Proposition 2.2.1. ��

Lemma 2.6.4. The group G1 contains Sym0(Z) as a normal subgroup. More-over, G1 is the semidirect product of Sym0(Z) with the infinite cyclic subgroupof G1 generated by T .

Proof. For all i ∈ Z, we have

T iST−i = (i i + 1), (2.1)

which shows that (i i + 1) ∈ G1. If i < j, we also have

(i j + 1) = (j j + 1)(i j)(j j + 1)

which, by induction on j, shows that (i j) ∈ G1 for all i, j ∈ Z with i < j.Since the transpositions (i j), with i, j ∈ Z and i < j, generate Sym0(Z) byCorollary C.2.4, it follows that G1 contains Sym0(Z). The fact that Sym0(Z)is normal in Sym(Z) (Proposition C.2.2) implies that Sym0(Z) is normalin G1. Let H denote the subgroup of G1 generated by T . It is clear thatH ∩ Sym0(Z) is reduced to the identity map IdZ. On the other hand, byusing (2.1), we see that every element g ∈ G1 may be written in the formg = hσ, where h ∈ H and σ ∈ Sym0(Z). Therefore, G1 is the semidirectproduct of Sym0(Z) and H. ��

Proof of Proposition 2.6.1. The group G1 contains Sym0(Z) as a subgroup byLemma 2.6.4. As the group Sym0(Z) is not residually finite by Lemma 2.6.3,we conclude that G1 is not residually finite by applying Proposition 2.2.1.

It remains to show that G1 is Hopfian. We start by observing that thereare exactly two surjective homomorphisms from G1 onto Z. Indeed, as G1 isthe semidirect product of Sym0(Z) with the subgroup generated by T (Lem-ma 2.6.4), every element g ∈ G1 can be uniquely written in the form g = T kσ,where k ∈ Z and σ ∈ Sym0(Z), and the map u : G1 → Z defined by u(g) = kis a surjective homomorphism. As all elements of Sym0(Z) have finite order, itimmediately follows that the only homomorphisms from G1 onto Z are u and−u. Now, let φ : G1 → G1 be a surjective homomorphism. Then u◦φ : G1 → Z

is a surjective homomorphism so that, by our preceding observation, we haveu◦φ = u or −u. As Ker(u) = Sym0(Z) and φ : G1 → G1 is onto, it follows thatKer(φ) ⊂ Sym0(Z) and φ(Sym0(Z)) = Sym0(Z). The simplicity of Sym+

0 (Z)(Theorem C.4.3) implies that Ker(φ) ∩ Sym+

0 (Z) is either equal to Sym+0 (Z)

or reduced to the identity. We cannot have Ker(φ) ∩ Sym+0 (Z) = Sym+

0 (Z),that is, Sym+

0 (Z) ⊂ Ker(φ), since this would imply that φ(Sym0(Z)) is eitherreduced to the identity or cyclic of order 2, which contradicts φ(Sym0(Z)) =Sym0(Z). Consequently, we have Ker(φ)∩Sym+

0 (Z) = {IdZ}. If Ker(φ) is not

2.6 Examples of Finitely Generated Groups Which Are Not Residually Finite 49

reduced to the identity, it follows that Ker(φ) is a normal subgroup of order2 of Sym0(Z). But this is impossible since Proposition C.2.3 combined withProposition C.3.2 implies that, if X is an infinite set, then every non-trivialelement in Sym0(X) has an infinite number of conjugates in Sym0(X). ThusKer(φ) must be reduced to the identity, that is, φ is injective. This showsthat G1 is Hopfian. ��

Our second example of a finitely generated but not residually finite groupwill also show that the class of residually finite groups is not closed underextensions. In other words, a group G admitting a normal subgroup N suchthat both N and G/N are residually finite may fail to be residually finite.

In order to construct it we consider first the group

H =⊕

i∈Z

Hi,

where each Hi is a copy of the alternating group Sym+5 . Let ψ : Z → Aut(H)

be the homomorphism defined by ψ(n) = αn, where α ∈ Aut(H) is theone-step shift given by

α(h) = (hi−1)i∈Z

for all h = (hi)i∈Z ∈ H.

Proposition 2.6.5. The semidirect product G2 = H �ψ Z is a finitely gen-erated but not residually finite group.

Proof. By definition of a semidirect product, G2 is the group with underlyingset H × Z and group operation given by

(h, n)(h′, n′) = (hαn(h′), n + n′)

for all (h, n), (h′, n′) ∈ H × Z.Let t be the element of G2 defined by t = (1H , 1) and identify each element

h ∈ H with the element (h, 0) ∈ G. Then H is a normal subgroup of G2 andthe quotient group G2/H is an infinite cyclic group generated by the class oft. One has

(h, n) = htn (2.2)

andtnht−n = αn(h) (2.3)

for all h ∈ H and n ∈ Z. It follows from (2.2) that G2 is generated by t andthe elements of H. In fact, since H is the direct sum of the groups Hi and

Hi = tiH0t−i (2.4)

by (2.3), we deduce that G2 is generated by t and the elements of H0. As thegroup H0 is finite, this shows that G is finitely generated.


Let us prove now that G2 is not residually finite. Suppose on the contrarythat G2 is residually finite. Let g be an element in H0 such that g �= 1H0 .The residual finiteness of G2 implies the existence of a finite group F and ahomomorphism φ : G2 → F such that φ(g) �= 1F . As H0 is a simple group, therestriction of φ to H0 is injective. Therefore, the group φ(H0) is isomorphicto H0 and hence to Sym+

5 . Let m = |F |. We have φ(tm) = φ(t)m = 1F . Thus,it follows from (2.4) that φ(Hm) = φ(H0). Since xy = yx for all x ∈ H0 andy ∈ Hm, this implies that φ(H0) is abelian. This contradicts the fact thatφ(H0) is isomorphic to Sym+

5 , which is not abelian. ��

Remark 2.6.6. Observe that the group H is residually finite by Corollary2.2.3. The quotient group G2/H is also residually finite since it is isomorphicto Z. This shows that an extension of a residually finite group by a residuallyfinite group need not to be residually finite.

2.7 Dynamical Characterization of Residual Finiteness

Let G be a group and let A be a set. Recall that the set AG = {x : G → A}is equipped with the prodiscrete topology and that G acts on AG by the leftshift defined by (1.2).

Theorem 2.7.1. Let G be a group. Then the following conditions are equiv-alent:

(a) the group G is residually finite;(b) for every set A, the set of points of AG which have a finite G-orbit is

dense in AG;(c) there exists a set A having at least two elements such that the set of points

of AG which have a finite G-orbit is dense in AG;(d) there exists a Hausdorff topological space X equipped with a continuous

and faithful action of G such that the set of points of X which have afinite G-orbit is dense in X.

We recall that an action of a group G on a set X is called faithful if 1G isthe only element of G fixing all points of X.

Lemma 2.7.2. Let G be a group and let A be a set having at least two ele-ments. Then the action of G on AG is faithful.

Proof. Let a and b be two distinct elements in A. Consider an element g0 inG such that g0 �= 1G. Let x ∈ AG be the configuration defined by x(g) = a ifg = 1G and x(g) = b otherwise. We have g0x �= x since g0x(1G) = x(g−1

0 ) = band x(1G) = a. Consequently, the action of G on AG is faithful. ��

Lemma 2.7.3. Let G be a residually finite group and let Ω be a finite subsetof G. Then there exists a normal subgroup of finite index K of G such that therestriction of the canonical homomorphism ρ : G → G/K to Ω is injective.

Notes 51

Proof. Consider the finite subset

S = {g−1h : g, h ∈ Ω and g �= h} ⊂ G.

Since G is residually finite, we can find, for every s ∈ S, a normal subgroupof finite index Ns ⊂ G such that s /∈ Ns. The set K = ∩s∈SNs is a normalsubgroup of finite index in G by Lemma 2.5.2. Let ρ : G → G/K be thecanonical homomorphism. If g and h are distinct elements in Ω, then g−1h /∈K and hence ρ(g) �= ρ(h). ��

Proof of Theorem 2.7.1. Suppose that G is residually finite. Let A be a setand let W be a neighborhood of a point x in AG. Let us show that W containsa configuration with finite G-orbit.

Consider a finite subset Ω ⊂ G such that

V (x,Ω) = {y ∈ AG : y|Ω = x|Ω} ⊂ W.

By Lemma 2.7.3, we can find a normal subgroup of finite index K ⊂ G suchthat the restriction to Ω of the canonical homomorphism ρ : G → G/K(=K\G) is injective. This implies that the map Φ : AG/K → AΩ defined byΦ(z) = (z ◦ ρ)|Ω is surjective. Thus we can find an element z0 ∈ AG/K

such that the configurations z0 ◦ ρ and x coincide on Ω, that is, such thatz0 ◦ ρ ∈ V (x,Ω). On the other hand, the configuration z0 ◦ ρ is K-periodicby Proposition 1.3.3 (observe that K\G = G/K as K is normal in G). AsK is of finite index in G, we deduce that the G-orbit of z0 ◦ ρ is finite.Thus W contains a configuration whose G-orbit is finite. This shows that (a)implies (b).

Implication (b) ⇒ (c) is trivial.The fact that (c) implies (d) follows from Proposition 1.2.1, Proposi-

tion 1.2.2, and Lemma 2.7.2.Let us show that (d) implies (a). Suppose that the group G acts con-

tinuously and faithfully on a Hausdorff topological space X and let Edenote the set of points of X whose G-orbit is finite. For x ∈ E, letStab(x) = {g ∈ G : gx = x} denote the stabilizer of x in G. Observe thatx ∈ E if and only if Stab(x) is of finite index in G. If E is dense in X, then∩x∈E Stab(x) = {1G} since the action of G on X is continuous and faithful.Thus, by Proposition 2.1.11 we have that G is residually finite. ��

Notes

A survey article on residually finite groups has been written by W. Magnus[Mag].

A group G is called linear if there exist an integer n ≥ 1 and a field K suchthat G is isomorphic to a subgroup of GLn(K). A theorem of A.I. Mal’cev[Mal1] asserts that every finitely generated linear group is residually finite.


An example of a finitely presented group which is residually finite but notlinear was recently given by C. Drutu and M. Sapir [DrS].

The residual finiteness of free groups was established by F. Levi [Lev].The proof presented in Sect. 2.3 is based on the fact that the matrices

(1 20 1

)

and(

1 02 1

)generate a free subgroup of GL2(Z). This last result is due to I.N.

Sanov [San]. The reader may find other proofs of the residual finiteness offree groups in [RobD, p. 158] and [MaKS, p. 116].

Hopfian groups are named after H. Hopf who used topological methodsto prove that fundamental groups of closed orientable surfaces are Hop-fian (see [MaKS, p. 415]) and raised the question of the existence of finitelygenerated non-Hopfian groups. The fact that every finitely generated resid-ually finite group is Hopfian (Theorem 2.4.3) was discovered by Mal’cev[Mal1]. The first example of a finitely generated non-Hopfian group was givenby B.H. Neumann [Neu2]. Shortly after, a finitely presented non-Hopfiangroup was found by G. Higman [Hig]. The simplest example of a finitelypresented non-Hopfian group is certainly provided by the Baumslag-Solitargroup BS(2, 3) = 〈x, y : yx2y−1 = x3〉, that is, the quotient of the free groupF2 based on two generators x and y by the smallest normal subgroup of F2

containing the element x−3yx2y−1 (see [BaS], [MaKS], [LS]). On the otherhand, the Baumslag-Solitar group BS(2, 4) = 〈x, y : yx2y−1 = x4〉 is anexample of a finitely presented Hopfian group which is not residually finite(the incorrect statement in [BaS] about the residual finiteness of BS(2, 4)was corrected by S. Meskin in [Mes]).

The residual finiteness of the automorphism group of a finitely generatedresidually finite group (Theorem 2.5.1) was proved by G. Baumslag in [Bau].

The group G1 of Sect. 2.6 was considered by Mal’cev in [Mal1]. Givengroups A and G, the wreath product A � G is the semidirect product H �ψ

G, where H = ⊕g∈GA and ψ : G → Aut(A) is the group homomorphismassociated with the action of G on H ⊂ AG induced by the G-shift (see[RobD], [Rot]). Thus, the group G2 of Sect. 2.6 is the wreath product G2 =Sym+

5 �Z. The proof of Proposition 2.6.5 shows that, if A is a finitely generatednonabelian simple group and G is a finitely generated infinite group, thenthe wreath product A �G is a finitely generated group which is not residuallyfinite. Examples of finitely generated nonabelian simple groups are providedby the (finite) groups Sym+

n , where n ≥ 5, or PSLn(K), where n ≥ 2 and Ka finite field having at least 4 elements, or one of the famous infinite finitelypresented simple Thompson groups T or V (see [CFP]).

Exercises

2.1. Show that every quotient of a divisible group is a divisible group.2.2. Show that every torsionfree divisible abelian group G is isomorphic to adirect sum of copies of Q. Hint: Prove that there is a natural Q-vector spacestructure on G.

Exercises 53

2.3. Let G be a group.(a) Show that it is possible to define a topology on G by taking as open

sets the subsets Ω ⊂ G which satisfy the following property: for each g ∈ Ωthere is a subgroup of finite index H ⊂ G such that gH ⊂ Ω. This topologyis called the profinite topology on G.

(b) Show that G is residually finite if and only if the profinite topology onG is Hausdorff.

2.4. Show that the class of residually finite groups is not closed under takingquotients. Hint: Any group is isomorphic to a quotient of a free group (seeCorollary D.4.2).

2.5. Let X be an infinite set. Show that the symmetric group Sym(X) is notresidually finite. Hint: Use Cayley’s theorem (Theorem C.1.2) to prove thatSym(X) contains a subgroup isomorphic to Q or use Lemma 2.6.3.

2.6. Let G be a residually finite group and let A be a finite set.(a) Show that there exists a canonical injective homomorphism of the

group ICA(G;A) (cf. Sect. 1.10) into the group∏

H Sym(Fix(H)), whereH runs over all finite index subgroups of G. Hint. Use the fact that theconfigurations with finite G-orbit are dense in AG (cf. Theorem 2.7.1).

(b) Deduce from (a) that ICA(G;A) is residually finite.

2.7. Let m be an integer such that |m| ≥ 2. Let G denote the quotient ofthe free group F2 on two generators x and y by the normal closure of thesingle element xyx−1y−m. Thus, G is the group given by the presentationG = 〈a, b : aba−1 = bm〉, where a and b denote the images in G of x and y bythe quotient homomorphism.

(a) Show that every element g ∈ G may be (not uniquely) written in theform g = aibjak for some i, j, k ∈ Z.

(b) Show that there is a unique group homomorphism φ : G → GL2(Q)which satisfies

φ(a) =(

m 00 1

)and φ(b) =

(1 10 1

).

(c) Use (a) to show that φ is injective and that the image of φ is thesubgroup φ(G) ⊂ GL2(Q) given by

φ(G) ={(

mn r0 1

): n ∈ Z, r ∈ Z[1/m]

},

where Z[1/m] denotes the set of all rationals r ∈ Q which can be written inthe form r = umv for some u, v ∈ Z.

(d) Use (c) to prove that G is torsionfree.(e) Prove that G is residually finite. Hint: You have to show that, for any

element g ∈ G \ {1G}, there exist a finite group F and a homomorphismψ : G → F such that ψ(g) �= 1F . Write g = aibjak, where i, j, k ∈ Z, and


treat first the case i + k �= 0 by using determinants. If i + k = 0, choose aprime number p which is not a divisor of m nor a divisor of j, and considerthe homomorphism ψ : G → GL2(Z/pZ) defined by

ψ(a) =(

m 00 1

)and ψ(b) =

(1 10 1

),

where n ∈ Z/pZ denotes the class of n ∈ Z modulo p.

2.8. Let m ≥ 2 be an integer. Denote by Z[1/m] the subgroup of the additivegroup Q consisting of all rationals r which can be written in the form r = kmn

for some k, n ∈ Z. Show that the group Z[1/m] is residually finite. Hint:Observe that Z[1/m] is isomorphic to a subgroup of the group G studied inExercise 2.7.

2.9. Let K be a field of characteristic p > 0. Show that the additive groupK is residually finite. Hint: Use a base of K seen as a Z/pZ vector space andapply Corollary 2.2.3.

2.10. Show that all elements of the additive group Q/Z have finite order butthat Q/Z is not residually finite.

2.11. Show that the multiplicative group Q∗ of nonzero rational numbers is

residually finite. Hint: Use the unique factorization of elements of Q∗ into

powers of prime numbers and apply Corollary 2.2.3.

2.12. Show that the automorphism group Aut(Q) of the additive group Q isisomorphic to the multiplicative group Q

∗.

2.13. Show that the additive group R is neither residually finite nor Hopfian.

2.14. Let p be a prime. Show that the additive group Zp of p-adic integers isnot Hopfian.

2.15. Show that the multiplicative group R∗ of nonzero real numbers is nei-

ther residually finite nor Hopfian.

2.16. Let F be a free group of finite rank n. Suppose that S is a finitegenerating subset of F of cardinality |S| ≤ n. Use the fact that F is Hopfian(Corollary 2.4.5) to prove that |S| = n and that S is a base for F .

2.17. (M. Hall [Hal-M-2]) Let G be a finitely generated group and let n ≥ 1be an integer. Show that G contains only a finite number of subgroups ofindex n. Hint: Use Lemma 2.1.10 and Lemma 2.4.4.

2.18. Show that the group GLn(Z) is finitely generated. Hint: For 1 ≤ i, j ≤n, let Eij denote the n×n matrix all of whose entries are 0 except the entrylocated on the ith row and the jth column which is equal to 1. Use Euclideandivision in Z to show that GLn(Z) is generated by the n2 matrices In − 2Eii,In + Eij , where 1 ≤ i, j ≤ n and i �= j.

Exercises 55

2.19. Show that GLn(Z) is Hopfian.

2.20. Give a direct proof of the fact that the automorphism group of a finitelygenerated abelian group G is residually finite (Corollary 2.5.4). Hint: Showthat there is a finite group F and an integer n ≥ 0 such that Aut(G) isisomorphic to F × GLn(Z).

2.21. Show that the group G1 considered in Sect. 2.6 contains no normalsubgroup H such that both H and G/H are residually finite.

2.22. Show that the group G2 considered in Sect. 2.6 is Hopfian.

2.23. A group G is called almost perfect if all nontrivial finite quotients of Gare nonabelian. Show that, if A is a finitely generated almost perfect nontriv-ial group and G is a finitely generated infinite group, then the wreath productgroup A � G is finitely generated but not residually finite. Hint: Follow theproof of Proposition 2.6.5.

2.24. Let G be a group. Denote by Nfq the set of all normal subgroups offinite index of G, partially ordered by reverse inclusion.

(a) Show that Nfq is a directed set.(b) For H,K ∈ Nfq with H ⊂ K, let ϕK,H : G/H → G/K denote the

canonical homomorphism. Show that the directed set Nfq together with thehomomorphisms ϕK,H form a projective system of groups. The limit of thisprojective system is called the profinite completion of the group G and isdenoted by G.

(c) Show that there is a canonical homomorphism η : G → G and that thekernel of η is the residual subgroup of G.

(d) Prove that G is residually finite if and only if the canonical homomor-phism η : G → G is injective.

Chapter 3

Surjunctive Groups

Surjunctive groups are defined in Sect. 3.1 as being the groups on which allinjective cellular automata with finite alphabet are surjective. In Sect. 3.2 it isshown that every subgroup of a surjunctive group is a surjunctive group andthat every locally surjunctive group is surjunctive. Every locally residuallyfinite group is surjunctive (Corollary 3.3.6). The class of locally residuallyfinite groups is quite large and includes in particular all finite groups, allabelian groups, and all free groups (a still wider class of surjunctive groups,namely the class of sofic groups, will be described in Chap. 7). In Sect. 3.4,given an arbitrary group Γ , we introduce a natural topology on the set of itsquotient groups. In Sect. 3.7, it is shown that the set of surjunctive quotientsis closed in the space of all quotients of Γ .

3.1 Definition

A set X is finite if and only if every injective map f : X → X is surjective.The definition given below is related to this characterization of finite sets.

Definition 3.1.1. A group G is said to be surjunctive if it satisfies the fol-lowing condition: if A is a finite set, then every injective cellular automatonτ : AG → AG is surjective (and hence bijective).

Remark 3.1.2. Given a group G and a finite set A, it follows from Theo-rem 1.8.1 that a map f : AG → AG is a cellular automaton if and only if f isG-equivariant (with respect to the G-shift) and continuous (with respect tothe prodiscrete topology on AG). Thus, the definition of a surjunctive groupmay be reformulated as follows: a group G is surjunctive if and only if, forany finite set A, every injective G-equivariant continuous map f : AG → AG

is surjective.

Proposition 3.1.3. Every finite group is surjunctive.


57

http://dx.doi.org/10.1007/978-3-642-14034-1_3

58 3 Surjunctive Groups

Proof. If G is a finite group and A is a finite set, then the set AG is finite.Therefore, every injective cellular automaton τ : AG → AG is surjective. ��

3.2 Stability Properties of Surjunctive Groups

Proposition 3.2.1. Every subgroup of a surjunctive group is surjunctive.

Proof. Suppose that H is a subgroup of a surjunctive group G. Let A bea finite set and let τ : AH → AH be an injective cellular automaton overH. Consider the cellular automaton τG : AG → AG over G obtained from τby induction (see Sect. 1.7). The fact that τ is injective implies that τG isinjective by Proposition 1.7.4(i). Since G is surjunctive, it follows that τG issurjective. By applying Proposition 1.7.4(ii), we deduce that τ is surjective.

��


(a) G is surjunctive;(b) every finitely generated subgroup of G is surjunctive.

Proof. The fact that (a) implies (b) follows from Proposition 3.2.1. Con-versely, let G be a group all of whose finitely generated subgroups are sur-junctive. Let A be a finite set and let τ : AG → AG be an injective cellularautomaton with memory set S. Let H denote the subgroup of G generatedby S and consider the cellular automaton τH : AH → AH obtained by re-striction of τ (see Sect. 1.7). The fact that τ is injective implies that τH isinjective by Proposition 1.7.4(i). As H is finitely generated, it is surjunc-tive by our hypothesis on G. It follows that τH is surjective. By applyingProposition 1.7.4(ii), we deduce that τ is also surjective. This shows that (b)implies (a). ��

If P is a property of groups (e.g. being finite, being nilpotent, being solv-able, being free, etc.), a group G is called locally P if all finitely generatedsubgroups of G satisfy P. With this terminology, Proposition 3.2.2 may berephrased by saying that the class of surjunctive groups and the class oflocally surjunctive groups coincide.

Corollary 3.2.3. Every locally finite group is surjunctive.

Proof. This immediately follows from Proposition 3.2.2 since every finitegroup is surjunctive by Proposition 3.1.3. ��

Example 3.2.4. Let X be a set and consider the group Sym0(X) consisting ofall permutations of X which have finite support (see Sect. C.2). The groupSym0(X) is locally finite. Indeed, if Σ is a finite subset of Sym0(X), then

3.3 Surjunctivity of Locally Residually Finite Groups 59

the subgroup H ⊂ Sym0(X) generated by Σ is finite since it is isomorphicto a subgroup of Sym(A), where A denotes the union of the supports ofthe elements of Σ. Consequently, the group Sym0(X) is surjunctive. Observethat Sym0(X) is infinite when X is infinite. This yields our first examplesof infinite surjunctive groups. Finally, note that it follows from Lemma 2.6.3that Sym0(X) is not residually finite whenever X is infinite.

3.3 Surjunctivity of Locally Residually Finite Groups

Theorem 3.3.1. Every residually finite group is surjunctive.

Let us first establish an important property of cellular automata with finitealphabet, namely the fact that they always have a closed image with respectto the prodiscrete topology on the set of configurations:

Lemma 3.3.2. Let G be a group and let A be a finite set. Let τ : AG → AG bea cellular automaton. Then the set τ(AG) is closed in AG for the prodiscretetopology.

Proof. Since A is finite, the space AG is compact by Tychonoff theorem (seeCorollary A.5.3). As τ is continuous by Proposition 1.4.8, we deduce thatτ(AG) is a compact subset of AG. This implies that τ(AG) is closed in AG

since AG is Hausdorff by Proposition 1.2.1. ��

The following example shows that Lemma 3.3.2 becomes false if we sup-press the hypothesis that the alphabet is finite.

Example 3.3.3. Consider the map τ : NZ → N

Z given by

τ(x)(n) = max(0, x(n) − x(n + 1))

for all x ∈ NZ and n ∈ Z. Clearly, τ is a cellular automaton over the group

Z and the alphabet N with memory set {0, 1}. Consider the configurationy : Z → N defined by y(n) = 1 if n ≥ 0 and y(n) = 0 otherwise. Let F be afinite subset of Z and choose an integer M ≥ 1 such that F ⊂ [−M+1, M−1].Consider the configuration xM : Z → N defined by xM (n) = max(0, M − n)if n ≥ 0 and xM (n) = M otherwise. Observe that yM = τ(xM ) is suchthat yM (n) = 1 if 0 ≤ n ≤ M − 1 and yM (n) = 0 otherwise. Therefore, theconfigurations yM and y coincide on [−M +1, M −1] and hence on F . Thus yis in the closure of τ(NZ) in N

Z. On the other hand, it is clear that y is not inthe image of τ . This shows that τ(NZ) is not closed in N

Z for the prodiscretetopology.

In the proof of the surjunctivity of residually finite groups, we shall alsouse the following result:


Lemma 3.3.4. Let G be a group. Suppose that G satisfies the following prop-erty: for each finite subset Ω ⊂ G, there exist a surjunctive group Γ and ahomomorphism φ : G → Γ such that the restriction of φ to Ω is injective.Then G is surjunctive.

Proof. Let A be a finite set and equip AG with its prodiscrete topology.Suppose that τ : AG → AG is an injective cellular automaton. Let us showthat τ is surjective, that is, τ(AG) = AG. Since τ(AG) is closed in AG byLemma 3.3.2, it suffices to prove that τ(AG) is dense in AG. Consider aconfiguration x ∈ AG and a neighborhood W of x in AG. Then there is afinite subset Ω ⊂ G such that

V (x,Ω) = {y ∈ AG : y|Ω = x|Ω} ⊂ W.

By our hypothesis, we may find a surjunctive group Γ and a homomorphismφ : G → Γ such that the restriction of φ to Ω is injective. Let K denote theimage of φ and let N denote its kernel. By Proposition 1.3.3, there is a bijec-tive map ψ : AK → Fix(N) given by ψ(z) = z ◦ φ for all z ∈ AK . Moreover,if we identify AK with Fix(N) via ψ, the restriction of τ to Fix(N) yields acellular automaton τ : AK → AK over the group K (see Proposition 1.6.1).Since the group Γ is surjunctive, the group K is surjunctive by Proposi-tion 3.2.1. We deduce that τ is surjective and hence τ(Fix(N)) = Fix(N).On the other hand, as the restriction of φ to Ω is injective, we may findz0 ∈ AK such that ψ(z0) = x|Ω. We then have

ψ(z0) ∈ Fix(N) ∩ V (x,Ω) = τ(Fix(N)) ∩ V (x,Ω).

This shows that W meets the image of τ . Thus τ(AG) is dense in AG. ��

Proof of Theorem 3.3.1. Let G be a residually finite group. If Ω is a finitesubset of G, then there exist, By Lemma 2.7.3, a finite group Γ and a ho-momorphism φ : G → Γ such that the restriction of φ to Ω is injective. Asfinite groups are surjunctive by Proposition 3.1.3, it follows that G satisfiesthe hypothesis of Lemma 3.3.4. Consequently, G is surjunctive. ��

Corollary 3.3.5. Every free group is surjunctive.

Proof. Free groups are residually finite by Theorem 2.3.1. ��

Corollary 3.3.6. Every locally residually finite group is surjunctive.

Proof. This immediately follows from Theorem 3.3.1 by using Proposi-tion 3.2.2. ��

Corollary 3.3.7. Every abelian group is surjunctive.

Proof. This is an immediate consequence of Corollary 3.3.6 since everyabelian group is locally residually finite by Corollary 2.2.4. ��

3.4 Marked Groups 61

When X is a finite set, every surjective map f : X → X is injective.Therefore, a surjective cellular automaton with finite alphabet over a finitegroup is necessarily injective. However, a surjective cellular automaton withfinite alphabet τ : AG → AG may fail to be injective when the group G isinfinite as shown by the following example.

Example 3.3.8. Take A = Z/2Z = {0, 1} and G = Z. Let τ : AZ → AZ be themap defined by τ(x)(n) = x(n) + x(n + 1) for all x ∈ AZ and n ∈ Z. Clearly,τ is a cellular automaton with memory set S = {0, 1} ⊂ Z and local definingmap μ : AS → A given by μ(y) = y(0) + y(1) for all y ∈ AS . The cellularautomaton τ is surjective. Indeed, given an element y ∈ AG, the configurationx : Z → Z/2Z defined by

x(n) =

⎧⎪⎨

⎪⎩

0 if n = 0,

y(0) + y(1) + · · · + y(n − 1) if n > 0,

y(n) + y(n + 1) + · · · + y(−1) if n < 0,

satisfies τ(x) = y. However τ is not injective since the two constant configu-rations have the same image under τ .

3.4 Marked Groups

Let Γ be a group.A Γ -quotient is a pair (G, ρ), where G is a group and ρ : Γ → G is a

surjective homomorphism. We define an equivalence relation on the class ofall Γ -quotients by declaring that two Γ -quotients (G1, ρ1) and (G2, ρ2) areequivalent when there exists a group isomorphism φ : G2 → G1 such that thefollowing diagram is commutative:

G2

φ∼=Γ

ρ2

ρ1

G1

that is, such that ρ1 = φ ◦ ρ2. An equivalence class of Γ -quotients is calleda Γ -marked group. Observe that two Γ -quotients (G1, ρ1) and (G2, ρ2) areequivalent if and only if Ker(ρ1) = Ker(ρ2). Thus, the set of Γ -marked groupsmay be identified with the set N (Γ ) consisting of all normal subgroups of Γ .

Let us identify the set P(Γ ) consisting of all subsets of Γ with the set{0, 1}Γ by means of the bijection from P(Γ ) onto {0, 1}Γ given by A �→ χA,


where χA : Γ → {0, 1} is the characteristic map of A ⊂ Γ . We equip the setP(Γ ) = {0, 1}Γ with its prodiscrete uniform structure and N (Γ ) ⊂ P(Γ )with the induced uniform structure. Thus, a base of entourages of N (Γ ) isprovided by the sets

VF = {(N1, N2) ∈ N (Γ ) ×N (Γ ) : N1 ∩ F = N2 ∩ F},

where F runs over all finite subsets of Γ . Intuitively, two normal subgroupsof Γ are “close” in N (Γ ) when their intersection with a large finite subset ofΓ coincide.

Proposition 3.4.1. Let Γ be a group. Then the space N (Γ ) of Γ -markedgroups is a totally disconnected compact Hausdorff topological space.

Proof. The space P(Γ ) = {0, 1}Γ is totally disconnected and Hausdorff byProposition 1.2.1. Moreover, P(Γ ) is compact by Corollary A.5.3 since it isa product of finite spaces. Thus, it suffices to show that N (Γ ) is closed inP(Γ ). To see this, observe that a subset E ∈ P(Γ ) is a normal subgroup ofΓ if and only if it satisfies

(1) 1Γ ∈ E;(2) αβ−1 ∈ E for all α, β ∈ E;(3) γαγ−1 ∈ E for all α ∈ E and γ ∈ Γ .

Denoting, for each γ ∈ Γ , by πγ : P(Γ ) = {0, 1}Γ → {0, 1} the projectionmap corresponding to the γ-factor, these conditions are equivalent to

(1’) π1Γ(E) = 1;

(2’) πα(E)πβ(E)(παβ−1(E) − 1) = 0 for all α, β ∈ Γ ;(3’) πα(E)(πγαγ−1(E) − 1) = 0 for all α, γ ∈ Γ .

As all projection maps are continuous on P(Γ ), this shows that N (Γ ) isclosed in P(Γ ). ��

Remark 3.4.2. If Γ is countable then the uniform structure on N (Γ ) is metriz-able. Indeed, the uniform structure on P(Γ ) = {0, 1}Γ is metrizable when Γis countable (cf. Remark 1.9.2).

Let P be a property of groups. A group Γ is called residually P if foreach element γ ∈ Γ with γ �= 1Γ , there exist a group Γ ′ satisfying P andan epimorphism φ : Γ → Γ ′ such that φ(γ) �= 1Γ ′ . Observe that every groupwhich satisfies P is residually P.

Proposition 3.4.3. Let Γ be a group and let P be a property of groups.Suppose that the class of groups which satisfy P is closed under taking finiteproducts and subgroups. Then the following conditions are equivalent:

(a) Γ is residually P;(b) there exists a net (Ni)i∈I in N (Γ ) which converges to {1Γ } such that

Γ/Ni satisfies P for all i ∈ I.

3.4 Marked Groups 63

Proof. Suppose that Γ is residually P. For all γ ∈ Γ \{1Γ } we can find a groupΓγ satisfying P and an epimorphism φγ : Γ → Γγ such that φγ(γ) �= 1Γγ . LetI ⊂ P(Γ ) be the directed set consisting of all finite subsets of Γ not containing1Γ partially ordered by inclusion. For i ∈ I we denote by φi =

∏γ∈i φγ : Γ →∏

γ∈i Γγ the homomorphism defined by φi(γ′) = (φγ(γ′))γ∈I for all γ′ ∈ Γ .Set Ni = ker(φi) and let us show that the net (Ni)i∈I in N (Γ ) converges to{1Γ }. Given a finite set F ⊂ Γ set iF = F \{1Γ }. Let f ∈ F \{1Γ } and i ∈ Isuch that iF ⊂ i. Then φi(f) �= 1Γi since φf (f) �= 1Γf

. Thus

f /∈ Ni and f /∈ {1Γ } ∩ F. (3.1)

It follows that if 1Γ /∈ F then Ni ∩ F = ∅ = {1Γ } ∩ F . On the other hand,if 1Γ ∈ F , from (3.1) we deduce Ni ∩ F = {1Γ } = {1Γ } ∩ F . In either caseswe have Ni ∩F = {1Γ } ∩F . We deduce that limi Ni = {1Γ }. Finally, by ourassumptions on P, we have that

∏γ∈i Γγ and its subgroup φi(Γ ) ∼= Γ/Ni

satisfy P for all i ∈ I. This shows (a) ⇒ (b).Suppose now that (b) holds. Let γ ∈ Γ \ {1Γ }. Consider the finite set

F = {γ}. Then there exists iF ∈ I such that NiF∩ F = {1Γ } ∩ F and

Γ/NiFsatisfy P. As {1Γ } ∩ F = ∅ we have that γ /∈ NiF

. In other words,if φF : Γ → Γ/NiF

denotes the canonical epimorphism, we have φF (γ) �=1Γ/NiF

. We deduce that Γ is residually P. This shows (b) ⇒ (a). ��

Note that in the above proposition, the assumptions on P are not neededfor the implication (b) ⇒ (a)

Let A be a set. Consider the set AΓ equipped with its prodiscrete uniformstructure and the Γ -shift action. For each N ∈ N (Γ ), let

Fix(N) = {x ∈ AΓ : γx = x for all γ ∈ N} ⊂ AΓ

denote the set of configurations which are fixed by N . Recall from Proposi-tion 1.3.6 that Fix(N) is a closed Γ -invariant subset of AΓ .

Let us equip P(AΓ ) with the Hausdorff-Bourbaki uniform structure asso-ciated with the prodiscrete uniform structure on AΓ (see Sect. B.4 for thedefinition of the Hausdorff-Bourbaki uniform structure on the set of subsetsof a uniform space).

Theorem 3.4.4. Let Γ be a group and let A be a set. Then the mapΨ : N (Γ ) → P(AΓ ) defined by Ψ(N) = Fix(N) is uniformly continuous.Moreover, if A contains at least two elements then Ψ is a uniform embed-ding.

Proof. Let N0 ∈ N (Γ ) and let W be an entourage of P(AΓ ). Let us showthat there exists an entourage V of N (Γ ) such that

Ψ(V [N0]) ⊂ W [Ψ(N0)]. (3.2)

This will prove that Ψ is continuous. By definition of the Hausdorff-Bourbakiuniform structure on P(AΓ ), there is an entourage T of AΓ such that


T = {(X,Y ) ∈ P(AΓ ) × P(AΓ ) : Y ⊂ T [X] and X ⊂ T [Y ]} ⊂ W. (3.3)

Since AΓ is endowed with its prodiscrete uniform structure, there is a finitesubset F ⊂ Γ such that

U = {(x, y) ∈ AΓ × AΓ : πF (x) = πF (y)} ⊂ T, (3.4)

where πF : AΓ → AF is the projection map. Consider now the finite subsetE ⊂ Γ defined by

E = F · F−1 = {γη−1 : γ, η ∈ F},

and the entourage V of N (Γ ) given by

V = {(N1, N2) ∈ N (Γ ) ×N (Γ ) : N1 ∩ E = N2 ∩ E}.

We claim that V satisfies (3.2). To prove our claim, suppose that N ∈ V [N0].Let x ∈ Fix(N). The fact that N ∩ E = N0 ∩ E implies that if γ and ηare elements of F with γ = νη for some ν ∈ N , then ν ∈ N0 and thereforex(γ) = x(η). Denoting by ρ0 : Γ → Γ/N0 the canonical epimorphism, wededuce that we may find an element x0 ∈ AΓ/N0 such that x(γ) = x0 ◦ ρ0(γ)for all γ ∈ F . We have x0 ◦ ρ0 ∈ Fix(N0) and (x, x0 ◦ ρ0) ∈ U . Since U ⊂ Tby (3.4), this shows that Fix(N) ⊂ T [Fix(N0)]. Therefore Ψ is continuous.As N (Γ ) is compact by Proposition 3.4.1, we deduce that Ψ is uniformlycontinuous by applying Theorem B.2.3.

Suppose now that A has at least two elements. Let us show that Ψ isinjective. Let N1, N2 ∈ N (Γ ). Fix two elements a, b ∈ A with a �= b andconsider the map x : Γ → A defined by x(γ) = a if γ ∈ N1 and x(γ) = botherwise. We clearly have x ∈ Fix(N1). Suppose that Ψ(N1) = Ψ(N2),that is, Fix(N1) = Fix(N2). Then for all ν ∈ N2, we have ν−1x = x sincex ∈ Fix(N1) = Fix(N2), and hence x(ν) = ν−1x(1Γ ) = x(1Γ ) = a. Thisimplies N2 ⊂ N1. By symmetry, we also have N1 ⊂ N2. Therefore N1 = N2.This shows that Ψ is injective. As N (Γ ) is compact and P(AΓ ) is Hausdorff,we conclude that Ψ is a uniform embedding by applying Proposition B.2.5.

��

3.5 Expansive Actions on Uniform Spaces

Let X be a uniform space and let Γ be a group acting on X. We consider thediagonal action of Γ on X × X defined by γ(x, y) = (γx, γy) for all γ ∈ Γand x, y ∈ X.

One says that the action of Γ on X is uniformly continuous if the orbitmap X → X, x �→ γx, is uniformly continuous for each γ ∈ Γ . This is

3.6 Gromov’s Injectivity Lemma 65

equivalent to saying that γ−1V is an entourage of X for all entourage V ofX and γ ∈ Γ .

The action of Γ on X is said to be expansive if there exists an entourageW0 of X such that ⋂

γ∈Γ

γ−1W0 = ΔX , (3.5)

where ΔX = {(x, x) : x ∈ X} denotes the diagonal in X × X. Equality (3.5)means that if x, y ∈ X satisfy (γx, γy) ∈ W0 for all γ ∈ Γ , then x = y. Anentourage W0 satisfying (3.5) is then called an expansivity entourage for theaction of Γ on X.

Remarks 3.5.1. (a) If a uniform space X admits an expansive uniformly con-tinuous action of a group Γ , then the topology on X must be Hausdorff.Indeed, if W0 is an expansivity entourage for such an action then ΔX isequal to the intersection of the entourages γ−1W0, γ ∈ Γ , by (3.5).

(b) Suppose that B is a base of the uniform structure on X. Then anaction of a group Γ on X is expansive if and only if it admits an expansivityentourage B0 ∈ B.

(c) Let (X, d) be a metric space. Then an action of a group Γ on X isexpansive if and only if there exists a real number ε0 > 0 with the followingproperty: if x, y ∈ X satisfy d(γx, γy) < ε0 for all γ ∈ Γ , then x = y. Suchan ε is called an expansivity constant for the action of Γ on X.

Our basic example of a uniformly continuous and expansive action is pro-vided by the following:

Proposition 3.5.2. Let G be a group and let A be a set. Then the G-shifton AG is uniformly continuous and expansive with respect to the prodiscreteuniform structure on AG.

Proof. Uniform continuity follows from the fact that G acts on AG by per-muting coordinates. Expansiveness is due to the fact that the action of G onitself by left multiplication is transitive. Indeed, consider the entourage W0

of AG defined by

W0 = {(x, y) ∈ AG × AG : x(1G) = y(1G)}.

Given g ∈ G, we have (x, y) ∈ g−1W0 if and only if x(g−1) = y(g−1). Thus⋂g∈G g−1W0 is equal to the diagonal in AG × AG. ��

3.6 Gromov’s Injectivity Lemma

The following result will be used in the next section to prove that surjunctivegroups define a closed subset of the space of Γ -marked groups for any group Γ .


Theorem 3.6.1 (Gromov’s injectivity lemma). Let X be a uniformspace endowed with a uniformly continuous and expansive action of a group Γ .Let f : X → X be a uniformly continuous and Γ -equivariant map. Supposethat Y is a subset of X such that the restriction of f to Y is a uniformembedding. Then there exists an entourage V of X satisfying the followingproperty: if Z is a Γ -invariant subset of X such that Z ⊂ V [Y ], then therestriction of f to Z is injective.

(We recall that the notation Z ⊂ V [Y ] means that for each z ∈ Z thereexists y ∈ Y such that (z, y) ∈ V .)

Proof. By expansivity of the action of Γ , there is an entourage W0 of X suchthat ⋂

γ∈Γ

γ−1W0 = ΔX . (3.6)

It follows from the axioms of a uniform structure that we can find a symmetricentourage S of X such that

S ◦ S ◦ S ⊂ W0. (3.7)

Since the restriction of f to Y is a uniform embedding, we can find an en-tourage T of X such that

(f(y1), f(y2)) ∈ T ⇒ (y1, y2) ∈ S (3.8)

for all y1, y2 ∈ Y . Let U be a symmetric entourage of X such that

U ◦ U ⊂ T. (3.9)

Since f is uniformly continuous, we can find an entourage E of X such that

(x1, x2) ∈ E ⇒ (f(x1), f(x2)) ∈ U (3.10)

for all x1, x2 ∈ X.Let us show that the entourage V = S ∩ E has the required property. So

let Z be a Γ -invariant subset of X such that Z ⊂ V [Y ] and let us show thatthe restriction of f to Z is injective.

Let z′ and z′′ be points in Z such that f(z′) = f(z′′). Since f is Γ -equivariant, we have

f(γz′) = f(γz′′) (3.11)

for all γ ∈ Γ . As the points γz′ and γz′′ stay in Z, the fact that Z ⊂ V [Y ]implies that there are points y′

γ and y′′γ in Y such that (γz′, y′

γ) ∈ V and(γz′′, y′′

γ ) ∈ V . Since V ⊂ E, it follows from (3.10) that (f(γz′), f(y′γ)) and

(f(γz′′), f(y′′γ )) are in U . As U is symmetric, we also have (f(y′

γ), f(γz′)) ∈ U .We deduce that (f(y′

γ), f(y′′γ )) ∈ U ◦ U ⊂ T by using (3.9) and (3.11). This

implies (y′γ , y′′

γ ) ∈ S by (3.8). On the other hand, we also have (γz′, y′γ) ∈ S

and (y′′γ , γz′′) ∈ S since V ⊂ S and S is symmetric. It follows that

3.7 Closedness of Marked Surjunctive Groups 67

(γz′, γz′′) ∈ S ◦ S ◦ S ⊂ W0

by (3.7). This gives us(z′, z′′) ∈

⋂

γ∈Γ

γ−1W0,

and hence z′ = z′′ by (3.6). Thus the restriction of f to Z is injective. ��

3.7 Closedness of Marked Surjunctive Groups

Theorem 3.7.1. Let Γ be a group. Then the set of normal subgroups N ⊂ Γsuch that the quotient group Γ/N is surjunctive is closed in N (Γ ).

Proof. Let N ∈ N (Γ ) and let (Ni)i∈I be a net in N (Γ ) converging to N .Suppose that the groups Γ/Ni are surjunctive for all i ∈ I. Let us show thatthe group Γ/N is also surjunctive.

Let A be a finite set and let τ : AΓ/N → AΓ/N be an injective cellularautomaton over the group Γ/N . Let S ⊂ Γ/N be a memory set for τ withassociated local defining map μ : AS → A. Choose a subset S ⊂ Γ such thatthe canonical epimorphism ρ : Γ → Γ/N gives a bijection ψ : S → S, and letπ : AS → A

eS denote the bijective map induced by ψ. Consider the cellularautomaton τ : AΓ → AΓ over Γ with memory set S and local defining mapμ = μ ◦ π−1 : A

eS → A. To simplify notation, let us set X = AΓ , f = τ ,Y = Fix(N) and Zi = Fix(Ni).

We claim that the hypotheses of Theorem 3.6.1 are satisfied by X, f andY . Indeed, we first observe that the action of Γ on X is uniformly continuousand expansive by Proposition 3.5.2. On the other hand, the cellular automa-ton f : X → X is uniformly continuous and Γ -equivariant by Theorem 1.9.1.Finally, the restriction of f to Y is injective since this restriction is conjugateto τ by Proposition 1.6.1. As Y is a closed subset of X and hence compact,it follows from Proposition B.2.5 that the restriction of f to Y is a uniformembedding. By applying Theorem 3.6.1, it follows that there exists an en-tourage V of X such that if Z is a Γ -invariant subset of X with Z ⊂ V [Y ],then the restriction of f to Z is injective.

Since the net (Zi)i∈I converges to Y for the Hausdorff-Bourbaki topologyon P(X) by Theorem 3.4.4, there is an element i0 ∈ I such that Zi ⊂ V [Y ]for all i ≥ i0. As the sets Zi are Γ -invariant by Proposition 1.3.6, it followsthat the restriction of f to Zi is injective for all i ≥ i0. On the other hand,f(Zi) ⊂ Zi and the restriction of f to Zi is conjugate to a cellular automatonτi : AΓ/Ni → AΓ/Ni over the group Γ/Ni for all i ∈ I by Proposition 1.6.1. Asthe groups Γ/Ni are surjunctive by our hypotheses, we deduce that f(Zi) =Zi for all i ≥ i0. Now, it follows from Proposition B.4.6 that the net (f(Zi))i∈I

converges to f(Y ) in P(X). Thus, the net (Zi)i∈I converges to both Y andf(Y ). As Y and f(Y ) are closed in X (by compactness of Y ), we deduce that


Y = f(Y ) by applying Proposition B.4.3. This shows that τ is surjective sinceτ is conjugate to the restriction of f to Y . Consequently, the group Γ/N issurjunctive. ��

Notes

Surjunctive groups were introduced by W. Gottschalk. The first resultson these groups are due to W. Lawton who proved in particular Proposi-tion 3.2.1, Corollary 3.2.3, Theorem 3.3.1, and Corollary 3.3.7 (see [Got],[Law]).

The characterization of infinite sets by the existence of injective but notsurjective self-maps is known as Dedekind’s definition of infinite sets.

Proposition 3.3.6 together with Mal’cev theorem [Mal1] which says thatevery finitely generated linear group is residually finite implies that everylinear group is surjunctive.

The problem of the existence of a non surjunctive group was raised byGottschalk in [Got] and remains open up to now. Note that to prove thatall groups are surjunctive it would suffice to prove that the symmetricgroup Sym(N) is surjunctive (this immediately follows from Proposition 3.2.1,Proposition 3.2.2, and the fact that every finitely generated group is count-able and hence isomorphic to a subgroup of Sym(N) by Cayley’s theorem).

In [CeC11] it is shown that if G is a non-periodic group, then for every infi-nite set A there exists a cellular automaton τ : AG → AG whose image τ(AG)is not closed in AG with respect to the prodiscrete topology (cf. Lemma 3.3.2and Example 3.3.3).

Theorem 3.6.1 is a uniform version of Lemma 4.H” in [Gro5]. The proofpresented in this chapter of the surjunctivity of limits of surjunctive groups(Theorem 3.7.1) closely follows Sect. 4 of [Gro5] (see also [CeC10]). There isanother proof based on techniques from model theory (see [Gro5] and [GlG]).

Exercises

3.1. Let K be an algebraically closed field. Show that every injective polyno-mial map f : K → K is surjective.

3.2. Show that every injective polynomial map f : R → R is surjective.

3.3. Show that the polynomial map f : Q → Q defined by f(x) = x3 isinjective but not surjective.

3.4. Show that every injective holomorphic map f : C → C is surjective.

3.5. Give an example of a real analytic map f : R → R which is injective butnot surjective.

Exercises 69

3.6. Let K be a field and let V be a vector space over K. Show that V isfinite-dimensional if and only if every injective endomorphism f : V → V issurjective.

3.7. Let R be a ring and let M be a left (or right) R-module. One says thatM is Artinian if every descending chain N0 ⊃ N1 ⊃ N2 ⊃ . . . eventuallystabilizes (i.e., there is an integer n0 ≥ 0 such that NN = Nn+1 for alln ≥ n0). Show that if M is Artinian then every injective endomorphismf : M → M is surjective. Hint: Consider the sequence of submodules definedby Nn = Im(fn) for n ≥ 0.

3.8. Let U = {z ∈ C : |z| = 1}. Show that every injective continuous mapf : U → U is surjective.

3.9. Let n ≥ 0 be an integer. An n-dimensional topological manifold is anonempty Hausdorff topological space X such that each point in X ad-mits a neighborhood homeomorphic to R

n. Show that if X is a compactn-dimensional topological manifold, then every injective continuous mapf : X → X is surjective. Hint: Use Brouwer’s invariance of domain to provethat f(X) is open in X.

3.10. Let P be a property of groups. Show that every subgroup of a locallyP group is itself locally P.

3.11. Let P be a property of groups. Let G be a group. Show that G is locallyP if and only if all its finitely generated subgroups are locally P.

3.12. Show that the additive group Q is locally cyclic.

3.13. Show that in a locally finite group every element has finite order.

3.14. Let G be an abelian group. Show that G is locally finite if and only ifevery element of G has finite order.

3.15. Show that every subgroup and every quotient of a locally finite groupis a locally finite group.

3.16. Let G be a group. Suppose that G contains a normal subgroup H suchthat both H and G/H are locally finite. Show that G is locally finite.

3.17. Let G be a group which is the limit of an inductive system of locallyfinite groups. Show that G is locally finite.

3.18. Show that the direct sum of any family of locally finite groups is alocally finite group.

3.19. Let G be a group. Let S denote the set consisting of all normal locallyfinite subgroups of G.

(a) Show that if H ∈ S and K ∈ S then HK ∈ S.(b) Show that M =

⋃H∈S H is a normal locally finite subgroup of G and

that every normal locally finite subgroup of G is contained in M .


3.20. Let G be a locally finite group and let A be a set. Show that everybijective cellular automaton τ : AG → AG is invertible.

3.21. Let G be a locally finite group and let A be a finite set. Show thatevery surjective cellular automaton τ : AG → AG is injective.

3.22. Let G be a locally finite group and let A be a set. Let τ : AG → AG

be a cellular automaton. Show that τ(AG) is closed in AG with respect tothe prodiscrete topology. Hint: First treat the case when G is finite, then useSect. 1.7 and Proposition A.4.3.

3.23. Let G be a group and let A be a nonzero abelian group. Suppose thats0 ∈ G is an element of infinite order. Show that the map τ : AG → AG

defined by τ(x)(g) = x(gs0) − x(g), for all x ∈ AG and g ∈ G, is a cellularautomaton which is surjective but not injective.

3.24. Let X be an infinite set. Show that the symmetric group Sym(X) isnot locally residually finite.

3.25. Let G be a group. Suppose that there exists a family (Ni)i∈I of normalsubgroups of G satisfying the following properties:

(1) for all i and j in I, there exists k in I such that Nk ⊂ Ni ∩ Nj ;(2)

⋂i∈I Ni = {1G};

(3) the group G/Ni is surjunctive for each i ∈ I.Show that G is surjunctive. Hint: Apply Lemma 3.3.4.

3.26. It follows from Theorem 3.3.1 that every residually finite group is sur-junctive. The goal of this exercise is to present an alternative proof of thisresult. Let G be a residually finite group and let A be a finite set. Letτ : AG → AG be an injective cellular automaton. Fix a family (Γi)i∈I ofsubgroups of finite index of G such that

⋂i∈I Γi = {1G} (the existence of

such a family follows from the residual finiteness of G).(a) Show that, for each i ∈ I, the set Fix(Γi) = {x ∈ AG : gx =

x for all g ∈ Γi} is finite.(b) Show that τ(Fix(Γi)) = Fix(Γi) for all i ∈ I.(c) Prove that

⋃i∈I Fix(Γi) is dense in AG and conclude.

3.27. Let X be a set. Let (Ai)i∈I be a net of subsets of X.(a) Show that ⋃

i

⋂

j≥i

Aj ⊂⋂

i

⋃

j≥i

Aj .

(b) Let B be a subset of X. Show that the net (Ai)i∈I converges to B withrespect to the prodiscrete topology on P(X) = {0, 1}X if and only if

B =⋃

i

⋂

j≥i

Aj =⋂

i

⋃

j≥i

Aj .

Exercises 71

3.28. Let G be a group and let A be a set. Let H be a subgroup of G.In Exercise 1.33, we associated with each subshift X ⊂ AH the subshiftX(G) = {x ∈ AG : xH

g ∈ X for all g ∈ G} ⊂ AG, where xHg = (gx)|H for

all x ∈ AG, h ∈ H and g ∈ G. Let σ : AH → AH be a cellular automatonand denote by σG : AG → AG the induced cellular automaton. Show thatif X,Y ⊂ AH are two subshifts such that σ(X) ⊂ Y , then the cellularautomaton σG|X(G) : X(G) → Y (G) is injective (resp. surjective) if and onlyif the cellular automaton σ|X : X → Y is injective (resp. surjective).

3.29. Let G be a group and let A be a finite set. Given a subshift Z ⊂ AG

we denote by Zf the set of all configurations in Z whose G-orbit is finite(cf. Example 1.3.1(c)). We say that Z is surjunctive if every injective cellularautomaton σ : Z → Z is surjective. Let X ⊂ AG be a subshift such that Xf

is dense in X.(a) Let τ : AG → AG be a cellular automaton. Consider the subshift Y =

τ(X). Show that Yf is dense in Y .(b) Deduce from (a) that X is surjunctive.

3.30. Let A be a finite set and let X ⊂ AZ be an irreducible subshift of finitetype.

(a) Show that Xf is dense in X.(b) Deduce from (a) and Exercise 3.29(b) that X is surjunctive (compare

with its stronger version in Exercise 6.36).(c) Let A = {0, 1} and consider the subshift of finite type Y ⊂ AZ defined

by Y = {x ∈ AZ : (x(n), x(n + 1)) �= (0, 1) for all n ∈ Z}. Show thatYf consists of the two constant configurations x0 and x1 and therefore it isclosed but not dense in Y .

(d) Show that the subshift Y is surjunctive.(e) Let A = {0, 1} and let Y as in (c). Consider the associated subshift

Z = Y (Z2) ⊂ AZ2

defined in Exercise 1.33. Show that Z is an irreduciblesubshift of finite type but that Zf is not dense in Z.

(f) Show that the subshift Z is surjunctive.

3.31. Show that the golden mean subshift is surjunctive. Hint: Use Exer-cise 1.39 and Exercise 3.30(b).

3.32. Show that the even subshift is surjunctive. Hint: Use Exercise 3.30.

3.33. A subshift of finite type which is not surjunctive (cf. [Weiss, Sect. 4]).Let A = {0, 1, 2} and consider the subshift of finite type X = X(Ω,A) ⊂ AZ

with memory set Ω = {0, 1} ⊂ Z and defining set of admissible words A ={00, 01, 11, 12, 22} ⊂ AΩ .

(a) Show that X is not irreducible.(b) Consider the cellular automaton σ : AZ → AZ with memory set S =

{0, 1} and local defining map μ : AS → A defined by


μ(y) =

{1 if y(0)y(1) = 01y(0) otherwise.

Show that σ(X) ⊂ X.(c) Show that the cellular automaton τ = σ|X : X → X is injective but

not surjective. Deduce that X is not surjunctive.

3.34. Let G be a group and let A be a finite set. Suppose that G contains anelement of infinite order and that A contains at least two distinct elements.

(a) Show that there exists a subshift X ⊂ AG which is not surjunctive.(b) Show that if, in addition, G is not infinite cyclic then there exists an

irreducible subshift X ⊂ AG which is not surjunctive. Hint: Use Exercise 3.33and Exercise 1.33(c).

3.35. Given a group Γ acting continuously on a topological space Z, a subsetM ⊂ Z is called a minimal set if M is a nonempty Γ -invariant subset of Zand the Γ -orbit of every point m ∈ M is dense in M .

Let G be a group and let A be a set. A subshift X ⊂ AG is called minimalif X is a minimal set with respect to the G-shift action on AG.

(a) Show that a subshift X ⊂ AG is minimal if and only if X �= ∅ andthere is no subshift Y in AG such that ∅ �= Y � X.

(b) Show that every minimal subshift X ⊂ AG is irreducible.(c) Show that the finite minimal subshifts in AG are precisely the finite

G-orbits.(d) Show that if X ⊂ AG is an infinite minimal subshift then the G-orbit

of every configuration x ∈ X is infinite.(e) Show that if X and Y are minimal subshifts in AG then either X = Y

or X ∩ Y = ∅.

3.36. Let G be a group and let A be a finite set. Let X,Y ⊂ AG be twosubshifts. Show that if X is nonempty and Y is minimal then every cellularautomaton τ : X → Y is surjective.

3.37. Let G be a group and let A be a finite set. Show that every minimalsubshift X ⊂ AG is surjunctive.

3.38. Let G be a group and let A be a finite set. Show that every nonemptysubshift X ⊂ AG contains a minimal subshift. Hint: Apply Zorn’s lemma tothe set of nonempty subshifts contained in X.

3.39. Let G be a group and let A be a set. A subset R ⊂ G is called syndeticif there exists a finite subset K ⊂ G such that the set Kg meets R for everyg ∈ G. A configuration x ∈ AG is called almost periodic if for every finitesubset Ω ⊂ G the set R(x,Ω) = {g ∈ G : (gx)|Ω = x|Ω} is syndetic in G.

(a) Show that if x ∈ AG is an almost periodic configuration then its orbitclosure X = Gx ⊂ AG is a minimal subshift. Hint: Suppose that the orbitclosure X = Gx of a configuration x ∈ AG is not minimal. Then there exists

Exercises 73

a nonempty subshift Y ⊂ X with x /∈ Y . This implies that there is a finitesubset Ω ⊂ G such that x|Ω �= y|Ω for all y ∈ Y . Let K be a finite subsetof G and consider the set Ω′ = K−1Ω. Choose an arbitrary configurationy0 ∈ Y . Since y0 ∈ X, there is an element g0 ∈ G such that (g0x)|Ω′ = y0|Ω′ .This implies that (kg0x)|Ω = (ky0)|Ω �= x|Ω for all k ∈ K. Thus Kg0 doesnot meet R(x,Ω) so that x is not almost-periodic.

(b) Suppose that the set A is finite. Show that if X ⊂ AG is a minimalsubshift, then every configuration x ∈ X is almost-periodic. Hint: Observethat if Ω ⊂ G is a finite subset and x ∈ X, then the open sets gV , whereV = {y ∈ AG : y|Ω = x|Ω} and g runs over G, cover X by minimality anduse the compactness of X.

(c) Suppose that the set A is finite. Show that a nonempty subshift X ⊂AG is minimal if and only if X is irreducible and every configuration x ∈ Xis almost-periodic.

3.40. Let A be a set. The language of a configuration x ∈ AZ is the subsetL(x) ⊂ A∗ consisting of all words w ∈ A∗ which can be written in the formw = x(n + 1)x(n + 2) . . . x(n + m) for some integers n ∈ Z and m ≥ 0.

(a) Let X ⊂ AZ be a subshift and let x ∈ X. Show that L(x) ⊂ L(X) andthat one has L(x) = L(X) if and only if the Z-orbit of x is dense in X.

(b) Show that a nonempty subshift X ⊂ AZ is minimal if and only if onehas L(x) = L(X) for all x ∈ X.

(c) Show that a configuration x ∈ AZ is almost periodic if and only ifit satisfies the following condition: for every word u ∈ L(x), there exists aninteger n = n(u) ≥ 0 such that u is a subword of any word v ∈ L(x) oflength n.

3.41. The Thue-Morse sequence. Let A = {0, 1}. The Thue-Morse sequenceis the map x : N → A defined by

x(n) =

{1 if the number of ones in the binary expansion of n is odd0 otherwise.

(a) Check that

x(0)x(1) . . . x(16) = 0110100110010110.

(b) Show that one has x(k + 3m2n) = x(k) for all k,m, n ∈ N such that0 ≤ k ≤ 2n.

(c) Show that x satisfies the recurrence relations⎧⎪⎨

⎪⎩

x(0) = 0x(2n) = x(n)x(2n + 1) = 1 − x(n)

and that these relations uniquely determine x.


(d) Consider the unique monoid homomorphism ϕ : A∗ → A∗ which satis-fies

ϕ(0) = 01 and ϕ(1) = 10.

Show that

ϕ(x(0)x(1) . . . x(2n − 1)) = x(0)x(1) . . . x(2n+1 − 1)

for all n ∈ N.(e) Consider the unique monoid isomorphism ι : A∗ → A∗ which satisfies

ι(0) = 1 and ι(1) = 0.

Show that

x(0)x(1) . . . x(2n+1 − 1) = x(0)x(1) . . . x(2n − 1)ι(x(0)x(1) . . . x(2n − 1))

for all n ∈ N.Note: The Thue-Morse sequence was first studied by E. Prouhet in a 1851

paper dealing with problems in number theory. It is an example of an auto-matic sequence (see [AlS, Sect. 5.1]).

3.42. The Morse subshift. Let A = {0, 1} and let L = {(x(n), x(n + 1),. . . , x(n + m)) : n, m ∈ N} ⊂ A∗, where x : N → A is the Thue-Morsesequence.

(a) Show that the set L satisfies the conditions (i) and (ii) in Exer-cise 1.36(b).

(b) Let X ⊂ AZ denote the unique subshift whose language is L(X) = L(cf. Exercise 1.36(c)). Show that X is an infinite minimal subshift (it is calledthe Morse subshift). Hint: Use Exercises 3.39(a), 3.40 and 3.41(b).

3.43. Toeplitz subshifts. Let A be a finite set. A configuration x ∈ AZ is saidto be a Toeplitz configuration if for every n ∈ Z there exists k ≥ 1 such thatx(n) = x(n + kr) for all r ∈ Z. In other words, x is a Toeplitz configurationif there exists a partition of Z into arithmetic progressions such that x isconstant on each element of the partition. Given a Toeplitz configurationx ∈ AZ and n ∈ Z we set

k(x, n) = min{k ≥ 1 : x(n) = x(n + kr) for all r ∈ Z}.

One says that a subshift X ⊂ AZ is a Toeplitz subshift if it is the orbit closureof some Toeplitz configuration x ∈ AZ.

(a) Let x ∈ AZ be a Toeplitz configuration. Show that the Z-orbit of x isfinite if and only if the set {k(x, n) : n ∈ Z} is finite.

(b) Show that every Toeplitz configuration x ∈ AZ is almost-periodic.Hint: Use Exercise 3.40.

(c) Show that every Toeplitz subshift X ⊂ AZ is minimal.

Exercises 75

3.44. Let A be a finite set and let X ⊂ AZ be a Toeplitz subshift.(a) Show that X is irreducible. Hint: Use Exercises 3.43(d) and 3.35.(b) Show that X is surjunctive. Hint: Use Exercises 3.43(d) and 3.37.

Chapter 4

Amenable Groups

This chapter is devoted to the class of amenable groups. This is a class ofgroups which plays an important role in many areas of mathematics such asergodic theory, harmonic analysis, representation theory, dynamical systems,geometric group theory, probability theory and statistics. As residually finitegroups, amenable groups generalize finite groups but there are residually fi-nite groups which are not amenable and there are amenable groups whichare not residually finite. An amenable group is a group whose subsets admitan invariant finitely additive probability measure (see Sect. 4.4). The class ofamenable groups contains in particular all finite groups, all abelian groupsand, more generally, all solvable groups (Theorem 4.6.3). It is closed under theoperations of taking subgroups, taking quotients, taking extensions, and tak-ing inductive limits (Sect. 4.5). The notion of a Følner net, roughly speaking,a net of almost invariant finite subsets of a group, is introduced in Sect. 4.7.Paradoxical decompositions are defined in Sect. 4.8. The Følner-Tarski the-orem (Theorem 4.9.1) asserts that amenability, existence of a Følner net,and non-existence of a paradoxical decomposition are three equivalent con-ditions for groups. Another characterization of amenable groups is given inSect. 4.10: a group is amenable if and only if every continuous affine action ofthe group on a nonempty compact convex subset of a Hausdorff topologicalvector space admits a fixed point (Corollary 4.10.2).

4.1 Measures and Means

Let E be a set. We shall denote by P(E) the set of all subsets of E.

Definition 4.1.1. A map μ : P(E) → [0, 1] is called a finitely additive prob-ability measure on E if it satisfies the following properties:

(1) μ(E) = 1,(2) μ(A ∪ B) = μ(A) + μ(B) for all A, B ∈ P(E) such that A ∩ B = ∅.


77

http://dx.doi.org/10.1007/978-3-642-14034-1_4

78 4 Amenable Groups

Example 4.1.2. Let F be a nonempty finite subset of E. Then the mapμF : P(E) → [0, 1] defined by

μF (A) =|A ∩ F ||F |

for all A ⊂ E, is a finitely additive probability measure on E.

Proposition 4.1.3. Let μ : P(E) → [0, 1] be a finitely additive probabilitymeasure on E. Then one has:

(i) μ(∅) = 0;(ii) μ(A ∪ B) = μ(A) + μ(B) − μ(A ∩ B);(iii) μ(A ∪ B) ≤ μ(A) + μ(B);(iv) A ⊂ B ⇒ μ(B \ A) = μ(B) − μ(A);(v) A ⊂ B ⇒ μ(A) ≤ μ(B);

for all A, B ∈ P(E).

Proof. We have μ(E) = μ(E ∪ ∅) = μ(E) + μ(∅), which implies (i). Forall A, B ∈ P(E), we have μ(A ∪ B) = μ(A) + μ(B \ A) and μ(B) = μ(A ∩B) + μ(B \ A), which gives (ii). Since μ(A ∩ B) ≥ 0, equality (ii) implies(iii). If A ⊂ B, we have μ(B) = μ(B \A) + μ(A), which implies (iv). Finally,inequality (v) follows from (iv) since μ(B \ A) ≥ 0. �

Consider now the real vector space �∞(E) consisting of all bounded func-tions x : E → R. Recall that �∞(E) is a Banach space for the norm ‖ · ‖∞defined by

‖x‖∞ = supa∈E

|x(a)|.

We equip �∞(E) with the partial ordering ≤ given by

x ≤ y ⇐⇒ (x(a) ≤ y(a) for all a ∈ E).

For each λ ∈ R, we shall also denote by λ the element of �∞(E) which isidentically equal to λ on E.

Definition 4.1.4. A mean on E is a linear map m : �∞(E) → R such that:

(1) m(1) = 1,(2) x ≥ 0 ⇒ m(x) ≥ 0

for all x ∈ �∞(E).

Example 4.1.5. Let S be a countable (finite or infinite) subset of E and letf : S → R such that:

(C1) f(s) > 0 for all s ∈ S;(C2)

∑s∈S f(s) = 1.

4.1 Measures and Means 79

Then the map mf : �∞(E) → R defined by

mf (x) =∑

s∈S

f(s)x(s)

is a mean on E.One says that a mean m on E has finite (resp. countable) support if there

exists a finite (resp. countable) subset S ⊂ E and a map f : S → R satisfyingconditions (C1) and (C2) above such that m = mf .

Proposition 4.1.6. Let m : �∞(E) → R be a mean on E. Then one has:

(i) m(λ) = λ,(ii) x ≤ y ⇒ m(x) ≤ m(y),(iii) infE x ≤ m(x) ≤ supE x,(iv) |m(x)| ≤ ‖x‖∞,

for all λ ∈ R and x, y ∈ �∞(E).

Proof. (i) By linearity, we have m(λ) = λm(1) = λ.(ii) If x ≤ y then m(y) − m(x) = m(y − x) ≥ 0 and thus m(x) ≤ m(y).(iii) We have infE x ≤ x ≤ supE x and therefore infE x ≤ m(x) ≤ supE x

by applying (i) and (ii).(iv) From (iii) we get −‖x‖∞ ≤ m(x) ≤ ‖x‖∞, that is, |m(x)| ≤ ‖x‖∞. �

Consider the topological dual of �∞(E), that is, the vector space (�∞(E))∗

consisting of all continuous linear maps u : �∞(E) → R.Recall that (�∞(E))∗ is a Banach space for the operator norm ‖ · ‖ defined

by‖u‖ = sup

‖x‖∞≤1

|u(x)| (4.1)

for all u ∈ (�∞(E))∗.

Proposition 4.1.7. Let m : �∞(E) → R be a mean on E. Then m ∈(�∞(E))∗ and ‖m‖ = 1.

Proof. By definition m is linear. Inequality (iv) in Proposition 4.1.6 shows thatm is continuous and satisfies ‖m‖ ≤ 1. Since m(1) = 1, we have ‖m‖ = 1. �

Let PM(E) (resp. M(E)) denote the set of all finitely additive probabilitymeasures (resp. of all means) on E. We are going to show that there is anatural bijection between the sets PM(E) and M(E). This natural bijection,which is analogous to the Riesz representation in measure theory, may be usedto view finitely additive measures on E as points in the dual space (�∞(E))∗

where classical techniques of functional analysis may be applied.For each subset A ⊂ E, we denote by χA the characteristic map of A,

that is, the map χA : E → R defined by χA(x) = 1 if x ∈ A and χA(x) = 0otherwise.


Let m ∈ M(E). Consider the map m : P(E) → R given by

m(A) = m(χA).

Observe that 0 ≤ χA ≤ 1 and therefore m(A) ∈ [0, 1] by Proposition 4.1.6(ii).We have m(E) = m(χE) = m(1) = 1. On the other hand, if A and B aredisjoint subsets of E, we have χA∪B = χA + χB and thus m(A ∪ B) =m(A) + m(B) by linearity of m. It follows that m ∈ PM(E).

Theorem 4.1.8. The map Φ : M(E) → PM(E) defined by Φ(m) = m isbijective.

The proof will be divided in several steps.Denote by E(E) the set of all maps x : E → R which take only finitely

many values. Observe that E(E) is the vector subspace of �∞(E) spanned bythe set of characteristic maps {χA : A ⊂ E}.

Lemma 4.1.9. The vector subspace E(E) is dense in �∞(E).

Proof. Let x ∈ �∞(E) and ε > 0. Set α = infE x and β = supE x. Choosean integer n ≥ 1 such that (β − α)/n < ε and set λi = α + i(β − α)/n fori = 1, 2, . . . , n. Consider the map y : E → R defined by

y(a) = min{λi : x(a) ≤ λi}

for all a ∈ E. The map y take its values in the set {λ1, . . . , λn}. Thus y ∈E(E). We have ‖x − y‖∞ ≤ (β − α)/n < ε by construction. Consequently,E(E) is dense in �∞(E). �

Let μ ∈ PM(E). Let us set, for all x ∈ E(E),

μ(x) =∑

λ∈R

μ(x−1(λ))λ. (4.2)

Observe that there is only a finite number of nonzero terms in the right-handside of (4.2) since x ∈ E(E) and μ(∅) = 0.

Lemma 4.1.10. Let x ∈ E(E). Suppose that there is a finite partition (Ai)i∈I

of E such that the restriction of x to Ai is constant equal to αi for each i ∈ I.Then one has

μ(x) =∑

i∈I

μ(Ai)αi.

Proof. Since μ is finitely additive, we have

∑

i∈I

μ(Ai)αi =∑

λ∈R

(∑

αi=λ

μ(Ai)

)λ =

∑

λ∈R

μ(x−1(λ))λ = μ(x).

�

4.1 Measures and Means 81

Lemma 4.1.11. The map μ : E(E) → R is linear and continuous.

Proof. Let x, y ∈ E(E) and ξ, η ∈ R. Denote by V (resp. W ) the set of valuestaken by x (resp. y). The subsets x−1(α) ∩ y−1(β), (α, β) ∈ V × W , form afinite partition of E. By applying Lemma 4.1.10, we get

μ(ξx + ηy) =∑

(α,β)∈V ×W

μ(x−1(α) ∩ y−1(β))(ξα + ηβ)

= ξ

⎛

⎝∑

(α,β)∈V ×W

μ(x−1(α) ∩ y−1(β))α

⎞

⎠

+

⎛

⎝η∑

(α,β)∈V ×W

μ(x−1(α) ∩ y−1(β))β

⎞

⎠

= ξμ(x) + ημ(y).

Consequently, μ is linear.Formula (4.2) implies infE x ≤ μ(x) ≤ supE x since

∑

λ∈R

μ(x−1(λ)) = μ(E) = 1.

It follows that |μ(x)| ≤ ‖x‖∞ for all x ∈ E(E). This shows that the linearmap μ is continuous. �

Lemma 4.1.12. Let X be a normed vector space and let Y be a dense vectorsubspace of X. Suppose that ϕ : Y → R is a continuous linear map. Thenthere exists a continuous linear map ϕ : X → R extending ϕ (that is, suchthat ϕ|Y = ϕ).

Proof. Let x ∈ X. Since Y is dense in X, we can find a sequence (yn)n≥0 ofpoints of Y which converges to x. The sequence (ϕ(yn)) is a Cauchy sequencesince

|ϕ(yp) − ϕ(yq)| = |ϕ(yp − yq)| ≤ ‖ϕ‖‖yp − yq‖

for all p, q ≥ 0. As R is complete, this sequence converges. Set ϕ(x) =limϕ(yn). If (y′

n) is another sequence of points in Y converging to x, then|ϕ(yn) − ϕ(y′

n)| ≤ ‖ϕ‖‖yn − y′n‖. Thus we have lim ϕ(yn) = limϕ(y′

n).This shows that ϕ(x) does not depend of the choice of the sequence (yn).The linearity of ϕ gives the linearity of ϕ by taking limits. We also get‖ϕ(x)‖ ≤ ‖ϕ‖‖x‖ for all x ∈ X by taking limits. This shows that the linearmap ϕ is continuous. If x ∈ Y , we have ϕ(x) = ϕ(x) since we can take as(yn) the constant sequence yn = x in this case. Thus ϕ extends ϕ. �

Proof of Theorem 4.1.8. Let μ ∈ PM(E). By Lemma 4.1.9, Lemma 4.1.11and Lemma 4.1.12, the map μ : E(E) → R can be extended to a continuouslinear map μ : �∞(E) → R. We have


μ(1) = μ(1) = μ(E) = 1.

If y ∈ E(E) and y ≥ 0, then μ(y) = μ(y) ≥ 0 by (4.2). Consider now anelement x ∈ �∞(E) such that x ≥ 0. Let (yn)n≥0 be a sequence of elementsof E(E) converging to x. Let us set zn = |yn|. Since zn ∈ E(E) and zn ≥ 0, wehave μ(zn) ≥ 0. On the other hand, the sequence (zn) converges to x since,by the triangle inequality, ‖x − zn‖∞ ≤ ‖x − yn‖∞. It follows that

μ(x) = limn→∞

μ(zn) ≥ 0.

Thus μ ∈ M(E).For every subset A ⊂ E, we have

μ(χA) = μ(χA) = μ(A).

Therefore Φ(μ) = μ. This proves that Φ is surjective.It remains to show that Φ is injective. To see this, consider m1, m2 ∈ M(E)

such that Φ(m1) = Φ(m2). This means that m1(χA) = m2(χA) for everysubset A ⊂ E. By linearity, this implies that m1 and m2 coincide on E(E).Since E(E) is dense in �∞(E) by Lemma 4.1.9, we deduce that m1 = m2 bycontinuity of m1 and m2. This shows that Φ is injective. �

4.2 Properties of the Set of Means

Let E be a set.The topology on (�∞(E))∗ associated with the operator norm ‖ · ‖ defined

in (4.1) is called the strong topology on (�∞(E))∗. Another topology that iscommonly used is the weak-∗ topology on (�∞(E))∗. Recall that the weak-∗topology on (�∞(E))∗ is by definition the smallest topology for which theevaluation map

ψx : (�∞(E))∗ → R

u �→ u(x)

is continuous for each x ∈ �∞(E) (see Sect. F.2).By Proposition 4.1.7, the set M(E) is contained in the unit sphere

{u : ‖u‖ = 1} ⊂ (�∞(E))∗.

The following result will play an important role in the sequel.

Theorem 4.2.1. The set M(E) is a convex compact subset of (�∞(E))∗ withrespect to the weak-∗ topology.

4.3 Measures and Means on Groups 83

Proof. Let m1, m2 ∈ M(E) and t ∈ [0, 1]. Then (tm1 + (1 − t)m2)(1) =tm1(1)+ (1− t)m2(1) = t+(1− t) = 1 and (tm1 +(1− t)m2)(x) = tm1(x)+(1− t)m2(x) ≥ 0 for all x ∈ �∞(E) such that x ≥ 0. Thus tm1 + (1− t)m2 ∈M(E). This shows that M(E) is convex. Equip now (�∞(E))∗ with theweak-∗ topology and suppose that (mi)i∈I is a net in M(E) converging tou ∈ (�∞(E))∗. Then, for every i ∈ I, we have ψ1(mi) = mi(1) = 1 andψx(mi) = mi(x) ≥ 0 for all x ∈ �∞(E) such that x ≥ 0. By taking limits,we get u(1) = ψ1(u) = 1 and u(x) = ψx(u) ≥ 0 for all x ∈ �∞(E) suchthat x ≥ 0. Thus u ∈ M(E). This shows that M(E) is closed in (�∞(E))∗

(Proposition A.2.1). As M(E) is contained in the unit ball of (�∞(E))∗,which is compact for the weak-∗ topology by the Banach-Alaoglu Theorem(Theorem F.3.1), it follows that M(E) is compact. �

4.3 Measures and Means on Groups

In this section, we shall see that the set of finitely additive measures and theset of means carry additional structure when the underlying set is a group.Indeed, in this case, the group naturally acts on both sets. Moreover, thereare involutions coming from the operation of taking inverses in the group.

More precisely, let G be a group.The group G naturally acts on the left and on the right on each of the sets

PM(G) and M(G) in the following way.Firstly, for μ ∈ PM(G) and g ∈ G, we define the maps gμ : P(G) → [0, 1]

and μg : P(G) → [0, 1] by

gμ(A) = μ(g−1A) and μg(A) = μ(Ag−1)

for all A ∈ P(G). One clearly has gμ ∈ PM(G) and μg ∈ PM(G). Moreover,it is straightforward to check that the map (g, μ) �→ gμ (resp. (μ, g) �→ μg)defines a left (resp. right) action of G on PM(G). Note that these two actionscommute in the sense that g(μh) = (gμ)h for all g, h ∈ G and μ ∈ PM(G).

On the other hand, recall that we introduced in Sect. 1.1 a left action ofG on R

G (the G-shift) by defining, for all g ∈ G and x ∈ RG, the element

gx ∈ RG by

gx(g′) = x(g−1g′) for all g′ ∈ G.

Similarly, we make G act on the right on RG by defining the element xg ∈ R

G

byxg(g′) = x(g′g−1) for all g′ ∈ G.

These two actions of G on RG are linear and commute. Moreover, the vec-

tor subspace �∞(G) ⊂ RG is left invariant by both actions. Observe that

‖gx‖∞ = ‖xg‖∞ = ‖x‖∞ for all g ∈ G and x ∈ �∞(G). Thus the left (resp.right) action of G on �∞(G) is isometric (and therefore continuous).


By duality, we also get a left and a right action of G on (�∞(G))∗. Moreprecisely, for g ∈ G and u ∈ (�∞(G))∗, we define the elements gu and ug by

gu(x) = u(g−1x) and ug(x) = u(xg−1),

for all x ∈ �∞(G). We have gu ∈ (�∞(G))∗ and ug ∈ (�∞(G))∗ since the leftand right actions of G on �∞(G) are linear and continuous.

Note that the set M(G) is invariant under both actions of G on (�∞(G))∗.

Proposition 4.3.1. The left (resp. right) action of G on M(G) is affine andcontinuous with respect to the weak-∗ topology on M(G).

Proof. It is clear that the left (resp. right) action of G on (�∞(G))∗ is linear.On the other hand, these actions are continuous if we equip (�∞(G))∗ withthe weak-∗ topology. Indeed, if we fix g ∈ G, the map u �→ gu is continuouson (�∞(G))∗ since, for each x ∈ �∞(G), the map u �→ gu(x) is the evalua-tion map at g−1x, which is continuous by definition of the weak-∗ topology.Similarly, the map u �→ ug is continuous. Since M(G) is a convex subset of(�∞(G))∗, we deduce that the restrictions of both actions to M(G) are affineand continuous. �

Given μ ∈ PM(G), we define the map μ∗ : P(G) → [0, 1] by

μ∗(A) = μ(A−1) for all A ∈ P(G).

It is clear that μ∗ ∈ PM(G) and that the map μ �→ μ∗ is an involution ofPM(G).

For x ∈ �∞(G), define x∗ ∈ �∞(G) by

x∗(g) = x(g−1) for all g ∈ G.

The map x �→ x∗ is an isometric involution of �∞(G). By duality, it gives anisometric involution u �→ u∗ of (�∞(G))∗ defined by

u∗(x) = u(x∗) for all x ∈ �∞(G).

Note that m∗ ∈ M(G) for all m ∈ M(G).

Proposition 4.3.2. Let g ∈ G, x ∈ �∞(G), μ ∈ PM(G) and u ∈ (�∞(G))∗.Then one has

(i) (gx)∗ = x∗g−1;(ii) (xg)∗ = g−1x∗;(iii) (gμ)∗ = μ∗g−1;(iv) (μg)∗ = g−1μ∗;(v) (gu)∗ = u∗g−1;(vi) (ug)∗ = g−1u∗.

4.4 Definition of Amenability 85

Proof. For every h ∈ G, we have

(gx)∗(h) = gx(h−1) = x(g−1h−1) = x∗(hg) = x∗g−1(h),

which gives (gx)∗ = x∗g−1. The proofs of the other properties are similar. �

Proposition 4.3.3. Let g ∈ G and m ∈ M(G). Then one has:

(i) gm = gm;(ii) mg = mg;(iii) m∗ = m∗.

We need the following result.

Lemma 4.3.4. Let g ∈ G and A ⊂ G. Then one has:

(i) (χA)∗ = χA−1 ;(ii) gχA = χgA.

Proof. (i) For h ∈ G one has (χA)∗(h) = χA(h−1) = 1 ⇔ h−1 ∈ A ⇔ h ∈A−1 ⇔ χA−1(h) = 1. This shows that (χA)∗ = χA−1 . (ii) For h ∈ G onehas (gχA)(h) = χA(g−1h) = 1 ⇔ g−1h ∈ A ⇔ h ∈ gA ⇔ χgA(h) = 1. Thisshows that gχA = χgA. �

Proof of Proposition 4.3.3. For every A ∈ P(G), we have, using Lem-ma 4.3.4(ii),

gm(A) = gm(χA) = m(g−1χA) = m(χg−1A) = m(g−1A) = gm(A).

Thus gm = gm. Similarly, we get mg = mg.On the other hand, for every A ∈ P(G), using Lemma 4.3.4(i) one obtains

m∗(A) = m(A−1) = m(χA−1) = m((χA)∗) = m∗(χA) = m∗(A)

and this shows that m∗ = m∗. �

Remark 4.3.5. Properties (i) and (ii) in Proposition 4.3.3 say that the bi-jective map Φ : M(G) → PM(G) given by m �→ m (see Theorem 4.1.8) isbi-equivariant.

4.4 Definition of Amenability

Let G be a group.A finitely additive probability measure μ ∈ PM(G) is called left-invariant

(resp. right-invariant) if μ is fixed under the left (resp. right) action of G onPM(G), that is, if it satisfies gμ = μ (resp. μg = μ) for all g ∈ G. One saysthat μ is bi-invariant if μ is both left and right-invariant.


Similarly, a mean m ∈ M(G) is called left-invariant (resp. right-invariant)if m is fixed under the left (resp. right) action of G on M(G). One says thatm is bi-invariant if m is both left and right invariant.

Proposition 4.4.1. Let μ ∈ PM(G). Then μ is left-invariant (resp. right-invariant) if and only if μ∗ is right-invariant (resp. left-invariant).

Proof. This immediately follows from Proposition 4.3.2(ii). �

Proposition 4.4.2. Let m ∈ M(G). Then m is left-invariant (resp. right-invariant) if and only if m∗ is right-invariant (resp. left-invariant).

Proof. This immediately follows from Proposition 4.3.2(iii). �

Proposition 4.4.3. Let m ∈ M(G). Then m is left-invariant (resp. right-invariant, resp. bi-invariant) if and only if the associated finitely additiveprobability measure m ∈ PM(G) is left-invariant (resp. right-invariant, resp.bi-invariant).

Proof. This immediately follows from assertions (i) and (ii) of Proposi-tion 4.3.3. �


(a) there exists a left-invariant finitely additive probability measureμ : P(G) → [0, 1] on G;

(b) there exists a right-invariant finitely additive probability measureμ : P(G) → [0, 1] on G;

(c) there exists a bi-invariant finitely additive probability measureμ : P(G) → [0, 1] on G;

(d) there exists a left-invariant mean m : �∞(G) → R on G;(e) there exists a right-invariant mean m : �∞(G) → R on G;(f) there exists a bi-invariant mean m : �∞(G) → R on G.

Proof. Equivalences (a) ⇔ (d), (b) ⇔ (e) and (c) ⇔ (f) immediately followfrom Proposition 4.4.3. The equivalence between (a) and (b) follows fromProposition 4.4.1. Implication (f) ⇒ (d) is trivial.

Thus it suffices to show (d) ⇒ (f) to conclude. Suppose that there existsa left-invariant mean m : �∞(G) → R. For each x ∈ �∞(G), define the mapx : G → R by

x(g) = m(xg) for all g ∈ G.

By Proposition 4.1.6(iv), we have

|x(g)| = |m(xg)| ≤ ‖xg‖∞ = ‖x‖∞.

Therefore x ∈ �∞(G) for all x ∈ �∞(G). Consider now the map M : �∞(G) →R defined by

4.4 Definition of Amenability 87

M(x) = m(x) for all x ∈ �∞(G).

Clearly M is a mean on G.Let h ∈ G and x ∈ �∞(G). For all g ∈ G, we have

hx(g) = m(hxg) = m(xg) = x(g),

since m is left-invariant. We deduce that hx = x. This implies

M(hx) = m(hx) = m(x) = M(x).

Consequently M is left-invariant. On the other hand, for all g ∈ G, we have

xh(g) = m((xh)g) = m(x(hg)) = x(hg) = h−1x(g).

It follows that xh = h−1x. Therefore, by using again the fact that m isleft-invariant, we get

M(xh) = m(xh) = m(h−1x) = m(x) = M(x).

Therefore the mean M is also right-invariant. This shows that (d) implies (f).�

Definition 4.4.5. A group G is called amenable if it satisfies one of theequivalent conditions of Proposition 4.4.4.

Proposition 4.4.6. Every finite group is amenable.

Proof. If G is a finite group, then the map μ : P(G) → [0, 1] defined by

μ(A) =|A||G| for all A ∈ P(G)

is a bi-invariant finitely additive probability measure on G. �

Theorem 4.4.7. The free group on two generators F2 is not amenable.

Proof. Suppose that there exists a left-invariant finitely additive probabilitymeasure μ : P(F2) → [0, 1]. Denote by a and b the canonical generators ofF2. Let A ⊂ F2 be the set of elements of F2 whose reduced form begins by anonzero (positive or negative) power of a. We have F2 = A ∪ aA and henceμ(F2) ≤ μ(A) + μ(aA) = 2μ(A). Since μ(F2) = 1, this implies

μ(A) ≥ 12. (4.3)

On the other hand, the subsets A, bA and b2A are pairwise disjoint. Thusone has μ(F2) ≥ μ(A) + μ(bA) + μ(b2A) = 3μ(A). This gives


μ(A) ≤ 13

contradicting (4.3). �

4.5 Stability Properties of Amenable Groups

Proposition 4.5.1. Every subgroup of an amenable group is amenable.

Proof. Let G be an amenable group and let H be a subgroup of G. Letμ : P(G) → [0, 1] be a left-invariant finitely additive probability measure onG. Choose a complete set of representatives of the right cosets of G mod-ulo H, that is, a subset R ⊂ G such that, for each g ∈ G, there exists aunique element (h, r) ∈ H × R satisfying g = hr. Let us check that the mapμ : P(H) → [0, 1] defined by

μ(A) = μ

(⋃

r∈R

Ar

)for all A ∈ P(H)

is a left-invariant finitely additive probability measure on H.Firstly, we have

μ(H) = μ

(⋃

r∈R

Hr

)= μ(G) = 1.

On the other hand, if A and B are disjoint subsets of H, then

μ(A ∪ B) = μ

(⋃

r∈R

(A ∪ B)r

)

= μ

((⋃

r∈R

Ar

)∪

(⋃

r∈R

Br

))

= μ

(⋃

r∈R

Ar

)+ μ

(⋃

r∈R

Br

)

= μ(A) + μ(B),

since the sets ⋃

r∈R

Ar and⋃

r∈R

Br

are disjoint.

4.5 Stability Properties of Amenable Groups 89

Finally, for all h ∈ H and A ∈ P(H), we have

μ(hA) = μ

(⋃

r∈R

hAr

)= μ

(h

⋃

r∈R

Ar

)= μ

(⋃

r∈R

Ar

)= μ(A).

This shows that H is amenable. �

Combining Proposition 4.5.1 with Theorem 4.4.7, we get:

Corollary 4.5.2. If G is a group containing a subgroup isomorphic to thefree group on two generators F2, then G is not amenable. �

Examples 4.5.3. (a) Let X be a set having at least two elements. Then thefree group F (X) based on X is not amenable. Indeed, the subgroup of F (X)generated by two distinct elements a, b ∈ X is isomorphic to F2.

(b) The group SL(n, Z) is not amenable for n ≥ 2. Indeed, as we have seenin Lemma 2.3.2, the subgroup of SL(2, Z) generated by the matrices

(1 20 1

)

and(

1 02 1

)is isomorphic to F2.

Proposition 4.5.4. Every quotient of an amenable group is amenable.

Proof. Let G be an amenable group and let H be a normal subgroup of G. Letρ : G → G/H denote the canonical homomorphism. Consider a left-invariantfinitely additive probability measure μ : P(G) → [0, 1] on G. Let us show thatthe map μ : P(G/H) → [0, 1] defined by

μ(A) = μ(ρ−1(A)) for all A ∈ P(G/H)

is a left-invariant finitely additive probability measure on G/H.Firstly, we have

μ(G/H) = μ((ρ−1(G/H)) = μ(G) = 1.

On the other hand, if A and B are disjoint subsets of G/H, then

μ(A∪B) = μ(ρ−1(A∪B)) = μ(ρ−1(A)∪ ρ−1(B)) = μ(ρ−1(A)) + μ(ρ−1(B))

since ρ−1(A) ∩ ρ−1(B) = ∅. Thus we have μ(A ∪ B) = μ(A) + μ(B).Finally, if g ∈ G and A ⊂ G/H, we have

μ(ρ(g)A) = μ(ρ−1(ρ(g)A)) = μ(gρ−1(A)) = μ(ρ−1(A)) = μ(A).

As ρ is surjective, it follows that μ is left-invariant with respect to G/H. Thisshows that G/H is amenable. �

In the next proposition we show that the class of amenable groups is closedunder the operation of taking extensions of amenable groups by amenablegroups.


Proposition 4.5.5. Let G be a group and let H be a normal subgroup ofG. Suppose that the groups H and G/H are amenable. Then the group G isamenable.

Proof. Since H is amenable, we can find a H-left-invariant mean m0 : �∞(H)→R. Let x ∈ �∞(G). Consider the map x : G/H → R defined by

x(g) = m0((g−1x)|H) for all g ∈ G,

where g = gH = Hg ∈ G/H denote the class of g modulo H. The map x iswell defined. Indeed, if g1 and g2 are elements of G such that g1 = g2, theng2 = g1h for some h ∈ H, and hence

m0((g−12 x)|H) = m0((h−1g−1

1 x)|H) = m0(h−1(g−11 x)|H) = m0((g−1

1 x)|H)

since m0 is H-left-invariant. Observe also that x ∈ �∞(G/H) since, by Propo-sition 4.1.6(iv),

|x(g)| = |m0((g−1x)|H)| ≤ suph∈H

|(g−1x)(h)| ≤ ‖g−1x‖∞ = ‖x‖∞

for all g ∈ G.As the group G/H is amenable, there exists a G/H-left-invariant mean

m1 : �∞(G/H) → R. Let us set, for each x ∈ �∞(G),

m(x) = m1(x).

Clearly m is a mean on G. Let us show that it is G-left-invariant.Let g ∈ G and x ∈ �∞(G). For all g′ ∈ G, we have

gx(g′) = m0((g′−1

gx)|H) = m0(((g−1g′)−1x)|H) = x(g−1g′) = x(g−1g′)

= gx(g′).

Thus gx = gx. As m1 is G/H-left-invariant, it follows that

m(gx) = m1(gx) = m1(gx) = m1(x) = m(x),

which shows that m is G-left-invariant.Therefore G is amenable. �

Corollary 4.5.6. Suppose that G1 and G2 are amenable groups. Then thegroup G = G1 × G2 is amenable.

Proof. The set H = {(g1, 1G2) : g1 ∈ G1} is a normal subgroup of G isomor-phic to G1 with quotient G/H isomorphic to G2. �

Remark 4.5.7. It immediately follows from the preceding corollary that everydirect product of a finite number of amenable groups is amenable. How-ever, a direct product of infinitely many amenable groups is not necessarilyamenable. For example, it follows from Theorem 2.3.1 and Corollary 2.2.6

4.5 Stability Properties of Amenable Groups 91

that there exists a (countable) family of finite, and hence amenable, groups(Gi)i∈I such that the group G =

∏i∈I Gi contains a subgroup isomorphic to

F2. Such a group G is not amenable by Corollary 4.5.2.

Corollary 4.5.8. Every virtually amenable group is amenable.

Proof. Let G be a virtually amenable group. Let H be an amenable subgroupof finite index of G. By Lemma 2.1.10, the set K = ∩g∈GgHg−1 is a normalsubgroup of finite index of G contained in H. The group K is amenableby Proposition 4.5.1. On the other hand the group G/K is finite and henceamenable. Consequently, G is amenable by Proposition 4.5.5. �

Our next goal is to show that an inductive limit of amenable groups isamenable. In the proof we shall use the following:

Lemma 4.5.9. Let G be a group. Suppose that there is a net (mi)i∈I inM(G) such that, for each g ∈ G, the net (gmi − mi)i∈I converges to 0 in(�∞(G))∗ for the weak-∗ topology. Then G is amenable.

Proof. Since M(G) is compact for the weak-∗ topology by Theorem 4.2.1, wemay assume, after taking a subnet if necessary, that the net (mi) convergesto a mean m ∈ M(G). Let g ∈ G. Since the left action of G on M(G) iscontinuous by Proposition 4.3.1, we deduce that, for every x ∈ �∞(G) thenet (gmi − mi)(x) = gmi(x) − mi(x) converges to 0. By taking limits, weget gm(x) − m(x) = 0. Therefore gm = m. This shows that the mean m isleft-invariant. Thus G is amenable. �

Proposition 4.5.10. Every group which is the limit of an inductive systemof amenable groups is amenable.

Proof. Let (Gi)i∈I be an inductive system of amenable groups and set G =lim−→Gi. Consider the family (Hi)i∈I of subgroups G defined by Hi = hi(Gi),where hi : Gi → G is the canonical homomorphism. As Hi is amenableby Proposition 4.5.4, we can find, for each i ∈ I, a Hi-left-invariant meanmi : �∞(Hi) → R. Consider the family mi : �∞(G) → R of means on G definedby

mi(x) = mi(x|Hi) for all x ∈ �∞(G).

Let g ∈ G. Since G = lim−→Gi, there exists i0(g) ∈ I such that g ∈ Hi for alli ≥ i0(g). For x ∈ �∞(G) and i ≥ i0(g), we have

(gmi − mi)(x) = gmi(x) − mi(x) = gmi(x|Hi) − mi(x|Hi) = 0,

since mi is Hi-left-invariant. Thus gmi−mi = 0 for all i ≥ i0(g). By applyingLemma 4.5.9, we deduce that G is amenable. �

Every group is the inductive limit of its finitely generated subgroups.Therefore we have:


Corollary 4.5.11. Every locally amenable group is amenable. �

Since every finite group is amenable, we obtain in particular the following:

Corollary 4.5.12. Every locally finite group is amenable. �

Example 4.5.13. Let X be a set. Then the group Sym0(X) consisting of allpermutations of X with finite support is locally finite (see Example 3.2.4).Consequently, the group Sym0(X) is amenable. The groups Sym0(X), whereX is a infinite set, are our first examples of infinite amenable groups. Re-call from Lemma 2.6.3 that Sym0(X) is not residually finite whenever X isinfinite.

Corollary 4.5.14. Let (Gi)i∈I be a family of amenable groups. Then theirdirect sum G =

⊕i∈I Gi is amenable.

Proof. Every finitely generated subgroup of G is a subgroup of some finiteproduct of the groups Gi and hence amenable by Corollary 4.5.6 and Propo-sition 4.5.1. Thus G is locally amenable and therefore amenable by Corol-lary 4.5.11. �

4.6 Solvable Groups

Theorem 4.6.1. Every abelian group is amenable.

Proof. Let G be an abelian group. Equip (�∞(G))∗ with the weak-∗ topology.By Theorem 4.2.1, the set M(G) is a nonempty convex compact subset of(�∞(G))∗. On the other hand, it follows from Proposition 4.3.1 that the actionof G on M(G) is affine and continuous (note that the left and the right actionscoincide since G is Abelian). By applying the Markov-Kakutani fixed-pointTheorem (Theorem G.1.1), we deduce that G has at least one fixed point inM(G). Such a fixed point is clearly a bi-invariant mean on G. This showsthat G is amenable. �

Let G be a group. Recall the following definitions. The commutator of twoelements h and k in G is the element [h, k] ∈ G defined by [h, k] = hkh−1k−1.If H and K are subgroups of G, we denote by [H,K] the subgroup of Ggenerated by all commutators [h, k], where h ∈ H and k ∈ K. Note that[H,K] ⊂ K if K is normal in G. Note also that [H,K] is normal in G if Hand K are both normal in G.

The subgroup D(G) = [G, G] is called the derived subgroup, or commutatorsubgroup, of G. The subgroup D(G) is normal in G and the quotient groupG/D(G) is abelian. Observe that G is abelian if and only if [G, G] = {1G}.A group is called metabelian if its derived subgroup is abelian.

The derived series of a group G is the sequence (Di(G))i≥0 of subgroupsof G inductively defined by D0(G) = G and Di+1(G) = D(Di(G)) for alli ≥ 0. One has

4.6 Solvable Groups 93

G = D0(G) ⊃ D1(G) ⊃ D2(G) ⊃ . . .

with Di+1(G) normal in Di(G) and Di(G)/Di+1(G) abelian for all i ≥ 0.The group G is said to be solvable if there is an integer i ≥ 0 such that

Di(G) = {1G}. The smallest integer i ≥ 0 such that Di(G) = {1G} is thencalled the solvability degree of G.

Examples 4.6.2. (a) A group is solvable of degree 0 if and only if it is reducedto the identity element.

(b) The solvable groups of degree 1 are the nontrivial abelian groups.(c) The solvable groups of degree 2 are the nonabelian metabelian groups.(d) Let K be a field. The affine group over K is the subgroup Aff(K) of

Sym(K) consisting of all permutations of K which are of the form x �→ ax+b,where a, b ∈ K and a �= 0. The group D1(Aff(K)) is the group of translationsx �→ x + b, b ∈ K. Since D1(Aff(K)) is abelian, we have Di(Aff(K)) ={1Aff(K)} for all i ≥ 2. Consequently, the group Aff(K) is solvable ofdegree 2.

(e) The alternating groups Sym+2 and Sym+

3 are abelian and hence solvableof degree 1. The group Sym+

4 is solvable of degree 2. For n ≥ 5, the alternatinggroup Sym+

n is simple (see Remark C.4.4) and therefore Di(Sym+n ) = Sym+

n

for all i ≥ 0. Thus Sym+n is not solvable for n ≥ 5.

(f) The symmetric group Symn is solvable of degree 1 for n = 2, solvableof degree 2 for n = 3, solvable of degree 3 for n = 4, and not solvable forn ≥ 5.

Theorem 4.6.3. Every solvable group is amenable.

Proof. We proceed by induction on the solvability degree i of the group. Fori = 0, the group is reduced to the identity element and there is nothing toprove. Suppose now that the statement is true for solvable groups of degreei for some i ≥ 0. Let G be a solvable group of degree i + 1. Then its derivedsubgroup D(G) is solvable of degree i and hence amenable by our inductionhypothesis. As G/D(G) is abelian and therefore amenable by Theorem 4.6.1,we deduce that G is amenable by applying Proposition 4.5.5. �

Remark 4.6.4. The alternating group Sym+5 is amenable since it is finite.

However, as mentioned above, Sym+5 is not solvable. An example of an infinite

amenable group which is not solvable is provided by the group Z × Sym+5 .

Let G be a group. The lower central series of G is the sequence (Ci(G))i≥0

of subgroups of G defined by C0(G) = G and Ci+1(G) = [Ci(G), G] forall i ≥ 0. An easy induction shows that Ci(G) is normal in G and thatCi+1(G) ⊂ Ci(G) for all i. The group G is said to be nilpotent if there is aninteger i ≥ 0 such that Ci(G) = {1G}. The smallest integer i ≥ 0 such thatCi(G) = {1G} is then called the nilpotency degree of G.


Example 4.6.5. Let R be a nontrivial commutative ring. The Heisenberg groupwith coefficients in R is the subgroup HR of GL3(R) consisting of all matricesof the form

M(x, y, z) =

⎛

⎝1 y z0 1 x0 0 1

⎞

⎠ (x, y, z ∈ R).

One easily checks that the center Z(HR) of HR consists of all matrices of theform M(0, 0, z), z ∈ R, and that Z(HR) is isomorphic to the additive group(R,+). Moreover, one has D(HR) = Z(HR). It follows that HR is nilpotentof degree 2.

Proposition 4.6.6. Every nilpotent group is solvable.

Proof. An easy induction yields Di(G) ⊂ Ci(G) for all i ≥ 0. �

Remark 4.6.7. We have seen in Example 4.6.2(d) that the affine group Aff(K)over a field K is solvable. However, the group G = Aff(K) is not nilpotent.Indeed, the lower central series satisfies Ci(G) = D(G) �= {1G} for all i ≥ 1.

From Theorem 4.6.3 and Proposition 4.6.6, we get:

Corollary 4.6.8. Every nilpotent group is amenable. �

4.7 The Følner Conditions


(a) for every finite subset K ⊂ G and every real number ε > 0, there exists anonempty finite subset F ⊂ G such that

|F \ kF ||F | < ε for all k ∈ K; (4.4)

(b) there exists a net (Fj)j∈J of nonempty finite subsets of G such that

limj

|Fj \ gFj ||Fj |

= 0 for all g ∈ G; (4.5)

(c) for every finite subset K ⊂ G and every real number ε > 0, there exists anonempty finite subset F ⊂ G such that

|F \ Fk||F | < ε for all k ∈ K; (4.6)

4.7 The Følner Conditions 95

(d) there exists a net (Fj)j∈J of nonempty finite subsets of G such that,

limj

|Fj \ Fjg||Fj |

= 0 for all g ∈ G. (4.7)

Proof. First observe that the equivalences (a) ⇔ (c) and (b) ⇔ (d) are obvi-ous. Indeed, for any finite subset F ⊂ G and any element k ∈ G, we have

|F \ kF ||F | =

|F−1 \ F−1k−1||F−1| (4.8)

since the map g �→ g−1 is bijective on G.Let us show that (a) implies (b). Suppose (a). Let J denote the set of all

pairs (K, ε), where K is a finite subset of G and ε > 0. We equip J with thepartial ordering ≤ defined by

(K, ε) ≤ (K ′, ε′) ⇔ (K ⊂ K ′ and ε ≥ ε′).

Note that (J,≤) is a lattice, that is, a partially ordered set in which any twoelements admit a supremum (also called a join) and an infimum (also calleda meet). Indeed, one has

sup{(K, ε), (K′, ε′)} = (K ∪ K ′, min(ε, ε′)) andinf{(K, ε), (K ′, ε′)} = (K ∩ K ′, max(ε, ε′)).

In particular, J is a directed set. By (a), for each j = (K, ε) ∈ J , there existsa nonempty finite subset Fj ⊂ G such that

|Fj \ kFj ||Fj |

< ε for all k ∈ K. (4.9)

Let us fix now some element g ∈ G and ε0 > 0. Set j0 = ({g}, ε0). If j = (K, ε)satisfies j ≥ j0, then g ∈ K and ε ≤ ε0. Thus we have

|Fj \ gFj ||Fj |

< ε0, (4.10)

for all j ≥ j0 by (4.9). This shows that the net (Fj)j∈J satisfies (4.5). Con-sequently, (a) implies (b).

Finally, let us show (b) ⇒ (a). Suppose (b). Let K ⊂ G be a finite subsetand let ε > 0. By (4.5), for every k ∈ K there exist j(k) ∈ J such that

|Fj \ kFj ||Fj |

< ε for all j ≥ j(k).

Since J is a directed set, we can find j0 ∈ J such that j0 ≥ j(k) for all k ∈ K.Then, taking F = Fj0 we have (4.6). This shows that (b) implies (a). �


Definition 4.7.2. One says that a group G satisfies the Følner conditions ifG satisfies one of the equivalent conditions of Proposition 4.7.1.

Given a group G, a net (Fj)j∈J of nonempty finite subsets of G whichsatisfies (4.5) (resp. (4.7)) is called a left (resp. right) Følner net for G.

Observe that Fj \ g−1Fj (resp. Fj \ Fjg−1) is the set of elements of Fj

that are moved out of Fj by left (resp. right) multiplication by g. Thus, a net(Fj)j∈J of nonempty finite subsets of G is a left (resp. right) Følner net ifand only if, for each g ∈ G, the proportion of elements of Fj that are movedout of Fj by left (resp. right) multiplication by g converges to 0.

Remarks 4.7.3. (a) It immediately follows from (4.8) that (Fj) is a left Følnernet if and only if (F−1

j ) is a right Følner net.(b) If (Fj)j∈J is a left Følner net for G and (aj)j∈J is a net of elements

of G indexed by the same directed set J , then the net (Fjaj) is also a leftFølner net. Indeed, we have

|Fjaj \ gFjaj ||Fjaj |

=|Fj \ gFj |

|Fj |

for all j ∈ J and g ∈ G, since right multiplication by aj gives a bijectionfrom Fj onto Fjaj and from Fj \ gFj onto Fjaj \ gFjaj . Similarly, if (Fj) isa right Følner net, then (ajFj) is also a right Følner net.

For commutative groups, the concepts of left and right Følner nets coincideso that one simply speaks of Følner nets in this case.

If the directed set J is isomorphic to N, one speaks of left and right Følnersequences.

Examples 4.7.4. (a) Let G be a finite group. Then the sequence (Fn)n∈N,where Fn = G for all n ∈ N, is a left (and right) Følner sequence sinceFn \ gFn = Fn \ Fng = G \ G = ∅ for all g ∈ G. Therefore G satisfies theFølner conditions.

(b) Let G = Z. Then the sequence (Fn)n∈N defined by

Fn = {0, 1, . . . , n − 1}

is a Følner sequence. Indeed, if g ∈ Z is fixed and n ≥ |g|, then

|Fn \ (Fn + g)||Fn|

=|g|n

which converges to 0 as n tends to infinity. This shows that Z satisfies theFølner conditions (see Fig. 4.1).

(c) More generally, if G = Zr (r ≥ 1), one checks that the sequence (Fn)n∈N

given byFn = {0, 1, . . . , n − 1}r

4.7 The Følner Conditions 97

Fig. 4.1 The Følner set F5 in Z, its translate F5 + 3 and F5 \ (F5 + 3)

is a Følner sequence for G (see Fig. 4.2). Thus, Zr satisfies the Følner condi-

tions.

Proposition 4.7.5. Let G be group. Then the following conditions are equiv-alent:

(a) G is countable and satisfies the Følner conditions;(b) G admits a left Følner sequence;(c) G admits a right Følner sequence.

Proof. (a) ⇒ (b). Suppose (a). As G is countable, we can find a sequence(Kn)n∈N of finite subsets of G such that Kn ⊂ Kn+1 for all n and G =

⋃n Kn.

Let (εn) be a sequence of positive real numbers converging to 0. By condition(a) in Proposition 4.7.1, we can find, for each n, a nonempty finite subsetFn ⊂ G such that

|Fn \ gFn||Fn|

< εn for all g ∈ Kn.

Given an element g ∈ G, one has g ∈ Kn for n large enough and the precedinginequality implies

limn→∞

|Fn \ gFn||Fn|

= 0.

This shows that (Fn) is a left Følner sequence for G.(b) ⇒ (a). Suppose that G admits a left Følner sequence (Fn)n∈N. Let us

show that G is countable. Set Kn = {ab−1 : a, b ∈ Fn}. Consider an elementg ∈ G. Then, for n large enough, we have

|Fn \ gFn||Fn|

< 1.

This implies Fn ∩ gFn �= ∅ and hence g ∈ Kn. It follows that G =⋃

n Kn.As Kn is finite for each n, we deduce that G is countable. This shows that(b) implies (a).

As observed above, (Fn) is a left Følner sequence if and only if (F−1n ) is a

right Følner sequence. This shows that conditions (b) and (c) are equivalent.�


Fig. 4.2 The Følner set F5 in Z2, its translate F5 + (3, 2) and F5 \ (F5 + (3, 2))

4.8 Paradoxical Decompositions

Definition 4.8.1. Let G be a group. A left paradoxical decomposition of G isa triple (K, (Ak)k∈K , (Bk)k∈K) where K is a finite subset of G, and (Ak)k∈K

and (Bk)k∈K are families of subsets of G indexed by K such that

G =

(∐

k∈K

kAk

)∐

(∐

k∈K

kBk

)=

∐

k∈K

Ak =∐

k∈K

Bk, (4.11)

where∐

means disjoint union.Similarly, a right paradoxical decomposition of G is a triple (K, (Ak)k∈K ,

(Bk)k∈K) where K is a finite subset of G, and (Ak)k∈K and (Bk)k∈K arefamilies of subsets of G indexed by K such that

G =

(∐

k∈K

Akk

)∐

(∐

k∈K

Bkk

)=

∐

k∈K

Ak =∐

k∈K

Bk. (4.12)

4.9 The Theorems of Tarski and Følner 99

By taking inverses in G, it is clear that (K, (Ak)k∈K , (Bk)k∈K) is a left par-adoxical decomposition of G if and only if (K−1, (A−1

k−1)k∈K−1 , (B−1k−1)k∈K−1)

is a right decomposition of G. This shows that G admits a left paradoxicaldecomposition if and only if it admits a right paradoxical decomposition. Ifthis is the case, we simply say that G admits a paradoxical decomposition.

Example 4.8.2. Consider the free group of rank two G = F2 freely generatedby a and b. Let us partition F2 into four subsets A+, A−, B+ and B− asfollows. We denote by A+ (resp. A−) the subset of F2 consisting of all elementswhose reduced form starts with a positive (resp. negative) power of a. Letalso B+ denote the subset of F2 consisting of all elements either of the formb−n for n = 0, 1, 2, . . ., or whose reduced form starts with a positive power ofb. Finally, let B− = F2 \ (A+ ∪A− ∪B+), so that B− consists of all elementsin F2 whose reduced form starts with, but is not equal to, a negative powerof b. Then we have

F2 = A+∐

A−∐

B+∐

B− = a−1A+∐

A− = b−1B+∐

B−. (4.13)

For example, if g = a2b ∈ F2, we have g ∈ A+ and, on the one hand,g = a−1(a3b) ∈ a−1A+ and, on the other, g = b−1(ba2b) ∈ b−1B+.

Thus, setting A1 = A−, Aa = a−1A+, Ab = ∅, B1 = B−, Ba = ∅ andBb = b−1B+, (4.13) becomes

F2 = A1

∐aAa

∐bAb

∐B1

∐aBa

∐bBb

= A1

∐Aa

∐Ab

= B1

∐Ba

∐Bb,

(4.14)

which, with K = {1, a, b}, is nothing but (4.11). This shows that F2 admitsa paradoxical decomposition.

4.9 The Theorems of Tarski and Følner

The Tarski alternative theorem asserts that a group is either amenable oradmits a paradoxical decomposition. The theorem of Følner says that a groupis amenable if and only if it satisfies the Følner conditions. Thus, we havethe following:

Theorem 4.9.1 (Tarski-Følner). Let G be a group. Then the followingconditions are equivalent.

(a) G is amenable;(b) G admits no paradoxical decompositions;(c) G satisfies the Følner conditions.


Theorem 4.9.1 directly follows from the equivalence of conditions (a), (b)and (g) contained in the theorem below:

Theorem 4.9.2. Let G be a group. Then the following conditions are equiv-alent:

(a) G is not amenable;(b) G does not satisfy the Følner conditions;(c) there exists a finite subset K ⊂ G such that |KF | ≥ 2|F | for every finite

subset F ⊂ G.;(d) there exists a finite subset K ⊂ G such that |FK| ≥ 2|F | for every finite

subset F ⊂ G;(e) there exist a 2-to-one surjective map ϕ : G → G and a finite subset K ⊂ G

such that g(ϕ(g))−1 ∈ K for all g ∈ G;(f) there exist a 2-to-one surjective map ϕ : G → G and a finite subset K ⊂ G

such that g−1ϕ(g) ∈ K for all g ∈ G;(g) G admits a paradoxical decomposition.

We recall that a surjective map f : X → Y from a set X onto a set Y issaid to be 2-to-one if every y ∈ Y has exactly two preimages in X.

Proof of Theorem 4.9.2. By taking inverses in G, it is clear that (c) ⇔ (d)and (e) ⇔ (f). Thus, it suffices to prove (a) ⇒ (b), (b) ⇒ (c), (c) ⇒ (e), (e)⇒ (g), and (g) ⇒ (a).

(a) ⇒ (b). Suppose that G satisfies the Følner conditions. Then G admitsa left Følner net (Fj)j∈J . Consider, for each j ∈ J , the mean with finitesupport mj : �∞(G) → R defined by

mj(x) =1

|Fj |∑

h∈Fj

x(h) for all x ∈ �∞(G).

Let g ∈ G. For every x ∈ �∞(G), we have

(gmj − mj)(x) = gmj(x) − mj(x)

= mj(g−1x) − mj(x)

=1

|Fj |

⎛

⎝∑

h∈Fj

x(gh) −∑

h∈Fj

x(h)

⎞

⎠

=1

|Fj |

⎛

⎝∑

h∈gFj

x(h) −∑

h∈Fj

x(h)

⎞

⎠

=1

|Fj |

⎛

⎝∑

h∈gFj\Fj

x(h) −∑

h∈Fj\gFj

x(h)

⎞

⎠ ,

4.9 The Theorems of Tarski and Følner 101

which gives, by using triangle inequality,

|(gmj − mj)(x)| ≤ 1|Fj |

⎛

⎝∑

h∈gFj\Fj

|x(h)| +∑

h∈Fj\gFj

|x(h)|

⎞

⎠

≤ 1|Fj |

(|gFj \ Fj | + |Fj \ gFj |)‖x‖∞.

As the sets gFj and Fj have the same cardinality, we have

|gFj \ Fj | = |Fj \ gFj |. (4.15)

Thus, we finally get

|(gmj − mj)(x)| ≤ 2|Fj \ gFj |

|Fj |‖x‖∞.

This implies thatlim

j(gmj − mj)(x) = 0,

since (Fj) is a left Følner net. Thus, the net (gmj − mj)j∈J converges to0 in (�∞(G))∗ for the weak-∗ topology. It follows that G is amenable byLemma 4.5.9. This shows that (a) implies (b).

(b) ⇒ (c). Suppose that G does not satisfy the Følner conditions. Then,there exist a finite subset K0 ⊂ G and ε0 > 0 such that for every nonemptyfinite subset F ⊂ G one has

|F \ k0F ||F | ≥ ε0 (4.16)

for some k0 = k0(F ) ∈ K0. Consider the set K1 = K0 ∪ {1G}. Let F be anonempty finite subset of G. Observe that K1F ⊃ F and K1F \F = K0F \F .Thus, we have

|K1F | − |F | = |K1F \ F |

= |K0F \ F |

≥ |k0F \ F |

= |F \ k0F | (since |F | = |k0F |)

≥ ε0|F | (by (4.16)),

which gives|K1F | ≥ (1 + ε0)|F |.


Choose n0 ∈ N such that (1 + ε0)n0 ≥ 2 and set K = Kn01 . Then, we have

|KF | = |Kn01 F | ≥ (1 + ε0)|Kn0−1

1 F | ≥ · · · ≥ (1 + ε0)n0 |F |

so that |KF | ≥ 2|F | for every finite subset F ⊂ G. This shows that (b)implies (c).

(c) ⇒ (e). Suppose that G satisfies condition (c), that is, there exists afinite subset K ⊂ G such that

|KF | ≥ 2|F | for every finite subset F ⊂ G. (4.17)

Consider the bipartite graph GK(G) = (G, G, E) (see Appendix H), wherethe set E ⊂ G × G of edges consists of all the pairs (g, h) such that g ∈ Gand h ∈ Kg.

We claim that GK(G) satisfies the Hall 2-harem conditions. Indeed, ifF is a finite subset of G, then, using the terminology for bipartite graphsintroduced in Sect. H.1, the right and left neighborhoods of F in GK(G)are the sets NR(F ) = KF and NL(F ) = K−1F respectively. Therefore, wehave

|NR(F )| = |KF | ≥ 2|F |,

by applying (4.17). On the other hand, if k ∈ K, then NL(F ) ⊃ k−1F sothat

|NL(F )| ≥ |k−1F | = |F | ≥ 12|F |.

This proves our claim. Thus, by virtue of the Hall harem Theorem (Theo-rem H.4.2), we deduce the existence of a perfect (1, 2)-matching M for GK(G).In other words, there exists a 2-to-one surjective map ϕ : G → G such that(ϕ(g), g) ∈ E, that is, g(ϕ(g))−1 ∈ K for all g ∈ G. This shows that (c)implies (e).

(e) ⇒ (g). Suppose (e), that is, there exist a 2-to-one surjective mapϕ : G → G and a finite set K ⊂ G such that

g(ϕ(g))−1 ∈ K for all g ∈ G. (4.18)

By the axiom of choice, we can find maps ψ1, ψ2 : G → G such that, for everyg ∈ G, the elements ψ1(g) and ψ2(g) are the two preimages of g for ϕ.

Observe that θ1(g) = ψ1(g)g−1 and θ2(g) = ψ2(g)g−1 belong to K forevery g ∈ G by (4.18). For each k ∈ K, define Ak and Bk by

Ak = {g ∈ G : θ1(g) = k} and Bk = {g ∈ G : θ2(g) = k}.

We have

G =∐

k∈K

Ak =∐

k∈K

Bk. (4.19)

4.10 The Fixed Point Property 103

On the other hand, observe that if g ∈ Ak then ψ1(g) = kg. Thus ψ1(G) =∐k∈K kAk. Similarly, we have ψ2(G) =

∐k∈K kBk. As G = ψ1(G)

∐ψ2(G),

we deduce that

G =

(∐

k∈K

kAk

)∐

(∐

k∈K

kBk

). (4.20)

Combining together (4.19) and (4.20), ,we deduce that (K, (Ak)k∈K , (Bk)k∈K)is a left paradoxical decomposition for G. This shows that (e) implies (g).

(g) ⇒ (a). Suppose that G admits a left paradoxical decomposition(K, (Ak)k∈K , (Bk)k∈K). If μ : P(G) → [0, 1] is a left invariant finitely ad-ditive probability measure on G, then (4.11) gives

1 = μ(G)

= μ

((∐

k∈K

kAk

)∐

(∐

k∈K

kBk

))

=∑

k∈K

μ(kAk) +∑

k∈K

μ(kBk)

=∑

k∈K

μ(Ak) +∑

k∈K

μ(Bk)

= μ(∐

k∈K

Ak) + μ(∐

k∈K

Bk)

= μ(G) + μ(G)= 2,

which is clearly absurd. Therefore G is not amenable. This shows that (g)implies (a). �

4.10 The Fixed Point Property

The following fixed point theorem is a generalization of the Markov-Kakutanitheorem (Theorem G.1.1).

Theorem 4.10.1. Let G be an amenable group acting affinely and contin-uously on a nonempty convex compact subset C of a Hausdorff topologicalvector space X. Then G fixes at least one point in C.

Proof. Let (Fj)j∈J be a left Følner net for G.Choose an arbitrary point x ∈ C and set

xj =1

|Fj |∑

h∈Fj

hx.


for each j ∈ J . Observe that xj ∈ C since C is convex. By compactness ofC, we can assume, after taking a subnet, that the net (xj)j∈J converges to apoint c ∈ C.

Let g ∈ G. For every j ∈ J , we have

gxj −xj = g

⎛

⎝ 1|Fj |

∑

h∈Fj

hx

⎞

⎠ − 1|Fj |

∑

h∈Fj

hx

=1

|Fj |∑

h∈Fj

ghx− 1|Fj |

∑

h∈Fj

hx (since the action of G on C is affine)

=1

|Fj |∑

h∈gFj

hx − 1|Fj |

∑

h∈Fj

hx,

which yields, after simplification,

gxj − xj =1

|Fj |∑

h∈gFj\Fj

hx − 1|Fj |

∑

h∈Fj\gFj

hx. (4.21)

Consider the points

yj =1

|gFj \ Fj |∑

h∈gFj\Fj

hx, and zj =1

|Fj \ gFj |∑

h∈Fj\gFj

hx.

Note that yj and zj belong to C by convexity of C.We have |Fj \ gFj | = |gFj \ Fj | since the sets Fj and gFj have the same

cardinality (cf. (4.15)). Setting

λj =|Fj \ gFj |

|Fj |=

|gFj \ Fj ||Fj |

,

equality (4.21) gives us

gxj − xj = λjyj − λjzj .

The net (λj) converges to 0 since (Fj) is a left Følner net. As C is compact,it follows that

limj

λjyj = limj

λjzj = 0,

by using Lemma G.2.2. Therefore, we have

gc − c = limj

(gxj − xj) = 0.

This shows that c is fixed by G. �

Notes 105


(a) the group G is amenable;(b) every continuous affine action of G on a nonempty convex compact subset

of a Hausdorff topological vector space admits a fixed point;(c) every continuous affine action of G on a nonempty convex compact subset

of a Hausdorff locally convex topological vector space admits a fixed point.

Proof. Implication (a) ⇒ (b) follows from the preceding theorem. Implication(b) ⇒ (c) is trivial. Finally,(c) ⇒ (a) follows from Theorem 4.2.1, Proposi-tion 4.3.1 and the fact that (�∞(G))∗ is a locally convex Hausdorff topologicalvector space for the weak-∗ topology (see Sect. F.2). �

Notes

The theory of amenable groups emerged from the study of the axiomaticproperties of the Lebesgue integral and the discovery of the Banach-Tarskiparadox at the beginning of the last century (see [Har3], [Pat], [Wag]). Thefirst definition of an amenable group, by the existence of an invariant finitelyadditive probability measure, is due to J. von Neumann in [vNeu1]. VonNeumann proved in particular that every abelian group is amenable (Theo-rem 4.6.1) and that an amenable group cannot contain a subgroup isomorphicto F2 (Corollary 4.5.2). The term amenable was introduced in the 1950s byM.M. Day, who played a central role in the development of the modern theoryof amenable groups by using means and applying techniques from functionalanalysis. Moreover, Day extended the notion of amenability to semigroups,for which one has to distinguish between right amenability and left amenabil-ity.

The question of the existence of a non-amenable group containing nosubgroup isomorphic to F2, which is called by some authors the von Neu-mann conjecture or Day’s problem, was answered in the affirmative by A.Yu.Ol’shanskii [Ols] who gave an example of a finitely generated non-amenablegroup all of whose proper subgroups are cyclic. Other examples of non-amenable groups with no subgroup isomorphic to F2 were found by S.I. Adyan[Ady] who showed that the free Burnside group B(m, n) is non-amenable form ≥ 2 and n odd with n ≥ 665. The free Burnside group B(m, n) is the quo-tient of the free group Fm by the subgroup of Fm generated by all n-powers,that is, all elements of the form wn for some w ∈ Fm. The order of everyelement of B(m, n) divide n. In particular, B(m, n) is a periodic group, thatis, a group in which every element has finite order. It is clear that a periodicgroup cannot contain a subgroup isomorphic to F2. A geometric method forconstructing finitely generated non-amenable periodic groups was describedby M. Gromov in [Gro3]. Examples of finitely presented non-amenable groups


which contain no subgroup isomorphic to F2 were given by A.Yu Ol’shanskiiand M. Sapir [OlS].

There is also a more general notion of amenability for locally compactgroups and actions of locally compact groups (see [Gre], [Pat]).

A group G is called polycyclic if it admits a finite sequence of subgroups

{1G} = H0 ⊂ H1 ⊂ H2 ⊂ · · · ⊂ Hn = G

such that Hi is normal in Hi+1 and Hi+1/Hi is a (finite or infinite) cyclicgroup for each 0 ≤ i ≤ n− 1. Clearly, every polycyclic group is solvable. It isnot hard to prove that every finitely generated nilpotent group is polycyclic. Itwas shown by L. Auslander [Aus] that every polycyclic group is isomorphicto a subgroup of SLn(Z) for some integer n ≥ 1. It follows in particularthat every polycyclic group is residually finite. This last result is due toK.A. Hirsch [Hir] (see also [RobD, p. 154]). P. Hall [Hall2] proved that everyfinitely generated metabelian group is residually finite and gave an exampleof a finitely generated solvable group of degree 3 which is not residually finite.

The equivalence between Følner conditions and amenability was estab-lished by E. Følner in [Føl]. The proof was later simplified by I. Namioka in[Nam]. The equivalence between amenability and the non-existence of a para-doxical decomposition is due to A. Tarski (see [Tar1], [Tar2] and [CGH1]).

Let G be a group. Given a left (or right) paradoxical decompositionP = (K, (Ak)k∈K , (Bk)k∈K) of G, the integer number c(P) = m + n, wherem = |{k ∈ K : Ak �= ∅}| and n = |{k ∈ K : Bk �= ∅}|, is called thecomplexity of P. Then the quantity T (G) = inf c(P), where the infimum istaken over all left (or right) paradoxical decompositions P of G, is calledthe Tarski number of G. One uses the convention that T (G) = ∞ if G ad-mits no paradoxical decompositions, that is, if G is amenable (cf. Theorem4.9.1). It was proved by B. Jonsson, a student of Tarski in the 1940s, that agroup G has Tarski number T (G) = 4 if and only if G contains a subgroupisomorphic to F2, the free group of rank 2. In [CGH1, CGH2] it was shownthat for the free Burnside groups B(m, n) with m ≥ 2 and n ≥ 665 odd onehas 6 ≤ T (B(m, n)) ≤ 14. The computations involve spectral analysis andCheeger-Buser type isoperimetric inequalities (see Sect. 6.10 and Sect. 6.12)and Adyan’s cogrowth estimates [Ady] for the free Burnside groups B(m, n)(see (6.117) in the Notes for Chap. 6).

The extension of the Markov-Kakutani fixed point theorem to amenablegroups (cf. Theorem 4.10.1) is due to Day [Day2].

Exercises

4.1. Let G be an infinite group and let μ : P(G) → [0, 1] be a left (or right)invariant finitely additive probability measure on G. Show that every finitesubset A ⊂ G satisfies μ(A) = 0.

Exercises 107

4.2. Let X be an infinite set. Prove that the symmetric group Sym(X) is notamenable. Hint: Show that Sym(X) contains a subgroup isomorphic to thefree group F2.

4.3. Show that the finitely generated and non residually finite groups G1

and G2 described in Sect. 2.6 are amenable. Hint: Each of these groups is thesemidirect product of a locally finite group and an infinite cyclic group.

4.4. Let G be a group.(a) Suppose that H and K are normal subgroups of G. Prove that HK

is a normal subgroup of G and that the groups HK/K and H/(H ∩ K) areisomorphic.

(b) Suppose that H and K are normal amenable subgroups of G. Provethat HK is a normal amenable subgroup of G.

(c) Show that the set of all normal amenable subgroups of G has a maximalelement for inclusion. This maximal element is called the amenable radical ofthe group G. Hint: Use (b) and Proposition 4.5.10 to show that the union ofall normal amenable subgroups of G is a normal amenable subgroup of G.

4.5. Let G be a group. Show that G is metabelian if and only if it containsa normal subgroup N such that the group G/N is abelian.

4.6. Let G be a finitely generated solvable group. Show that if all elementsof G have finite order then G is finite. Hint: Use induction on the solvabilitydegree of G.

4.7. Show that every subgroup of a solvable (resp. nilpotent) group is solvable(resp. nilpotent).

4.8. Show that every quotient of a solvable (resp. nilpotent) group is solvable(resp. nilpotent).

4.9. Let G be a group containing a normal subgroup H such that both Hand G/H are solvable. Show that G is solvable.

4.10. Show that the direct product of two nilpotent groups is a nilpotentgroup.

4.11. Show that a semidirect product of two nilpotent groups may fail to benilpotent. Hint: Take for example the symmetric group Sym3, which is thesemidirect product of two cyclic groups.

4.12. Let G be a group. Show that G is solvable if and only if there is a finitesequence

{1G} = H0 ⊂ H1 ⊂ H2 ⊂ · · · ⊂ Hn = G

of subgroups of G such that Hi is normal in Hi+1 and Hi+1/Hi is abelian forall 0 ≤ i ≤ n − 1.


4.13. Let G be a group and i ≥ 0. Show that the group Ci(G)/Ci+1(G) iscontained in the center of G/Ci+1(G).

4.14. Show that every nontrivial nilpotent group has a nontrivial center.

4.15. Let G be a nilpotent group of nilpotency degree d ≥ 1. Show that thequotient group G/Cd−1(G) is nilpotent of nilpotency degree d − 1.

4.16. Let G be a finite group whose order is a power of a prime number.Show that G is nilpotent. Hint: Prove that the center of G is nontrivial byconsidering the action of G on itself by conjugation and then proceed byinduction.

4.17. Let G be a group. Denote by Nnq (resp. Nsq, resp. Naq) the set of allnormal subgroups N ⊂ G such that the quotient group G/N is nilpotent(resp. solvable, resp. amenable). The sets Nnq, Nsq and Naq are partiallyordered by reverse inclusion.

(a) Show that if N1 and N2 are two elements of Nnq (resp. Nsq, resp. Naq),then N1 ∩N2 is an element of Nnq (resp. Nsq, resp. Naq). Hint: observe thatif ρ1 : G → G/N1 and ρ2 : G → G/N2 are the canonical homomorphisms,then the map ψ : G → G/N1 × G/N2 defined by ψ(g) = (ρ1(g), ρ2(g)) is ahomomorphism whose kernel is N1 ∩ N2.

(b) Show that Nnq (resp. Nsq, resp. Naq) gives rise to a projective sys-tem of groups in a natural way. The limit of this projective system is calledthe pronilpotent completion (resp. prosolvable completion, resp. proamenablecompletion) of the group G and is denoted by Gn (resp. Gs, resp. Ga).

(c) Show that there is a canonical homomorphism G → Gn (resp. G → Gs,resp. G → Ga) and that this homomorphism is injective if and only if G isresidually nilpotent (resp. residually solvable, resp. residually amenable).

4.18. Recall that a group G is called polycyclic if it admits a finite sequenceof subgroups

{1G} = H0 ⊂ H1 ⊂ H2 ⊂ · · · ⊂ Hn = G

such that Hi is normal in Hi+1 and Hi+1/Hi is a (finite or infinite) cyclicgroup for each 0 ≤ i ≤ n − 1.

(a) Show that every polycyclic group is finitely generated.(b) Show that every subgroup of a polycyclic group is polycyclic.(c) Deduce from (a) and (b) that every subgroup of a polycyclic group is

finitely generated.

4.19. The lamplighter group is the wreath product L = (Z/2Z) �Z. Thus, L isthe semidirect product of the groups H = ⊕n∈ZAn and Z, where An = Z/2Z

for all n ∈ Z and Z acts on H by the Z-shift.(a) Show that the group L is metabelian (and therefore solvable) and

residually finite.

Exercises 109

(b) Prove that L is finitely generated. Hint: Show that L is generated bythe two elements s and t corresponding respectively to the nontrivial elementof A0 and to the canonical generator of Z.

(c) Show that L is not polycyclic. Hint: Observe that H is not finitelygenerated and use Exercise 4.18(c).

4.20. Show that every finite solvable group is polycyclic.

4.21. Let m be an integer such that |m| ≥ 2. Let G be the group given bythe presentation G = 〈a, b : aba−1 = bm〉. Use the results in Exercise 2.7to prove that the commutator subgroup [G, G] is isomorphic to the additivegroup Z[1/m] and that the quotient group G/[G, G] is infinite cyclic.

4.22. Let G be a locally finite group. Let S denote the directed set consistingof all finitely generated subgroups of G partially ordered by inclusion. Provethat the net (H)H∈S is a Følner net for G.

4.23. Show that the sequence (Fn)n∈N, where Fn consists of all rational num-bers of the form k/n! with k ∈ N and k ≤ (n + 1)!, is a Følner sequence forthe additive group Q.

4.24. Let G = HZ denote the integral Heisenberg group (cf. Example 4.6.5).For each integer n ≥ 0, define the subset Fn ⊂ G by

Fn =

⎧⎨

⎩

⎛

⎝1 x z0 1 y0 0 1

⎞

⎠ ∈ G : 1 ≤ x ≤ n, 1 ≤ y ≤ n, 1 ≤ z ≤ n2

⎫⎬

⎭ .

Show that the sequence (Fn)n≥0 is a Følner sequence for G.

4.25. Let G be a group. Show that G is amenable if and only if the followingcondition holds: for every finite subset K ⊂ G and every ε > 0, there existsa finite subset F ⊂ G such that |KF | < (1 + ε)|F |.

4.26. Let G be a countable amenable group. Show that G admits a left (resp.right) Følner sequence (Fn)n∈N which satisfies G =

⋃n∈N

Fn and Fn ⊂ Fn+1

for all n ∈ N.

4.27. Let (K, (Ak)k∈K , (Bk)k∈K) be a left (or right) paradoxical decomposi-tion of a group G. Show that |K| ≥ 3.

4.28. Let G be a group and H ⊂ G a subgroup. Let (K, (Ak)k∈K , (Bk)k∈K)be a left paradoxical decomposition of H and T ⊂ G a set of representa-tives for the right cosets of H in G. For each k ∈ K set A′

k =∐

t∈T Aktand B′

k =∐

t∈T Bkt. Show that (K, (A′k)k∈K , (B′

k)k∈K) is a left paradoxicaldecomposition of G.


4.29. Let T (G) ∈ N ∪ {∞} denote the Tarski number of a group G.(a) Show that T (G) ≥ 4 for all groups G.(b) Show that T (G) = 4 if and only if G contains a subgroup isomorphic

to F2. Hint: Use Example 4.8.2 and the Klein Ping-Pong theorem (Theo-rem D.5.1).

(c) Let H be a subgroup of a group G. Show that T (G) ≤ T (H).(d) Let N be a normal subgroup of a group G. Show that T (G) ≤ T (G/N).(e) Let G be a group. Show that there exists a finitely generated subgroup

H ⊂ G such that T (H) = T (G).(f) Let G be a group. Suppose that all elements of G have finite order.

Show that T (G) ≥ 6.

Chapter 5

The Garden of Eden Theorem

The Garden of Eden Theorem gives a necessary and sufficient condition forthe surjectivity of a cellular automaton with finite alphabet over an amenablegroup. It states that such an automaton is surjective if and only if it is pre-injective. As the name suggests it, pre-injectivity is a weaker notion thaninjectivity. It means that any two configurations which have the same imageunder the automaton must be equal if they coincide outside a finite subsetof the underlying group (see Sect. 5.2). We shall establish the Garden ofEden theorem in Sect. 5.8 by showing that both the surjectivity and the pre-injectivity are equivalent to the maximality of the entropy of the image of thecellular automaton. The entropy of a set of configurations with respect to aFølner net of an amenable group is defined in Sect. 5.7. Another importanttool in the proof of the Garden of Eden theorem is a notion of tiling for groupsintroduced in Sect. 5.6. The Garden of Eden theorem is used in Sect. 5.9 toprove that every residually amenable (and hence every amenable) group issurjunctive. In Sect. 5.10 and Sect. 5.11, we give simple examples showingthat both implications in the Garden of Eden theorem become false over afree group of rank two. In Sect. 5.12 it is shown that a group G is amenableif and only if every surjective cellular automaton with finite alphabet overG is pre-injective. This last result gives a characterization of amenability interms of cellular automata.

5.1 Garden of Eden Configurations and Garden of EdenPatterns

Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton.A configuration y ∈ AG is called a Garden of Eden configuration for τ if yis not in the image of τ . Thus the surjectivity of τ is equivalent to the non-existence of Garden of Eden configurations.


111

http://dx.doi.org/10.1007/978-3-642-14034-1_5

112 5 The Garden of Eden Theorem

The biblical terminology “Garden of Eden” (a peaceful place where we willnever return) comes from the fact that one often regards a cellular automatonτ : AG → AG from a dynamical viewpoint. This means that one thinks of aconfiguration as evolving with time according to τ : if x ∈ AG is the config-uration at time t = 0, 1, 2, . . ., then τ(x) is the configuration at time t + 1.A configuration x ∈ AG \ τ(AG) is called a Garden of Eden configurationbecause it may only appear at time t = 0.

A pattern p : Ω → A is called a Garden of Eden pattern for τ if there isno configuration x ∈ AG such that τ(x)|Ω = p. It follows from this definitionthat if p : Ω → G is a Garden of Eden pattern for τ , then any configurationy ∈ AG such that y|Ω = p is a Garden of Eden configuration for τ . Thus,the existence of a Garden of Eden pattern implies the existence of Garden ofEden configurations, that is, the non-surjectivity of τ . It turns out that theconverse is also true when the alphabet set is finite:

Proposition 5.1.1. Let G be a group and let A be a finite set. Let τ : AG →AG be a cellular automaton. Suppose that τ is not surjective. Then τ admitsa Garden of Eden pattern.

Proof. We know that the set τ(AG) is closed in AG for the prodiscrete topol-ogy by Lemma 3.3.2. It follows that the set AG \ τ(AG) is open in AG.Therefore, if y ∈ AG is a Garden of Eden configuration for τ , we may find afinite subset Ω ⊂ G such that

V (y,Ω) = {x ∈ AG : x|Ω = y|Ω} ⊂ AG \ τ(AG).

In other words, every configuration extending y|Ω is not in τ(AG), that is,y|Ω is a Garden of Eden pattern for τ . ��

5.2 Pre-injective Maps

Let G be a group and let A be a set.Two configurations x1, x2 ∈ AG are called almost equal if the set {g ∈

G : x1(g) �= x2(g)} is finite. It is clear that being almost equal defines anequivalence relation on the set AG.

Given a subset X ⊂ AG and a set Z, a map f : X → Z is called pre-injective if it satisfies the following condition: if two configurations x1, x2 ∈ Xare almost equal and such that f(x1) = f(x2), then x1 = x2. It immediatelyfollows from this definition that the injectivity of f implies its pre-injectivity.The converse is trivially true when the group G is finite. However, a pre-injective map f : AG → Z may fail to be injective when G is infinite.

Examples 5.2.1. (a) Let us take G = Z and A = Z/3Z. Consider the cellularautomaton τ : AG → AG defined by τ(x)(n) = x(n−1)+x(n)+x(n+1) for all

5.2 Pre-injective Maps 113

x ∈ AG and n ∈ G. Then τ is pre-injective. Indeed, suppose that x1, x2 ∈ AG

are two configurations such that the set Ω = {n ∈ G : x1(n) �= x2(n)} is anonempty finite subset of Z. Let n0 denote the largest element in Ω. Thenτ(x1)(n0 + 1) �= τ(x2)(n0 + 1) and hence τ(x1) �= τ(x2). This sows that τis pre-injective. However, τ is not injective since the constant configurationsc0, c1 ∈ AG given by c0(n) = 0 and c1(n) = 1 for all n ∈ Z have the sameimage c0 by τ .

(b) Let G = Z2 and A = {0, 1}. Consider the cellular automaton τ : AG →

AG associated with the Game of Life (cf. Example 1.4.3(a)). Let x1 ∈ AG

be the configuration with no live cells (x1(g) = 0 for all g ∈ G) and letx2 ∈ AG be the configuration with only one live cell at the origin (x2(g) = 1if g = (0, 0) and x2(g) = 0 otherwise). Then x1 and x2 are almost equal andone has τ(x1) = τ(x2) = x1. Therefore τ is not pre-injective.

(c) Let G be a group and let A = {0, 1}. Let S be a finite subset of Ghaving at least 3 elements. Let τ : AG → AG be the majority action cellularautomaton associated with G and S (cf. Example 1.4.3(c)).

Let x0 ∈ AG be the configuration defined by x0(g) = 0 for all g ∈ G.Let x1 ∈ AG be the configuration defined by x1(1G) = 1 and x1(g) = 0 ifg �= 1G. One has τ(x0) = τ(x1) = x0 and {g ∈ G : x0(g) �= x1(g)} = {1G}.Consequently, τ is not pre-injective.

The operations of induction and restriction of cellular automata with re-spect to a subgroup of the underlying group have been introduced in Sect. 1.7.It turns out that pre-injectivity, like injectivity and surjectivity (see Propo-sition 1.7.4), is preserved by these operations. More precisely, we have thefollowing result (which will not be used in the proof of the Garden of Edentheorem given below):

Proposition 5.2.2. Let G be a group and let A be a set. Let H be a sub-group of G and let τ ∈ CA(G, H; A). Let τH ∈ CA(H;A) denote the cellularautomaton obtained by restriction of τ to H. Then, τ is pre-injective if andonly if τH is pre-injective.

Proof. First recall from (1.16) the factorizations

AG =∏

c∈G/H

Ac and τ =∏

c∈G/H

τc, (5.1)

where τc : Ac → Ac satisfies τc(x|c) = (τ(x))|c for all x ∈ Ac.Suppose that τ is pre-injective. We want to show that τH is pre-injective.

So let x, y ∈ AH be almost equal configurations over H such that τH(x) =τH(y). Let us fix an arbitrary element a0 ∈ A and extend x and y to config-urations x and y in AG by setting

x(g) =

{x(g) if g ∈ H,

a0 otherwiseand y(g) =

{y(g) if g ∈ H,

a0 otherwise


for all g ∈ G. Note that the configurations x and y are almost equal since{g ∈ G : x(g) �= y(g)} = {h ∈ H : x(h) �= y(h)}. By construction, x|c = y|cfor all c ∈ G/H \ {H}, while x|H = x and y|H = y. From (5.1), we deducethat τ(x) = τ(y). It follows that x = y, by pre-injectivity of τ . This impliesthat x = x|H equals y|H = y. This shows that τH is pre-injective.

Conversely, suppose that τH is pre-injective. Let x, y ∈ AG be almost equalconfigurations over G such that τ(x) = τ(y). For each g ∈ G, consider theconfigurations xg, yg ∈ AH defined by xg(h) = x(gh) and yg(h) = y(gh) forall h ∈ H. Observe that the configurations xg and yg are almost equal sincex and y are almost equal. On the other hand, we have xg = φ∗

g(x|c) and yg =φ∗

g(y|c), where c = gH ∈ G/H and φ∗g : Ac → AH is the bijective map defined

by φ∗g(u)(h) = u(gh) for all u ∈ Ac. As τ(x) = τ(y), we have τc(x|c) = τc(y|c)

by (5.1). We deduce that τH(xg) = τH(yg) since φ∗g conjugates τc and τH by

Proposition 1.18. Consequently, we have xg = yg for all g ∈ G by the pre-injectivity of τH . This implies that x = y. Therefore, τ is pre-injective. ��

5.3 Statement of the Garden of Eden Theorem

The Garden of Eden theorem gives a necessary and sufficient condition forthe surjectivity of a cellular automaton with finite alphabet over an amenablegroup. Its name comes from the fact that the surjectivity of a cellular au-tomaton is equivalent to the absence of Garden of Eden configurations (seeSect. 5.1).

Theorem 5.3.1 (The Garden of Eden theorem).Let G be an amenable group and let A be a finite set. Let τ : AG → AG be

a cellular automaton. Then one has

τ is surjective ⇐⇒ τ is pre-injective.

The proof of Theorem 5.3.1 will be given in Sect. 5.8 (see Theorem 5.8.1).Let us first present some applications of this theorem.

Examples 5.3.2. (a) Let G = Z and A = Z/3Z. We have seen in Exam-ple 5.2.1(a) that the cellular automaton τ : AG → AG defined by τ(x)(n) =x(n − 1) + x(n) + x(n + 1) is pre-injective. Since Z is amenable (cf. Theo-rem 4.6.1), it follows from the Garden of Eden theorem that τ is surjective.In fact, a direct proof of the surjectivity of τ is not difficult (see Exercise 5.5).

(b) Let G = Z2 and A = {0, 1}. Consider the cellular automaton τ : AG →

AG associated with the Game of Life. We have seen in Example 5.2.1(b)that τ is not pre-injective. The group Z

2 is amenable by Theorem 4.6.1. Byapplying the Garden of Eden theorem, we deduce that τ is not surjective (seeSect. 5.13 for a direct proof).

5.4 Interiors, Closures, and Boundaries 115

(c) Let G be a group and let A = {0, 1}. Let S be a finite subset of Ghaving at least 3 elements. Let τ : AG → AG be the majority action cellularautomaton associated with G and S.

We have seen in Example 5.2.1(c) that τ is not pre-injective. Thus itfollows from the Garden of Eden theorem that if G is amenable then τ is notsurjective. We shall see in Sect. 5.10 that if G is the free group F2, then τ issurjective.

5.4 Interiors, Closures, and Boundaries

Let G be a group.Let E and Ω be subsets of G. The E-interior Ω−E and the E-closure Ω+E

of Ω are the subsets of G defined respectively by

Ω−E = {g ∈ G : gE ⊂ Ω} and

Ω+E = {g ∈ G : gE ∩ Ω �= ∅}

(see Figs. 5.1–5.2). Observe that

Ω−E =⋂

e∈E

Ωe−1 (5.2)

andΩ+E =

⋃

e∈E

Ωe−1 = ΩE−1. (5.3)

Fig. 5.1 The E-interior Ω−E of a set Ω. Here, Ω ⊂ R2, E = {e, f} with e = (2, 1) and

f = (−1,−1)


Fig. 5.2 The E-closure Ω+E of a set Ω. Here, Ω ⊂ R2, E = {e, f} with e = (2, 1) and

f = (−1,−1)

The E-boundary of Ω is the subset ∂E(Ω) of G defined by

∂E(Ω) = Ω+E \ Ω−E

(see Fig. 5.3).

Examples 5.4.1. (a) If E = ∅, then Ω−E = G and Ω+E = ∂E(Ω) = ∅.(b) If E = {1G}, then Ω−E = Ω+E = Ω and ∂E(Ω) = ∅.(c) If a ∈ G and E = {1G, a}, then Ω−E = Ω ∩ Ωa−1, Ω+E = Ω ∪ Ωa−1,

so that

∂E(Ω) = (Ω ∪ Ωa−1) \ (Ω ∩ Ωa−1) = (Ω \ Ωa−1) ∪ (Ωa−1 \ Ω) (5.4)

is the symmetric difference between the sets Ω and Ωa−1 (see Fig. 5.4).(d) Let us take G = Z

2 and E = {−1, 0, 1}2. Let a, b, c, d ∈ Z and considerthe rectangle Ω = [a, b] × [c, d] = {(x, y) ∈ Z

2 : a ≤ x ≤ b, c ≤ y ≤ d}. Thenone has

Ω−E = [a+1, b−1]× [c+1, d−1] and Ω+E = [a−1, b+1]× [c−1, d+1]

(see Fig. 5.5).

Here are some general properties of the sets Ω−E , Ω+E and ∂E(Ω) which weshall frequently use in the sequel.

Proposition 5.4.2. Let G be a group. Let E, E1, E2 and Ω be subsets of G.Then the following hold:

(i) (G \ Ω)−E = G \ Ω+E;(ii) (G \ Ω)+E = G \ Ω−E;


Fig. 5.3 The E-boundary ∂E(Ω) of a set Ω. Here, Ω ⊂ R2, E = {e, f} with e = (2, 1)

and f = (−1,−1)

Fig. 5.4 The E-interior Ω−E = Ω ∩ Ωa−1, the E-closure Ω+E = Ω ∪ Ωa−1, and theE-boundary ∂E(Ω) = (Ω ∪ Ωa−1) \ (Ω ∩ Ωa−1) of a set Ω when E = {1G, a}

Fig. 5.5 The E-interior Ω−E , the E-closure Ω+E , and the E-boundary ∂E(Ω) of a rect-

angle Ω ⊂ Z2. Here, Ω = [a, b] × [b, c] and E = {−1, 0,−1}2

(iii) if a ∈ E, then Ω−E ⊂ Ωa−1 ⊂ Ω+E;(iv) if 1G ∈ E, then Ω−E ⊂ Ω ⊂ Ω+E;(v) if E is nonempty and Ω is finite, then Ω−E is finite;(vi) if E and Ω are both finite, then Ω+E and ∂E(Ω) are finite;


(vii) if E1 ⊂ E2, then Ω−E2 ⊂ Ω−E1 , Ω+E1 ⊂ Ω+E2 and ∂E1(Ω) ⊂ ∂E2(Ω);(viii) if h ∈ G, then h(Ω−E) = (hΩ)−E, h(Ω+E) = (hΩ)+E and h(∂E(Ω)) =

∂E(hΩ).

Proof. (i) By definition, we have g ∈ G\Ω+E if and only if gE does not meetΩ, that is, if and only if g ∈ (G \ Ω)−E .

(ii) By replacing Ω by G \ Ω in (i) we get Ω−E = G \ (G \ Ω)+E whichgives (ii) after taking complements.

(iii) If a ∈ E, then Ω−E ⊂ Ωa−1 by (5.2) and Ωa−1 ⊂ Ω+E by (5.3).(iv) Assertion (iii) gives (iv) by taking a = 1G.(v) If a ∈ E and Ω is finite, then |Ω−E | ≤ |Ωa−1| = |Ω| by (iii).(vi) If E and Ω are both finite, then |Ω+E | ≤ |Ω||E| by (5.3) so that Ω+E

is finite. The set ∂E(Ω) is then also finite since it is contained in Ω+E .(vii) The first two statements follow immediately from (5.2) and (5.3),

respectively, and imply that

∂E1(Ω) = Ω+E1 \ Ω−E1 ⊂ Ω+E2 \ Ω−E1 ⊂ Ω+E2 \ Ω−E2 = ∂E2(Ω).

(viii) By using (5.2) we have

h(Ω−E) = h ∩e∈E Ωe−1 = ∩e∈EhΩe−1 = (hΩ)−E ,

which gives the first statement. Similarly, from (5.3) we get

h(Ω+E) = h(ΩE−1) = (hΩ)E−1 = (hΩ)+E .

Finally, we have

h(∂E(Ω)) = h(Ω+E \ Ω−E) = hΩ−E − hΩ−E = ∂E(hΩ).

��

Proposition 5.4.3. Let G be a group and let A be a set. Let τ : AG → AG

be a cellular automaton with memory set S. Let x and x′ be elements of AG.Suppose that there is a subset Ω of G such that x and x′ coincide on Ω (resp.on G \ Ω). Then the configurations τ(x) and τ(x′) coincide on Ω−S (resp.on G \ Ω+S).

Proof. Suppose that x and x′ coincide on Ω. If g ∈ Ω−S , then gS ⊂ Ω andtherefore τ(x)(g) = τ(x′)(g) by Lemma 1.4.7. It follows that τ(x) and τ(x′)coincide on Ω−S .

Suppose now x and x′ coincide on G \Ω. Then τ(x) and τ(x′) coincide on(G \Ω)−S = G \Ω+S by the first part of the proof and Proposition 5.4.2(i).

��

Proposition 5.4.4. Let G be a group and let (Fj)j∈J be a net of nonemptyfinite subsets of G. Then the following conditions are equivalent:


(a) the net (Fj)j∈J is a right Følner net for G;(b) one has

limj

|∂E(Fj)||Fj |

= 0 for every finite subset E ⊂ G.

Proof. (b) ⇒ (a). Suppose (b). Let g ∈ G and take E = {1G, g−1}. By (5.4),we have

Fj \ Fjg ⊂ ∂E(Fj),

and hence|Fj \ Fjg| ≤ |∂E(Fj)|.

Therefore, property (b) implies

limj

|Fj \ Fjg||Fj |

= 0.

This shows that (Fj) is a right Følner net for G.(a) ⇒ (b). Let E be a finite subset of G. By (5.2) and (5.3) we have

∂E(Fj) =

(⋃

a∈E

Fja−1

)\

(⋂

b∈E

Fjb−1

)

=

(⋃

a∈E

Fja−1

)⋂

(G \

⋂

b∈E

Fjb−1

)

=

(⋃

a∈E

Fja−1

)⋂

(⋃

b∈E

(G \ Fjb−1)

)

=⋃

a,b∈E

(Fja−1 \ Fjb

−1).

This implies|∂E(Fj)| ≤

∑

a,b∈E

|Fja−1 \ Fjb

−1|. (5.5)

Now observe that, for all a, b ∈ E, we have

|Fja−1 \ Fjb

−1| = |Fj \ Fjb−1a|,

since right multiplication by a is bijective on G. Therefore, inequality (5.5)gives us

|∂E(Fj)||Fj |

≤ |E|2 maxg∈K

|Fj \ Fjg||Fj |

,

where K is the finite subset of G defined by K = {b−1a : a, b ∈ E}. Thisshows that (a) implies (b). ��



(a) G is amenable;(b) for every finite subset E ⊂ G and every real number ε > 0, there exists a

nonempty finite subset F ⊂ G such that

|∂E(F )||F | < ε. (5.6)

Proof. Suppose first that G is amenable. It follows from the Tarski-Følnertheorem (Theorem 4.9.1) and Proposition 4.7.1 that there exists a rightFølner net (Fj)j∈J in G. By Proposition 5.4.4 we have that limj

|∂E(Fj)||Fj | = 0

for every finite subset E ⊂ G. Thus, given ε > 0 and a finite subset S ⊂ G,there exists j0 ∈ J such that |∂E(Fj)|

|Fj | < ε for all j ≥ j0. Taking F = Fj0 wededuce (5.6). This shows (a) ⇒ (b).

Conversely, suppose (b). Let J denote the set of all pairs (E, ε), where Eis a finite subset of G and ε > 0. We equip J with the partial ordering ≤defined by

(E, ε) ≤ (E′, ε′) ⇔ (E ⊂ E′ and ε′ ≤ ε).

Then J is a directed set. By (b), for every j = (E, ε) ∈ J , there exists anonempty finite subset Fj ⊂ G such that

|∂E(Fj)||Fj |

< ε. (5.7)

Let us show that

limj

|∂E(Fj)||Fj |

= 0 for every finite subset E ⊂ G. (5.8)

Fix a finite subset E0 ⊂ G and ε0 > 0. Let j ∈ J and suppose that j ≥j0, where j0 = (E0, ε0) ∈ J . By virtue of Proposition 5.4.2(vii) we have∂E0(Fj) ⊂ ∂E(Fj) so that, from (5.7), we deduce that

|∂E0(Fj)||Fj |

≤ |∂E(Fj)||Fj |

< ε ≤ ε0.

This shows (5.8). From Proposition 5.4.4 and Proposition 4.7.1 we deducethat G satisfies the Følner conditions. Thus G is amenable by virtue of theTarski-Følner theorem (Theorem 4.9.1). ��

5.5 Mutually Erasable Patterns 121

5.5 Mutually Erasable Patterns

In this section, we give a characterization of pre-injective cellular automatabased on the notion of mutually erasable patterns. This leads to an equiv-alent definition of pre-injectivity which is frequently used in the literature.However, the material contained in this section will not be used in the proofof the Garden of Eden theorem so that the reader who is only interested inthis proof may go directly to the next section.

Let G be a group and let A be a set.Let Z be a set and let f : AG → Z be a map. Two distinct patterns

p1, p2 : Ω → Z with the same support Ω ⊂ G are called mutually erasable(with respect to f) if they satisfy the following condition: if x1, x2 ∈ AG areconfigurations such that x1|Ω = p1, x2|Ω = p2 and x1|G\Ω = x2|G\Ω, thenf(x1) = f(x2).

Example 5.5.1. Let G = Z2 and A = {0, 1}. Consider the cellular automaton

τ : AG → AG associated with the Game of Life (see Example 1.4.3(a)). LetΩ = {−1, 0, 1}2 be the 3× 3 square in Z

2 centered at the origin and considerthe pattern p1 (resp. p2) with support Ω defined by p1(g) = 0 for all g ∈ Ω(resp. p2(g) = 1 if g = (0, 0) and p2(g) = 0 otherwise). Then it is clear thatp1 and p2 are mutually erasable patterns for τ (see Fig. 5.6).

Fig. 5.6 Two mutually erasable patterns for the Game of Life (recall that ◦ denotes a

dead cell, while • denotes a live cell)

Suppose that p1 and p2 are mutually erasable patterns for a map f : AG →Z. Let Ω denote their common support and consider two configurations x1

and x2 which coincide outside Ω and such that x1|Ω = p1 and x2|Ω = p2.Then x1 and x2 are almost equal and f(x1) = f(x2). On the other hand,x1 �= x2 since p1 �= p2 and therefore f is not pre-injective. This shows that apre-injective map f : AG → Z admits no mutually erasable patterns. It turnsout that for cellular automata, the converse is true:


Proposition 5.5.2. Let G be a group and let A be a set. Let τ : AG → AG bea cellular automaton. Then τ is pre-injective if and only if it does not admitmutually erasable patterns.

Proof. It remains only to show that if τ is not pre-injective then it admitsmutually erasable patterns. So let us assume that τ is not pre-injective. Thismeans that there exist configurations x1, x2 ∈ AG satisfying τ(x1) = τ(x2)such that Σ = {g ∈ G : x1(g) �= x2(g)} is a nonempty finite subset of G.Let S be a memory set for τ such that S = S−1 and 1G ∈ S. Consider thefinite sets S2 = {s1s2 : s1, s2 ∈ S} and Ω = Σ+S2

. Let us show that thepatterns p1 = x1|Ω and p2 = x2|Ω are mutually erasable. First observe thatp1 �= p2 since Σ ⊂ Ω. Suppose now that y1, y2 ∈ AG are two configurationscoinciding outside Ω such that y1|Ω = p1 and y2|Ω = p2. Then y1 and y2

coincide outside Σ since p1 and p2 coincide on Ω \ Σ. This implies

τ(y1)(g) = τ(y2)(g) for all g ∈ G \ Σ+S (5.9)

by Proposition 5.4.3. On the other hand, for i = 1, 2, the configurations yi

and xi coincide on Ω by construction. Thus, it follows from Proposition 5.4.3that τ(yi) and τ(xi) coincide on Ω−S . As τ(x1) = τ(x2), we deduce that theconfigurations τ(y1) and τ(y2) coincide on Ω−S . Now observe that Σ+S ⊂Ω−S . Indeed, if g ∈ Σ+S , that is, gs0 ∈ Σ for some s0 ∈ S, then gs ∈ Ω forall s ∈ S since gss−1s0 = gs0 ∈ gsS2 ∩ Σ. It follows that τ(y1) and τ(y2)coincide on Σ+S . Combining this with (5.9), we conclude that τ(y1) = τ(y2).This sows that p1 and p2 are mutually erasable patterns. ��

5.6 Tilings

Let G be a group.Let E and E′ be subsets of G. A subset T ⊂ G is called an (E, E′)-tiling

of G if the sets gE, g ∈ T are pairwise disjoint and if the sets gE′, g ∈ Tcover G. In other words, T ⊂ G is an (E, E′)-tiling if and only if the followingconditions are satisfied:

(T1) g1E ∩ g2E = ∅ for all g1, g2 ∈ T such that g1 �= g2;(T2) G =

⋃g∈T gE′.

Examples 5.6.1. (a) In the additive group R, the set Z is a ([0, 1[, [0, 1[)-tilingand the set [0, 1] is a (2Z, Z)-tiling.

(b) If G is a group and H is a subgroup of G, then every complete set ofrepresentatives for left cosets of G modulo H is an (H,H)-tiling.

Remark 5.6.2. Let G be a group and let E and E′ be subsets of G. If T is an(E, E′)-tiling of G and if E1 and E′

1 are subsets of G such that E1 ⊂ E andE′ ⊂ E′

1, then it is clear that T is also an (E1, E′1)-tiling of G.

5.6 Tilings 123

The Zorn lemma may be used to prove the existence of (E, E′)-tilings forany subset E of G and for E′ ⊂ G “large enough”. More precisely, we havethe following:

Proposition 5.6.3. Let G be a group and let E be a nonempty subset of G.Let E′ = {g1g

−12 : g1, g2 ∈ E}. Then there is an (E, E′)-tiling T ⊂ G.

Proof. Consider the set S consisting of all subsets S ⊂ G such that the sets(gE)g∈S are pairwise disjoint. Observe that S is not empty since {1G} ∈ S.On the other hand, the set S, partially ordered by inclusion, is inductive.Indeed, if S ′ is a totally ordered subset of S, then the set M =

⋃S∈S′ S

belongs to S and is an upper bound for S ′. By applying Zorn’s lemma, wededuce that S admits a maximal element T . The sets (gE)g∈T are pairwisedisjoint since T ∈ S. On the other hand, consider an arbitrary element h ∈ G.By maximality of T , we can find g ∈ T such that the set hE meets gE. Thisimplies h ∈ gE′. This shows that the sets (gE′)g∈T cover G. Consequently,T is an (E, E′)-tiling of G. ��

Proposition 5.6.4. Let G be an amenable group and let (Fj)j∈J be a rightFølner net for G. Let E and E′ be finite subsets of G and suppose that T ⊂ Gis an (E, E′)-tiling of G. Let us set, for each j ∈ J ,

Tj = T ∩ F−Ej = {g ∈ T : gE ⊂ Fj}.

Then there exist a real number α > 0 and an element j0 ∈ J such that

|Tj | ≥ α|Fj | for all j ≥ j0.

Proof. After possibly replacing E′ by E ∪ E′, we can assume that E ⊂ E′.Let us set T+

j = T ∩ F+E′

j = {g ∈ T : gE′ ∩ Fj �= ∅}. As the sets gE′,g ∈ T+

j , cover Fj , we have |Fj | ≤ |T+j | · |E′|, which gives

|T+j |

|Fj |≥ 1

|E′| (5.10)

for all j ∈ J . Observe now that

T+j \ Tj =

(T ∩ F+E′

j

)\

(T ∩ F−E

j

)

= T ∩ (F+E′

j \ F−Ej )

⊂ T ∩ (F+E′

j \ F−E′

j )⊂ T ∩ ∂E′(Fj)⊂ ∂E′(Fj)

where the first inclusion follows from E ⊂ E′. Thus we have

|∂E′(Fj)| ≥ |T+j | − |Tj |.


Using (5.10), we then deduce

|Tj ||Fj |

≥|T+

j ||Fj |

− |∂E′(Fj)||Fj |

≥ 1|E′| −

|∂E′(Fj)||Fj |

.

Hence we have|Tj ||Fj |

≥ α =1

2|E′|for j large enough, by Proposition 5.4.4 (see Fig. 5.7). ��

Fig. 5.7 The set T7 = T ∩F−E7 ⊂ Z

2, where E = {(0, 0), (1, 0)}, E′ = {0, 1}× {0, 1} and

T = (2Z + 1)× (2Z + 1) ⊂ Z2 is an (E, E′)-tiling in Z

2. Note that|T7||F7|

= 1264

= 316

≥ 216

=18

= 12|E′| = α

5.7 Entropy 125

5.7 Entropy

In this section, G is an amenable group, F = (Fj)j∈J is a right Følner netfor G, and A is a finite set.

For E ⊂ G, we denote by πE : AG → AE the canonical projection (restric-tion map). We thus have πE(x) = x|E for all x ∈ AG.

Definition 5.7.1. Let X ⊂ AG. The entropy entF (X) of X with respect tothe right Følner net F = (Fj)j∈J is defined by

entF (X) = lim supj

log |πFj (X)||Fj |

.

Here are some immediate properties of entropy.

Proposition 5.7.2. One has

(i) entF (AG) = log |A|;(ii) entF (X) ≤ entF (Y ) if X ⊂ Y ⊂ AG;(iii) entF (X) ≤ log |A| for all X ⊂ AG.

Proof. (i) If X = AG, then, for every j, we have πFj (X) = AFj and therefore

log |πFj (X)||Fj |

=log |A||Fj |

|Fj |=

|Fj | log |A||Fj |

= log |A|.

Thus we have entF (X) = log |A|.(ii) If X ⊂ Y , then πFj (X) ⊂ πFj (Y ) and hence |πFj (X)| ≤ |πFj (Y )| for

all j. This implies entF (X) ≤ entF (Y ).(iii) This follows immediately from (i) and (ii). ��

An important property of cellular automata is the fact that applying acellular automaton to a set of configurations cannot increase the entropy ofthe set. More precisely, we have the following:

Proposition 5.7.3. Let τ : AG → AG be a cellular automaton and let X ⊂AG. Then one has

entF (τ(X)) ≤ entF (X).

Proof. Let Y = τ(X). Let S ⊂ G be a memory set for τ . After replacingS by S ∪ {1G}, we can assume that 1G ∈ S. Let Ω be a finite subset of G.Observe first that τ induces a map

τΩ : πΩ(X) → πΩ−S (Y )

defined as follows. If u ∈ πΩ(X), then

τΩ(u) = (τ(x))|Ω−S ,

where x is an element of X such that x|Ω = u. Note that the fact that τΩ(u)does not depend on the choice of such an x follows from Proposition 5.4.3.


Clearly τΩ is surjective. Indeed, if v ∈ πΩ−S (Y ), then there exists x ∈ Xsuch that (τ(x))|Ω−S = v. Then, setting u = πΩ(x) we have, by construction,τΩ(u) = v. Therefore, we have

|πΩ−S (Y )| ≤ |πΩ(X)|. (5.11)

Observe now that Ω−S ⊂ Ω, since 1G ∈ S (cf. Proposition 5.4.2(iv)). ThusπΩ(Y ) ⊂ πΩ−S (Y ) × AΩ\Ω−S

. This implies

log |πΩ(Y )| ≤ log |πΩ−S (Y ) × AΩ\Ω−S

|= log |πΩ−S (Y )| + log |AΩ\Ω−S

|= log |πΩ−S (Y )| + |Ω \ Ω−S | log |A|≤ log |πΩ(X)| + |Ω \ Ω−S | log |A|,

by (5.11). As Ω \ Ω−S ⊂ ∂S(Ω), we deduce that

log |πΩ(Y )| ≤ log |πΩ(X)| + |∂S(Ω)| log |A|.

By taking Ω = Fj , this gives us

log |πFj (Y )||Fj |

≤log |πFj (X)|

|Fj |+

|∂S(Fj)||Fj |

log |A|.

Since

limj

|∂S(Fj)||Fj |

= 0

by Proposition 5.4.4, we finally get

entF (Y ) = lim supj


≤ lim supj

log |πFj (X)||Fj |

= entF (X).

��

It follows from Proposition 5.7.2 that the maximal value for the entropy ofa subset X ⊂ AG is log |A|. The following result gives a sufficient conditionon X which implies that its entropy is strictly less than log |A|.

Proposition 5.7.4. Let X ⊂ AG. Suppose that there exist finite subsets Eand E′ of G and an (E, E′)-tiling T ⊂ G such that πgE(X) � AgE for allg ∈ T . Then one has entF (X) < log |A|.

Proof. For each j ∈ J , let us define, as in Proposition 5.6.4 (see Fig. 5.7), thesubset Tj ⊂ T by Tj = T ∩ F−E

j = {g ∈ T : gE ⊂ Fj} and set

F ∗j = Fj \

∐

g∈Tj

gE

(see Fig. 5.8). By hypothesis, we have

|πgE(X)| ≤ |AgE | − 1 = |A||gE| − 1 for all g ∈ T. (5.12)

5.7 Entropy 127

Fig. 5.8 The set F ∗7 = F7 \

‘

g∈T7gE = F7 \T7E ⊂ Z

2, where E and T are as in Fig. 5.7(note that here F7 is the whole square, while F ∗

7 ⊂ F7 consists of its •-points)

AsπFj (X) ⊂ AF∗

j ×∏

g∈Tj

πgE(X),

we get

log |πFj (X)| ≤ log |AF∗j ×

∏

g∈Tj

πgE(X)|

= |F ∗j | log |A| +

∑

g∈Tj

log |πgE(X)|

≤ |F ∗j | log |A| +

∑

g∈Tj

log(|A||gE| − 1) (by (5.12))

= |F ∗j | log |A| +

∑

g∈Tj

|gE| log |A| +∑

g∈Tj

log(1 − |A|−|gE|)

= |Fj | log |A| + |Tj | log(1 − |A|−|E|),


since|Fj | = |F ∗

j | +∑

g∈Tj

|gE| and |gE| = |E|.

By setting c = − log(1 − |A|−|E|) (note that c > 0), this gives us

log |πFj (X)| ≤ |Fj | log |A| − c|Tj | for all j ∈ J.

Now, by Proposition 5.6.4, there exist α > 0 and j0 ∈ J such that |Tj | ≥ α|Fj |for all j ≥ j0. Thus

log |πFj (X)||Fj |

≤ log |A| − cα for all j ≥ j0.

This implies that

entF X = lim supj

log |πFj (X)||Fj |

≤ log |A| − cα < log |A|.

��

Recall from Sect. 1.1 that G acts on the left on AG by the shift (g, x) �→ gxdefined by gx(g′) = x(g−1g′) for g, g′ ∈ G and x ∈ AG.

Corollary 5.7.5. Let X be a G-invariant subset of AG. Suppose that thereexists a finite subset E ⊂ G such that πE(X) � AE. Then one has entF (X) <log |A|.

Proof. Let E′ = {g1g−12 : g1, g2 ∈ E}. By Proposition 5.6.3, we may find

an (E, E′)-tiling T ⊂ G. Since πE(X) � AE and X is G-invariant, we haveπgE(X) � AgE for all g ∈ G. This implies entF (X) < log |A| by Proposi-tion 5.7.4. ��

5.8 Proof of the Garden of Eden Theorem

The purpose of this section is to establish the following:

Theorem 5.8.1. Let G be an amenable group and let A be a finite set. LetF = (Fj)j∈J be a right Følner net for G. Let τ : AG → AG be a cellularautomaton. Then the following conditions are equivalent:

(a) τ is surjective;(b) entF (τ(AG)) = log |A|;(c) τ is pre-injective.

Note that this will prove the Garden of Eden theorem (Theorem 5.3.1)since the Garden of Eden theorem asserts the equivalence of conditions (a)and (c) in Theorem 5.8.1.

5.8 Proof of the Garden of Eden Theorem 129

We divide the proof of Theorem 5.8.1 into several lemmas. In these lemmas,it is assumed that the hypotheses of Theorem 5.8.1 are satisfied: G is anamenable group, F = (Fj)j∈J is a right Følner net for G, A is a finite set,and τ : AG → AG is a cellular automaton.

Lemma 5.8.2. Suppose that τ is not surjective. Then one has entF (τ(AG)) <log |A|.

Proof. By Proposition 5.1.1, τ admits a Garden of Eden pattern. This meansthat there is a finite subset E ⊂ G such that πE(τ(AG)) � AE . The set τ(AG)is G-invariant since τ is G-equivariant by Proposition 1.4.4. We deduce thatentF (τ(AG)) < log |A| by applying Corollary 5.7.5. ��

Lemma 5.8.3. Suppose that

ent(τ(AG)) < log |A|. (5.13)

Then τ is not pre-injective.

Proof. Let S be a memory set for τ such that 1G ∈ S. Let Y = τ(AG). Wehave F−S

j ⊂ Fj ⊂ F+Sj by Proposition 5.4.2(iv) and therefore F+S

j \ Fj ⊂∂S(Fj). As πF+S

j(Y ) ⊂ πFj (Y ) × AF+S

j \Fj , it follows that

log |πF+Sj

(Y )| ≤ log |πFj (Y )| + |F+Sj \ Fj | log |A|

≤ log |πFj (Y )| + |∂S(Fj)| log |A|.

This implies

log |πF+Sj

(Y )||Fj |

≤log |πFj (Y )|

|Fj |+

|∂S(Fj)||Fj |

log |A|. (5.14)

As

ent(Y ) = lim supj


< log |A|

by hypothesis, and

limj

|∂S(Fj)||Fj |

= 0

by Proposition 5.4.4, we deduce from inequality (5.14) that there exists j0 ∈ Jsuch that

log |πF+Sj0

(Y )||Fj0 |

< log |A|. (5.15)

Let us fix an arbitrary element a0 ∈ A and denote by Z the finite set ofconfigurations z ∈ AG such that z(g) = a0 for all g ∈ G\Fj0 . Inequality (5.15)gives us


|πF+Sj0

(Y )| < |A||Fj0 | = |Z|.

Observe that τ(z1) and τ(z2) coincide outside Fj+S0

for all z1, z2 ∈ Z. Thus

|τ(Z)| = |πFj+S0

(τ(Z))| ≤ |πFj+S0

(Y )| < |Z|.

This implies that we may find distinct configurations z1, z2 ∈ Z such thatτ(z1) and τ(z2). Since z1 and z2 coincide outside the finite set Fj0 , this showsthat τ is not pre-injective. ��

Lemma 5.8.4. Suppose that τ is not pre-injective. Then one has

entF (τ(AG)) < log |A|. (5.16)

Proof. Since τ is not pre-injective, we may find two configurations x1, x2 ∈AG satisfying τ(x1) = τ(x2) such that the set

Ω = {g ∈ G : x1(g) �= x2(g)}

is a nonempty finite subset of G. Observe that, for each h ∈ G, the config-urations hx1 and hx2 satisfy τ(hx1) = τ(hx2) (since τ is G-equivariant byProposition 1.4.4) and {g ∈ G : hx1(g) �= hx2(g)} = hΩ.

Let S be a memory set for τ such that 1G ∈ S. Then the set

R = {s−1s′ : s, s′ ∈ S}

is finite and we have 1G ∈ R. Let E = Ω+R. By Proposition 5.6.3, we mayfind a finite subset E′ ⊂ G and an (E, E′)-tiling T ⊂ G. Consider the subsetZ ⊂ AG consisting of all configurations z ∈ AG such that

z|hE �= (hx1)|hE for all h ∈ T.

Observe that, for each h ∈ T , we have

πhE(Z) � AhE

since (hx1)|hE /∈ πhE(Z). We deduce that entF (Z) < log |A| by applyingProposition 5.7.4. As entF (τ(Z)) ≤ entF (Z) by Proposition 5.7.3, this implies

entF (τ(Z)) < log |A|. (5.17)

Thus, to establish inequality (5.16), it suffices to prove that τ(AG) = τ(Z).To see this, consider an arbitrary configuration x ∈ AG and let us show thatthere is a configuration z ∈ Z such that τ(x) = τ(z). Let

T ′ = {h ∈ T : x|hE = (hx1)|hE}.

5.9 Surjunctivity of Locally Residually Amenable Groups 131

Let z ∈ AG be the configuration defined by

z(g) =

{hx2(g) if there is h ∈ T ′ such that g ∈ hE,

x(g) otherwise.

Notice that the configuration z is obtained from x by modifying the valuestaken by x only on the subsets of the form hΩ, where h ∈ T ′, (since, as wehave seen above, hx1 and hx2 coincide outside hΩ).

By construction, we have z ∈ Z.Let g ∈ G. Let us shows that τ(x)(g) = τ(z)(g).Suppose first that gS does not meet any of the sets hΩ, h ∈ T ′. Then

we have z|gS = x|gS . We deduce that τ(z)(g) = τ(x)(g) by applyingLemma 1.4.7.

Suppose now that there is an element h ∈ T ′ such that gS meets hΩ. Thismeans that there exists an element s0 ∈ S such that gs0 ∈ hΩ. For eachs ∈ S, we have gss−1s0 = gs0 ∈ hΩ. As s−1s0 ∈ R, this implies

gs ∈ (hΩ)+R = hΩ+R (by Proposition 5.4.2(viii))= hE.

We deduce that gS ⊂ hE. Thus we have τ(x)(g) = τ(hx1)(g) since x|hE =hx1|hE . Similarly, by applying Lemma 1.4.7, we get τ(z)(g) = τ(hx2)(g),since z and hx2 coincide on hE. As τ(hx1) = τ(hx2), we deduce thatτ(x)(g) = τ(z)(g).

Thus τ(z) = τ(x). This shows that τ(AG) = τ(Z) and completes the proofof the lemma. ��

Proof of Theorem 5.8.1. If τ is surjective, then τ(AG) = AG and henceentF (τ(AG)) = entF (AG) = log |A|. Thus (a) implies (b). Since the converseimplication follows from Lemma 5.8.2, we deduce that conditions (a) and (b)are equivalent. The fact that (c) implies (b) follows from Lemma 5.8.3 andthe converse implication follows from Lemma 5.8.4. Thus, conditions (b) and(c) are also equivalent. ��

5.9 Surjunctivity of Locally Residually AmenableGroups

The notion of a residually finite group was introduced in Chap. 2. Moregenerally, if P is a property of groups, a group G is called residually P iffor each element g ∈ G with g �= 1G, there exist a group Γ satisfying P andan epimorphism φ : G → Γ such that φ(g) �= 1Γ . Observe that every groupwhich satisfies P is residually P.


According to the preceding definition, a group G is called residuallyamenable if for each element g ∈ G with g �= 1G, there exist an amenablegroup Γ and an epimorphism φ : G → Γ such that φ(g) �= 1Γ . Note that,as every subgroup of an amenable group is amenable, it is not necessary torequire that the homomorphism φ is surjective in this definition. Observe alsothat every subgroup of a residually amenable group is residually amenableand that the fact that every finite group is amenable (Proposition 4.4.6)implies that every residually finite group is residually amenable.

Theorem 5.9.1. Every residually amenable group is surjunctive.

Let us first establish the following:

Lemma 5.9.2. Let G be a residually amenable group and let Ω be a finitesubset of G. Then there exist an amenable group Γ and a homomorphismρ : G → Γ such that the restriction of ρ to Ω is injective.

Proof. Consider the finite subset S ⊂ G defined by

S = {g−1h : g, h ∈ Ω and g �= h}.

Since G is residually amenable, we can find, for each s ∈ S, an amenablegroup Λs and a homomorphism φs : G → Λs such that φs(s) �= 1Λs . Let usshow that the group

Γ =∏

s∈S

Λs

and the homomorphism ρ : G → Γ given by

ρ =∏

s∈S

φs

have the required properties. The fact that the group Γ is amenable followsfrom Corollary 4.5.6. On the other hand, suppose that g and h are distinctelements of Ω. Then s = g−1h ∈ S and φs(g) �= φs(h) since (φs(g))−1φs(h) =φs(g−1h) = φs(s) �= 1Λs . This implies ρ(g) �= ρ(h). Therefore, the restrictionof ρ to Ω is injective. ��

Proof of Theorem 5.9.1. Since every injective cellular automaton is pre-injective, the Garden of Eden theorem (Theorem 5.3.1) implies that everyamenable group is surjunctive. By applying Lemma 3.3.4 and Lemma 5.9.2,it follows that every residually amenable group is surjunctive. ��

As an immediate consequence of Theorem 5.9.1 and Proposition 3.2.2, weobtain the following:

Corollary 5.9.3. Every locally residually amenable group is surjunctive. ��

5.11 A Pre-injective but Not Surjective Cellular Automaton over F2 133

5.10 A Surjective but Not Pre-injective CellularAutomaton over F2

The Garden of Eden theorem (Theorem 5.3.1) implies that every surjectivecellular automaton with finite alphabet over an amenable group is necessarilypre-injective. In this section, we give an example of a surjective but not pre-injective cellular automaton with finite alphabet over the free group F2.

Let G = F2 be the free group on two generators a and b. Let S ={a, b, a−1, b−1} and A = {0, 1}. Consider the cellular automaton τ : AG → AG

defined by

τ(x)(g) =

⎧⎪⎨

⎪⎩

1 if∑

s∈S x(gs) > 2,

0 if∑

s∈S x(gs) < 2,

x(g) if∑

s∈S x(gs) = 2.

Thus τ is the majority action automaton associated with G and S (see Ex-ample 1.4.3(c)). We have already seen in Example 5.2.1(c) that τ is notpre-injective.

Let us show that τ is surjective. Let y ∈ AG be an arbitrary configuration.We may construct a configuration x ∈ AG such that y = τ(x) in the followingway. Consider the map ψ : G\{1G} → G which associates to each g ∈ G\{1G}the element of G obtained by suppressing the last factor in the reduced formof g. Thus if g = s1s2 . . . sn with si ∈ S for 1 ≤ i ≤ n and sisi+1 �= 1G for1 ≤ i ≤ n − 1, then ψ(g) = s1s2 . . . sn−1. Let x ∈ AG be the configurationdefined by x(g) = y(ψ(g)) for all g ∈ G \ {1G} and x(1G) = 0. Then, for eachg ∈ G, the configuration x takes the value y(g) at (at least) three of the fourelements gs, s ∈ S. it follows that τ(x) = y. Thus τ is surjective.

More generally, let G be a group containing a free subgroup H based ontwo elements a and b. Let τ : {0, 1}G → {0, 1}G be the majority action cellu-lar automaton over G associated with the set S = {a, b, a−1, b−1}. We haveseen in Example 5.2.1(c) that τ is not pre-injective. Consider its restrictionτH : {0, 1}H → {0, 1}H . Note that τH is the majority action cellular automa-ton over H associated with S. We have seen above that τH is surjective. Itfollows from Proposition 1.7.4(ii) that τ is also surjective. Thus, every groupcontaining a free subgroup of rank two admits a cellular automaton with fi-nite alphabet which is surjective but not pre-injective. This will be extendedto all non-amenable groups in Sect. 5.12.

5.11 A Pre-injective but Not Surjective CellularAutomaton over F2

It follows from the Garden of Eden theorem (Theorem 5.3.1) that every pre-injective cellular automaton with finite alphabet over an amenable group is


necessarily surjective. In this section, we give examples of cellular automatawith finite alphabet over the free group F2 which are pre-injective but notsurjective.

Let G = F2 be the free group on two generators a and b. Let H be anontrivial abelian group. Our alphabet will be the group A = H×H. We shalluse additive notation for the group operations on H and A. Let p1, p2 : A → Abe the group endomorphisms defined respectively by p1(u) = (h1, 0) andp2(u) = (h2, 0) for all u = (h1, h2) ∈ A. Let us equip the Cartesian productAG =

∏g∈G A with its natural Abelian group structure. Consider the map

τ : AG → AG given by

τ(x)(g) = p1(x(ga)) + p2(x(gb)) + p1(x(ga−1)) + p2(x(gb−1)) (5.18)

for all g ∈ G and x ∈ AG.It is clear that τ is a cellular automaton over the group G and the alphabet

A with memory set S = {a, b, a−1, b−1} and a group endomorphism of AG.

Proposition 5.11.1. The cellular automaton τ : AG → AG defined by (5.18)is pre-injective but not surjective.

Proof. The image of τ is contained in (H × {0})G� (H × H)G = AG. Thus

τ is not surjective.Let us show that τ is pre-injective. Suppose not. Then there exist con-

figurations x1, x2 ∈ AG satisfying τ(x1) = τ(x2) such that the set Ω ={g ∈ G : x1(g) �= x2(g)} is a nonempty finite subset of G. The configura-tion x0 = x1 − x2 ∈ AG satisfies τ(x0) = τ(x1) − τ(x2) = 0 and one hasΩ = {g ∈ G : x0(g) �= 0}. Consider an element g0 ∈ Ω whose reduced formhas maximal length, say n0. Then x0(g0) is a nonzero element (h0, k0) of A.

If h0 �= 0, take s0 ∈ {a, a−1} such that the reduced form of g0s0 has lengthn0 + 1. Then, for each s ∈ S \ {s−1

0 }, the length of the reduced form of g0s0sis n0 + 2 and hence x(g0s0s) = 0. By applying 5.18, we deduce that

τ(x0)(g0s0) = p1(x0(g0)) = (h0, 0) �= 0,

which contradicts the fact that τ(x0) = 0.If h0 = 0, then k0 �= 0 and we proceed similarly by taking s0 ∈ {b, b−1}

such that the reduced form of g0s0 has length n0 + 1. This gives us

τ(x0)(g0s0) = p2(x0(g0)) = (k0, 0) �= 0,

which yields again a contradiction. This shows that τ is pre-injective. ��

If we take for H a finite abelian group of cardinality |H| = n ≥ 2 (e.g.,the group H = Z/nZ), this gives us a pre-injective but not surjective cellularautomaton over F2 whose alphabet is finite of cardinality n2.

From Proposition 1.7.4 we deduce the following:

5.12 A Characterization of Amenability in Terms of Cellular Automata 135

Proposition 5.11.2. Let G be a group containing a free subgroup of ranktwo. Then there exist a finite set A and a cellular automaton τ : AG → AG

which is pre-injective but not surjective. ��

5.12 A Characterization of Amenability in Terms ofCellular Automata

In Sect. 5.10, we gave an example of a cellular automaton with finite alphabetover the free group F2 which is surjective but not pre-injective. In fact, theexistence of such an automaton holds for any non-amenable group:

Theorem 5.12.1. Let G be a non-amenable group. Then there exists a finiteset A and a cellular automaton τ : AG → AG which is surjective but notpre-injective.

Proof. Since G is non-amenable, it follows from Theorem 4.9.2 that thereexist a 2-to-one surjective map ϕ : G → G and a finite subset S ⊂ G suchthat

(ϕ(g))−1g ∈ S for all g ∈ G. (5.19)

Our alphabet will be the Cartesian product A = S ×S. Let us fix some totalorder ≤ on S and an arbitrary element s0 ∈ S. Define the map μ : AS → Aby

μ(y) =

⎧⎪⎨

⎪⎩

(s′, t′) if there exists a unique element (s, t) ∈ S × S with s < t

such that y(s) = (s, s′) and y(t) = (t, t′), where s′, t′ ∈ S,

(s0, s0) otherwise,(5.20)

for all y ∈ AS .Let us show that the cellular automaton τ : AG → AG with memory set S

and local defining map μ has the required properties.We fist observe that S has at least two elements since otherwise (5.19)

would imply that ϕ is bijective. Let s1 ∈ S such that s1 �= s0. Consider theconfigurations x0, x1 ∈ AG, where x0 is defined by x0(g) = (s0, s0) for allg ∈ G, and x1 is defined by x1(g) = (s0, s0) if g �= 1G and x1(1G) = (s0, s1).The configurations x0 and x1 are almost equal since they differ only at 1G.On the other hand, it is clear that x0 and x1 have the same image, namelyx0, by τ . Thus τ is not pre-injective.

We use the properties of ϕ to prove that τ is surjective. Let x ∈ AG bean arbitrary configuration. Let us show that there is a configuration z ∈ AG

such that x = τ(z). We construct z in the following way. Let u : G → S andv : G → S be the maps defined by x(g) = (u(g), v(g)) for all g ∈ G. Foreach g ∈ G, there are exactly two elements sg, tg ∈ S such that sg < tg andϕ(gsg) = ϕ(gtg) = g. Let us set z(gsg) = (sg, u(g)) and z(gtg) = (tg, v(g)).


Observe that z : G → A is well defined and that the value of z at g ∈ Gis either ((ϕ(g))−1g, u(ϕ(g)) or ((ϕ(g))−1g, v(ϕ(g)). It immediately followsfrom the definition of τ that x = τ(z). This shows that τ is surjective. ��

Combining Theorem 5.12.1 with Theorem 5.3.1, we obtain the followingcharacterization of amenable groups in terms of cellular automata:


(a) G is amenable;(b) every surjective cellular automaton with finite alphabet over G is pre-

injective.��

5.13 Garden of Eden Patterns for Life

Let τ : {0, 1}Z2 → {0, 1}Z

2denote the cellular automaton associated with the

Game of Life (see Example 1.4.3(a)). As it was observed in Example 5.3.2(b),it follows from the Garden of Eden theorem that τ is not surjective. Thus,Proposition 5.1.1 implies that τ admits Garden of Eden patterns. The purposeof this section is to present a direct proof of the existence of such patterns.This construction will provide a concrete illustration of the ideas underly-ing the proof that surjectivity implies pre-injectivity in the Garden of Edentheorem.

Given an integer n ≥ 1, one says that a subset Ω ⊂ Z2 is a square of size

n × n in Z2 if there exist p, q ∈ Z such that

Ω = {p, p + 1, . . . , p + n − 1} × {q, q + 1, . . . , q + n − 1}.

Proposition 5.13.1. Let τ : {0, 1}Z2 → {0, 1}Z

2be the cellular automaton

associated with the Game of Life. Then every square Ω ⊂ Z2 of size n × n

with n ≥ 3 × 109 is the support of a Garden of Eden pattern for τ .

Proof. Let n ≥ 1 be an integer. Consider a square Cn ⊂ Z2 of size 5n × 5n.

Let Dn ⊂ Cn be the square of size (5n− 2)× (5n− 2) which is the S-interiorof Dn for S = {−1, 0, 1}2 ⊂ Z

2. Thus Dn is obtained from Cn by removingthe 20n − 4 points of Z

2 located on the (usual) boundary of Cn.Let us set Xn = {0, 1}Cn and Yn = {0, 1}Dn . The map τ induces a map

τn : Xn → Yn defined as follows. If u ∈ Xn, we set τn(u) = (τ(x))|Dn , wherex ∈ {0, 1}Z

2satisfies x|Cn = u (the fact that (τ(x))|Dn does not depend of

the choice of x follows from Proposition 5.4.3 since S is a memory set for τand Dn is the S-interior of Cn). We have

|Yn| = 2(5n−2)2 = 225n2−20n+4.

Let us divide the square Cn into n2 squares of size 5 × 5 (see Fig. 5.10).

5.13 Garden of Eden Patterns for Life 137

Fig. 5.9 A square Cn ⊂ Z2 of size 5n × 5n and its S-interior Dn ⊂ Cn, a square of size

(5n − 2) × (5n − 2) where S = {−1, 0, 1}2 ⊂ Z2; here n = 3

There are 225 maps from each square of size 5 × 5 to the set {0, 1}. Nowobserve that if the restriction of an element u ∈ Xn to one of these 5 × 5squares is identically 0, then we may replace the 0 value at the center of this5 × 5 square by 1 without changing τn(u). We deduce that

|τn(Xn)| ≤ (225 − 1)n2= (2log2(2

25−1))n2= 2(25+log2(1−2−25))n2

.

Therefore we have |τn(Xn)| < |Yn| if

(25 + log2(1 − 2−25))n2 < 25n2 − 20n + 4.

This inequality is equivalent to

−n2 log2(1 − 2−25) − 20n + 4 > 0,

which is verified if and only if

n >2

− log2(1 − 2−25)(5 +

√25 + log2(1 − 2−25)),

that is, if and only ifn ≥ 465 163 744.


Fig. 5.10 The square Cn ⊂ Z2 of size 5n×5n is divided into n2 squares of size 5×5; here,

as in Fig. 5.9, n = 3

For such values of n, the map τn is not surjective, i.e., there exists a Gardenof Eden pattern whose support is Dn. This shows the existence of a Gardenof Eden pattern of size

2 325 818 718 × 2 325 818 718.��

Notes

According to M. Gardner (see [Gar-2, p.230]), it was J. Tuckey who intro-duced the term “Garden of Eden” in the theory of cellular automata. Thefirst contribution to the Garden of Eden theorem goes back to E.F. Moore[Moo] who proved that every surjective cellular automaton with finite alpha-bet over Z

2 is pre-injective. The converse implication was established shortlyafter by J. Myhill [Myh]. This is the reason why the Garden of Eden theo-rem for Z

2 is often referred to as the Moore-Myhill theorem. The next stepin the proof of the Garden of Eden theorem was done by A. Machı andF. Mignosi [MaM] who extended it to finitely generated groups of subexpo-nential growth (see Chap. 6 for the definition of finitely generated groups ofsubexponential growth). Then, the Garden of Eden theorem was proved for

Exercises 139

all finitely generated amenable groups by A. Machı, F. Scarabotti and thefirst author [CMS1]. The general case may be reduced to the case of finitelygenerated groups by considering the restriction of the cellular automaton tothe subgroup generated by a memory set and applying Proposition 1.7.4(ii)and Proposition 5.2.2 (see [CeC8]). The proof based on Følner nets which ispresented in this chapter is more direct.

The term “pre-injective” was introduced by M. Gromov in the appendixof [Gro5].

It follows from a result due to D.S. Ornstein and B. Weiss (see [OrW],[Gro6], [Kri]) that the lim sup appearing in the definition of entropy (Defini-tion 5.7.1) is in fact a true limit and is independent of the particular choiceof the right Følner net F in the group.

Versions of the Garden of Eden theorem for cellular automata over certainclasses of subshifts (closed invariant subsets of the full shift) may be foundin the appendix of [Gro5] and in two papers of F. Fiorenzi [Fio1], [Fio2]. Seealso [CFS].

The first examples of pre-injective (resp. surjective) cellular automatawhich are not surjective (resp. not pre-injective) where described by D.E.Muller (unpublished class notes). The underlying group was the modu-lar group G= PSL(2, Z) = (Z/2Z) ∗ (Z/3Z) (note that G contains the freegroup F2).

Theorem 5.12.1 is due to L. Bartholdi [Bar].The computation in Sect. 5.13 is taken from [BCG, Page 828]. The size

of the corresponding Garden of Eden pattern is far from being optimal. Thefirst explicit example of a Garden of Eden pattern for the cellular automatonassociated with Conway’s Game of Life was found by R. Banks in 1971.Banks’ pattern is supported by a rectangle of size 33 × 9 and has 226 alivecells. The smallest known Garden of Eden pattern for Life, found by N.Beluchenko on September 2009, has as support a square 11 × 11 and bears69 alive cells. It was proved that there exist no Garden of Eden patterns forLife with support contained in a rectangle of size 6 × 5.

Exercises

5.1. Let G be a group and let A be a finite set. Let τ : AG → AG be a cellularautomaton admitting a memory set M ⊂ G such that |M | = 1. Show thatthe following conditions are equivalent:

(i) τ is pre-injective;(ii) τ is injective;(iii) τ is surjective.

5.2. Let G be a group and let A be a finite set. Let τ : AG → AG be a non-surjective cellular automaton. Show that there are uncountably many Gardenof Eden configurations in AG.


5.3. Life on Z. Let A = {0, 1} and consider the cellular automaton τ : AZ →AZ with memory set S = {−1, 0, 1} and local defining map μ : AS → A givenby

μ(y) =

{1 if

∑s∈S y(s) = 2

0 otherwise

for all y ∈ AS .(a) Show that τ is not pre-injective.(b) Deduce from (a) that τ is not surjective either. Hint: Use Theorem 4.6.1

and the Garden of Eden Theorem (cf. Exemple 5.3.2(c)).(c) It follows from Proposition 5.1.1 that τ admits a Garden of Eden

pattern. Check that the map p : {0, 1, 2, 3, 4, 5, 6, 7, 8} → A defined by p(0) =p(1) = p(3) = p(5) = p(8) = 1 and p(2) = p(4) = p(6) = p(7) = 0 is a Gardenof Eden pattern for τ .

5.4. Let G = Z, A = {0, 1} and let τ : AG → AG be the majority actioncellular automaton (cf. Example 1.4.3(c)). We have seen in Example 5.2.1(c)that τ is not pre-injective so that, by amenability of the group G (cf. The-orem 4.6.1) and the Garden of Eden Thoerem, τ is not surjective either (cf.Exemple 5.3.2(c)). It follows from Proposition 5.1.1 that τ admits a Gar-den of Eden pattern. Check that the map p : {1, 2, 3, 4, 5} → A defined byp(1) = p(3) = p(4) = 0 and p(2) = p(5) = 1 is a Garden of Eden patternfor τ .

5.5. Let (A,+) be an abelian group (not necessarily finite). Let τ : AZ → AZ

be the cellular automaton defined by τ(x)(n) = x(n − 1) + x(n) + x(n + 1)for all x ∈ AZ and n ∈ Z. Show that τ is surjective. Hint: Given an arbitraryconfiguration y ∈ AZ, construct a configuration x ∈ AZ such that x(0) =x(1) = 0A and τ(x) = y.

5.6. Take G = Z and A = Z/2Z = {0, 1}. Let x1 ∈ AG be the configurationdefined by x1(n) = 1 for all n ∈ Z. Let x2 ∈ AG be the configuration definedby x2(0) = 1 and x2(n) = 0 for all n ∈ Z \ {0}. Let f : AG → AG be the mapdefined by f(x2) = x1 and f(x) = x for all x ∈ AG \ {x2}. Let g : AG → AG

be the map defined by g(x)(n) = x(n + 1) + x(n) for all n ∈ Z and x ∈ AG.Verify that the maps f and g are both pre-injective but that the map g ◦ fis not pre-injective.

5.7. Let G be group and let A be a set. Suppose that σ : AG → AG andτ : AG → AG are pre-injective cellular automata. Show that the cellularautomaton τ ◦ σ is pre-injective.

5.8. Let G be a locally finite group and let A be a set. Show that everypre-injective cellular automaton τ : AG → AG is injective.

5.9. Give a direct proof of the Garden of Eden theorem (Theorem 5.3.1) inthe case when the group G is locally finite.

Exercises 141

5.10. Life in a tree. Let G = F4 denote the free group based on four genera-tors a, b, c, d. Let A = {0, 1}. Consider the cellular automaton τ : AG → AG

with memory set

S = {1G, a, b, c, d, a−1, b−1, c−1, d−1}

and local defining map μ : AS → A given by

μ(y) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

1 if

⎧⎪⎨

⎪⎩

∑s∈S y(s) = 3

or∑s∈S y(s) = 4 and y((0, 0)) = 1,

0 otherwise

for all y ∈ AS (cf. Example 1.4.3(a)). Show that τ is surjective but notpre-injective.

5.11. Let G be a group. Let E, F and Ω be subsets of G.(a) Show that (Ω−E)−F = Ω−FE and (Ω+E)+F = Ω+FE .(b) Let G = Z, E = {−2,−1, 0, 1, 2}, F = {−1, 0, 1} and Ω = {−10,−9, . . . ,

−1, 0, 1, . . . , 9, 10}. Check that ∂F (∂E(Ω)) = {−13,−12,−9,−8, 8, 9, 12, 13}and ∂FE(Ω) = {−13,−12,−11,−10,−9,−8, 8, 9, 10, 11, 12, 13}. Deduce that,in general, one has ∂F (∂E(Ω)) �= ∂EF (Ω).

5.12. Let G be a group and let E ⊂ G. Determine the sets ∂E(∅) and ∂E(G).

5.13. Let G be a group. Let E and Ω be subsets of G. Show that ∂E(G\Ω) =∂E(Ω).

5.14. Let G be a group. Let E, Ω1 and Ω2 be subsets of G. Show that∂E(Ω1 ∪ Ω2) ⊂ ∂E(Ω1) ∪ ∂E(Ω2). Give an example showing that one mayhave ∂E(Ω1 ∪ Ω2) �= ∂E(Ω1) ∪ ∂E(Ω2).

5.15. Let G be a group. Let E, Ω1 and Ω2 be subsets of G. Show that∂E(Ω1 \ Ω2) ⊂ ∂E(Ω1) ∪ ∂E(Ω2). Give an example showing that one mayhave ∂E(Ω1 \ Ω2) �= ∂E(Ω1) ∪ ∂E(Ω2).

5.16. Let G be a group. Let E and Ω be subsets of G. Show that Ω−gE =Ω−Eg−1, Ω+gE = Ω+Eg−1, and ∂gE(Ω) = ∂E(Ω)g−1 for all g ∈ G.

5.17. Let G be a group. Show that a subset R ⊂ G is syndetic (cf. Exer-cise 3.39) if and only if there exists a finite subset S ⊂ G such that the setΩ = R−1 satisfies Ω+S = G.

5.18. Let G be a group. Let E and Ω be subsets of G. Show that Ω ⊂(Ω+E−1

)−E .

5.19. Let G be a group and let A be a set. Let τ : AG → AG be a cellularautomaton with memory set M . Let Ω be a subset of G. Show that if twoconfigurations x1, x2 ∈ AG coincide on Ω+M−1

then the configurations τ(x1)and τ(x2) coincide on Ω.


5.20. Let (A,+) be a finite abelian group and let ϕ : A × A → A be a map.Let τ : AZ → AZ be the map defined by τ(x)(n) = x(n)+ϕ(x(n+1), x(n+2))for all n ∈ Z and x ∈ AZ.

(a) Show that τ is a cellular automaton over the group Z and the alpha-bet A.

(b) Show that τ is pre-injective.(c) Show that τ is surjective.

5.21. Let G = Z and A = {0, 1}. Verify that, among the 16 cellular automataτ : AZ → AZ with memory set S = {0, 1} ⊂ Z, there are exactly 6 of themwhich are surjective, 4 of them which are injective, 4 of them which arebijective, and 10 of them which are neither surjective nor injective. Hint: TheGarden of Eden Theorem may help.

5.22. Topological entropy. Let G be an amenable group and let F = (Fj)j∈J

be a right Følner net for G. If X is a compact topological space equipped witha continuous action of G, the topological entropy hF (X,G), 0 ≤ hF (X,G) ≤∞, of X is defined as follows. Suppose that U = (Ui)i∈I is an open coverof X. We denote by N(U) the smallest integer n ≥ 0 such that there is afinite subset I0 ⊂ I of cardinality n such that

⋃i∈I0

Ui = X (note that thereexists such a finite subset I0 ⊂ I by compactness of X). Given a nonemptysubset F ⊂ G, we define the open cover UF = (Wα)α∈IF indexed by the setIF = {α : F → I} by setting

Wα =⋂

g∈F

gUα(g)

for all α ∈ IF . Finally, we set

hF (U) = lim supj

log N(UFj )|Fj |

andhF (X,G) = sup

UhF (U),

where U ranges over all open covers of X.Observe that it is clear from this definition that if X and Y are two

compact topological spaces, each equipped with a continuous action of G,such that there exists a G-equivariant homeomorphism f : X → Y , then onehas hF (X,G) = hF (Y,G).

(a) Let X be a compact space and let U = (Ui)i∈I and V = (Vk)k∈K betwo open covers of X. One says that the open cover V is finer than U if, foreach k ∈ K, there exists i ∈ I such that Vk ⊂ Ui. Show that if the open coverV is finer than the open cover U then one has N(U) ≤ N(V).

(b) Let X be a compact space and let U = (Ui)i∈I be an open cover of Xwhich forms a partition of X (i.e., such that Ui1 ∩ Ui2 = ∅ for all i1, i2 ∈ Iwith i1 �= i2). Show that N(U) = |{i ∈ I : Ui �= ∅}|.

Exercises 143

(c) Let A be a finite set and let X ⊂ AG be a subshift. Show that if X isequipped with the action of G induced by the G-shift on AG, then one has

hF (X,G) = entF (X).

Hint: Consider the open cover T = (Ta)a∈A of X defined by Ta = {x ∈ X :x(1G) = a} and deduce from (b) that N(TFj ) = |πFj (X)|, where πFj : AG →AFj denotes the projection map. This gives hF (X,G) ≥ hF (T ) = entF (X).To prove hF (X,G) ≤ entF (X), observe that if U = (Ui)i∈I is an arbitraryopen cover of X, then there exists a finite subset Ω ⊂ G such that the opencover TΩ is finer than U . Apply (a) to get N(UFj ) ≤ |πFjΩ(X)| and finallyuse the fact that F is a right Følner net to conclude.

Note: It can be shown by using a result due to Ornstein and Weiss (cf.the notes above) that the topological entropy hF (X,G) of a compact spaceX equipped with a continuous action of an amenable group G is in factindependent of the choice of the right Følner net F .

5.23. Let G be an amenable group and let F = (Fj)j∈J be a right Følnernet for G. Let A and B be finite sets. Suppose that X ⊂ AG and Y ⊂ BG

are two subshifts such that there exists a bijective continuous G-equivariantmap ϕ : X → Y . Show that entF (X) = entF (Y ). Hint: Use Exercise 5.22.

5.24. Let G be an amenable group, F a right Følner net for G, and A a finiteset. Let X be a subset of AG and let X denote the closure of X in AG forthe prodiscrete topology. Show that one has entF (X) = entF (X).

5.25. Let G be an amenable group and let (Fj)j∈J be a right Følner net forG. Let A be a finite set. Suppose that X (resp. Y ) is a subset of AG havingat least two distinct elements. Show that entF (X∪Y ) ≤ entF (X)+entF (Y ).

5.26. Let G be an amenable group, A a finite set, F be a right Følner netfor G, and X ⊂ AG a subshift. Let F be a nonempty finite subset of G andconsider the subshift X [F ] ⊂ BG, where B = AF (cf. Exercise 1.34). Showthat entF (X [F ]) = entF (X).

5.27. Let G be a group, A be a set and X ⊂ AG a subshift. Let also H ⊂ Gbe a subgroup of G and T ⊂ G a complete set of representatives for the rightcosets of H in G, and consider the subshift X(H,T ) ⊂ BH , where B = AH\G

(cf. Exercise 1.35). Suppose that G and H are isomorphic and denote byφ : G → H an isomorphism. Let F = (Fj)j∈J be a right Følner net in G sothat F ′ = (φ(Fj))j∈J is a right Følner net in G. Show that entF (X(H,T )) =entF ′(X).

5.28. Let G = Z and A = {0, 1}. Set Fn = {0, 1, . . . , n − 1} and recallfrom Example 4.7.4(b) that F = (Fn)n∈N is a Følner sequence in Z. Showthat if X ⊂ AZ is either the even subshift considered in Exercise 1.38 orthe golden mean subshift considered in Exercise 1.39 then entF (X) = log ϕ,where ϕ = 1+

√5

2 is the golden number.


5.29. Entropy of the Morse subshift. Let A = {0, 1} and let x ∈ AN andX ⊂ AZ denote the Thue-Morse sequence (cf. Exercises 3.41) and the Morsesubshift (cf. Exercise 3.42) respectively.

(a) Show that |Ln(X)| ≤ 8n for all n ≥ 1. Hint: Let n ≥ 1 and denoteby k the unique integer such that 2k−1 ≤ n < 2k. Consider the words u =x(0)x(1) · · ·x(2k − 1) and v = ι(u) (cf. Exercise 3.41(e)). Observe that anyword w ∈ Ln(X) is necessarily a subword of one of the words uv, vv, vu, oruu. Altogether, this gives at most 4 · 2k = 8 · 2k−1 ≤ 8n distinct possibilitiesfor w.

(b) Let F = (Fn)n≥1 denote the Følner sequence of Z where Fn ={0, 1, . . . , n − 1}. Deduce from (a) that entF (X) = 0.

5.30. Let A = {0, 1, 2} and set Y = {y ∈ AZ : y(g) ∈ {1, 2} for all g ∈ Z}and X = Y ∪{x0}, where x0 ∈ AZ denotes the constant configuration definedby x0(g) = 0 for all g ∈ Z.

(a) Show that X and Y are subshifts of finite type of AZ and that one hasthe strict inclusion Y � X.

(b) Show that X is not irreducible.(c) Check that X and Y have the same entropy entF (X) = entF (Y ) =

log 2 with respect to the Følner sequence F = (Fn)n∈N of Z given by Fn ={0, 1, . . . , n − 1}.

5.31. Let G be an amenable group, A a finite set, X ⊂ AG a strongly irre-ducible subshift, and Y ⊂ AG a nonempty subshift such that Y � X. Letalso F = (Fj)j∈J be a right Følner net in G. For a subset Ω ⊂ G we denoteby πΩ : AG → AΩ the projection (restriction) map.

(a) Show that there exists a finite subset Ω0 ⊂ G such that πΩ0(Y ) �

πΩ0(X).(b) Let Δ ⊂ G be a finite subset such that 1G ∈ Δ and X is Δ-irreducible.

Let E = Ω+Δ0 and set ξ = |πE(X)|−1 and E′ = EE−1 = {ab−1 : a, b ∈ E}.

By virtue of Proposition 5.6.3 we can find an (E, E′)-tiling T ⊂ G. For j ∈ Jset Tj = {g ∈ T : gE ⊂ Fj}. Show that ξ|πFj (X)| ≤ |πFj\gE(X)| for allg ∈ Tj .

(c) Let p ∈ πΩ0(X) \ πΩ0(Y ) (cf. (a)) and let x ∈ X such that x|Ω0 = p.For g ∈ G set pg = (gx)|gΩ0 ∈ AgΩ0 . Show that pg ∈ πgΩ0(X) \ πgΩ0(Y ).

(d) For j ∈ J and g ∈ Tj denote by πFj

gΩ0: πFj (X) → πgΩ0(X) the projec-

tion (restriction) map and let pg ∈ πgΩ0(X)\πgΩ0(Y ) (cf. (c)). Using (b) andthe Δ-irreducibility of X show that |πFj (X)\(πFj

gΩ0)−1(pg)| ≤ (1−ξ)|πFj (X)|

for all j ∈ J and g ∈ Tj .(e) For j ∈ J denote by πFj (X)∗ the set of patterns p ∈ πFj (X) such

that πFj

gΩ0(p) �= pg for all g ∈ Tj . Observe that πFj (X)∗ = πFj (X) \

⋃g∈Tj

(πFj

gΩ0)−1(pg) and using the Δ-irreducibility of X and an inductive ar-

gument based on (d), show that

|πFj (X)∗| ≤ (1 − ξ)|Tj ||πFj (X)|.

Exercises 145

(f) Observe that πFj (Y ) ⊂ πFj (X)∗ and deduce from (e) that log |πFj (Y )| ≤|Tj | log(1 − ξ) + log |πFj (X)|.

(g) By Proposition 5.6.4 there exists a real number α > 0 and an elementj0 ∈ J such that |Tj | ≥ α|Fj | for all j ≥ j0. Deduce from (f) that entF (Y ) ≤α log(1 − ξ) + entF (X).

(h) Deduce from (g) that entF (Y ) < entF (X).

5.32. Let G be an amenable group and let A be a finite set. Let F = (Fj)j∈J

be a right Følner net in G. Let X ⊂ AG be a nonempty strongly irreduciblesubshift. Show that if X is not minimal then one has entF (X) > 0. Hint:Show that, with the notation introduced in Exercise 5.31, one has entF (X) ≥α log 2.

5.33. Let G be a group and let A be a finite set. Let Δ be a finite subsetof G and let X ⊂ AG be a Δ-irreducible subshift. Suppose that (Ωi)i∈I is a(possibly infinite) family of (possibly infinite) subsets of G such that

Ω+Δi ∩

⎛

⎝⋃

k∈I\{i}Ωk

⎞

⎠ = ∅ for all i ∈ I.

Let also (xi)i∈I be a family of configurations in X.Denote by Pf (G) the set of all finite subsets of G. For each Λ ∈ Pf (G) let

X(Λ) ⊂ X denote the set consisting of all configurations in X which coincidewith xi on Λ ∩ Ωi for all i ∈ I.

(a) Show that X(Λ) is closed in X for each Λ ∈ Pf (G).(b) Show that if we fix Λ ∈ Pf (G), then the subsets Ψi = Λ ∩ Ωi are all

contained in Λ and satisfy

Ψ+Δi ∩

⎛

⎝⋃

k∈I\{i}Ψk

⎞

⎠ = ∅ for all i ∈ I.

(c) Fix Λ ∈ Pf (G) and set IΛ = {i ∈ I : Λ ∩ Ωi �= ∅}. Show that IΛ

is finite. Then, applying induction on the cardinality of IΛ, show that, byΔ-irreducibility of X, one has X(Λ) �= ∅.

(d) Show that X(Λ1)∩X(Λ2)∩ · · · ∩X(Λn) = X(Λ1 ∪Λ2 ∪ · · · ∪Λn) anddeduce that X(Λ1)∩X(Λ2)∩· · ·∩X(Λn) �= ∅ for all Λ1, Λ2, . . . , Λn ∈ Pf (G).

(e) Deduce from (d) that the family (X(Λ))Λ∈Pf (G) has a nonempty in-tersection.

(f) Let x ∈ X be such that x ∈ X(Λ) for each finite subset Λ ⊂ G (whoseexistence is guaranteed by (e)). Show that x coincides with xi on Ωi for alli ∈ I.

5.34. Let G be an amenable group and let A be a finite set. Let F = (Fj)j∈J

be a right Følner net in G. Let X ⊂ AG be a (possibly minimal) stronglyirreducible subshift containing at least two distinct configurations x0 and x1.


(a) Show that there exists a finite subset D ⊂ G such that (gx0)|gD �=(gx1)|gD for all g ∈ G.

(b) Let Δ be a finite subset of G such that X is Δ-irreducible and 1G ∈ Δ.Set E = D+Δ. By Proposition 5.6.3 there exists an (E, E′)-tiling T ⊂ G forsome finite subset E′ ⊂ G. Consider the subset Z ⊂ X consisting of all theconfigurations z ∈ X such that, for all g ∈ T , one has either z|gD = (gx0)|gD

or z|gD = (gx1)|gD. Applying Exercise 5.33 to the family (gD)g∈T show that,given any map ι : T → {0, 1}, there exists a configuration x ∈ X such thatx|gD = (gxι(g))|gD for all g ∈ T .

(c) Deduce that |ZFj | ≥ 2|Tj | for all j ∈ J , where, Tj = {g ∈ T : gE ⊂ Fj}.(d) By Proposition 5.6.4 there exists a real number α > 0 and an element

j0 ∈ J such that |Tj | ≥ α|Fj | for all j ≥ j0. Deduce from (c) that entF (Z) ≥α log 2.

(e) Deduce that entF (X) > 0.

5.35. Let G be a group and let A be a finite set. Let X ⊂ AG be a stronglyirreducible subshift. Suppose that Δ is a finite subset of G such that X isΔ-irreducible. Show that if Ω1 and Ω2 are (possibly infinite) subsets of Gsuch that Ω+Δ

1 ∩Ω2 = ∅, then, given any two configurations x1 and x2 in X,there exists a configuration x ∈ X which coincides with x1 on Ω1 and withx2 on Ω2. Hint: For each finite subset Ω ⊂ G, consider the subset X(Ω) ⊂ Xconsisting of all configurations in X which coincide with x1 on Ω ∩ Ω1 andwith x2 on Ω ∩Ω2. Observe that the family formed by the sets X(Ω), whereΩ runs over all finite subsets of G is a family of closed subsets of X havingthe finite intersection property and use the compactness of X.

5.36. (cf. Lemma 5.8.3) Let G be an amenable group, F = (Fj)j∈J a rightFølner net for G, and A a finite set. Let τ : X → Y be a cellular automa-ton, where X,Y ⊂ AG are subshifts such that X is strongly irreducible andentF (Y ) < entF (X).

(a) Let S ⊂ G be a memory set for τ . Up to enlarging the subset S ifnecessary, we can also suppose that 1G ∈ S and that X is S-irreducible.Deduce from the hypothesis entF (Y ) < entF (X) that there exists j0 ∈ Jsuch that |π

F+S2j0

(Y )| < |πFj0(X)|.

(b) Fix an arbitrary configuration x0 ∈ X and consider the finite subsetZ ⊂ X consisting of all configurations z ∈ X which coincide with x0 on G \F+S

j0. Show that πFj0

(Z) = πFj0(X). Hint: Use the result of Exercise 5.35(a)

by taking Ω1 = Fj0 and Ω2 = G \ F+Sj0

.

(c) Use Proposition 5.4.3 to show that τ(z) coincide with τ(x0) on G\F+S2

j0for all z ∈ Z.

(d) Deduce from (c) that |τ(Z)| = |πF+S2

j0

(τ(Z))|.(e) Using the fact that τ(Z) ⊂ Y , deduce from (c), (a) and (b) that there

exist configurations z1 �= z2 in Z such that τ(z1) = τ(z2).(f) Deduce from (e) that τ is not pre-injective.

Exercises 147

5.37. Let G be an amenable group, F = (Fj)j∈J a right Følner net for G,and A a finite set. Let X,Y ⊂ AG be two strongly irreducible subshifts suchthat entF (X) = entF (Y ). Show that every pre-injective cellular automatonτ : X → Y is surjective. Hint: Use the results of Exercise 5.31 and Exer-cise 5.36.

5.38. Let G be an amenable group and let A be a finite set. Let X ⊂ AG

be a strongly irreducible subshift. Show that every pre-injective cellular au-tomaton τ : X → X is surjective. Hint: Use Exercise 5.37 with Y = X.

5.39. Let G be an amenable group and let A be a finite set. Show that everystrongly irreducible subshift X ⊂ AG is surjunctive. Hint: This immediatelyfollows from Exercise 5.38.

5.40. (cf. Lemma 5.8.4) Let G be an amenable group, F = (Fj)j∈J a rightFølner net for G, and A a finite set. Let X ⊂ AG be a strongly irreduciblesubshift of finite type and let τ : X → AG be a cellular automaton which isnot pre-injective. Let x1, x2 ∈ X such that τ(x1) = τ(x2) and Ω = {g ∈ G :x1(g) �= x2(g)} is a nonempty finite subset of G. Let S be a memory set forboth τ and X. Up to enlarging the subset S if necessary, we can also supposethat 1G ∈ S and that X is S-irreducible. Set E = Ω+(S−1S) = {ωs−1s′ :ω ∈ Ω, s, s′ ∈ S} and E′ = EE−1 = {ab−1 : a, b,∈ E}. By Proposition 5.6.3there exists an (E, E′)-tiling T ⊂ G. Let Z ⊂ X be the set consisting of allconfigurations z in X such that z|tE �= (tx1)|tE for all t ∈ T .

(a) Using the S-irreducibility of X, show that entF (Z) < entF (X). Hint:Use the arguments in Exercise 5.31.

(b) From (a) and Proposition 5.7.3 deduce that entF (τ(Z)) ≤ entF (X).(c) Using the fact that S is a memory set for X, show that for every x ∈ X

the configuration z defined by

z(g) =

{tx2(g) if there is t ∈ T such that g ∈ tE and x|tE = tx1|tE ,x(g) otherwise.

satisfies z ∈ Z and τ(z) = τ(x) (cf. the end of the proof of Lemma 5.8.4).(d) Deduce from (c) that τ(Z) = τ(X).(e) Deduce from (b) and (d) that entF (τ(X)) < entF (X).

5.41. Let G be an amenable group, F = (Fj)j∈J a right Følner net for G,and A a finite set. Let X,Y ⊂ AG be two subshifts such that X is stronglyirreducible of finite type and entF (X) = entF (Y ). Show that every surjec-tive cellular automaton τ : X → Y is pre-injective. Hint: Use the result ofExercise 5.40.

5.42. The Garden of Eden theorem for strongly irreducible subshifts of finitetype. (cf. [Fio2, Theorem 4.7]). Let G be an amenable group, A a finite set,X ⊂ AG a strongly irreducible subshift of finite type, and τ : X → X acellular automaton. Show that τ is pre-injective if and only if it is surjective.


Hint: Combine together the results from Exercise 5.41 with Y = X andExercise 5.38.

5.43. Let G be a group. Let F be a nontrivial finite abelian group and setA = F ×F . Suppose that G contains an element h0 ∈ G of infinite order (e.g.G = Z). Denote by H ⊂ G the infinite cyclic subgroup generated by h0. Foreach x ∈ AG, let x1, x2 : G → F be the maps defined by x(g) = (x1(g), x2(g))for all g ∈ G. Consider the subset X ⊂ AG consisting of all configurationsx ∈ AG which satisfy x2(gh0) = x2(g) for any g ∈ G. In other words, X isformed by all the configurations x ∈ AG such that x2 is constant on each leftcoset of H in G.

(a) Show that X is a subshift of finite type.(b) Consider the map τ : X → X defined by τ(x) = (x1 + x2, 0) for all

x = (x1, x2) ∈ X (here 0 denotes the zero configuration, i.e., the constantconfiguration whose value at each element of G is equal to the identity elementof F ). Show that τ is a cellular automaton.

(c) Show that the cellular automaton τ defined in (b) is pre-injective butnot surjective.

(d) Show that if H is of infinite index in G (this is the case, for example,when G = Z

2) then the subshift X is irreducible.

5.44. Let G be a group and let A be a set. Let H be a subgroup of G.Suppose that σ : AH → AH is a cellular automaton and let σG : AG → AG

be the induced cellular automaton (cf. Sect. 1.7). Let X ⊂ AH be a subshiftand denote by X(G) ⊂ AG the associated subshift defined in Exercise 1.33.Suppose that σ(X) ⊂ X. Show that the cellular automaton σG|X(G) : X(G) →X(G) is pre-injective if and only if the cellular automaton σ|X : X → X ispre-injective.

5.45. Let A = {0, 1}. Let x0, x1 ∈ AZ denote the two constant configurationsdefined by x0(n) = 0 and x1(n) = 1 for all n ∈ Z.

(a) Show that X = {x0, x1} is a subshift of finite type.(b) Show that the map τ : X → X given by τ(x0) = τ(x1) = x0 is a

cellular automaton which is pre-injective but neither surjective nor injective.

5.46. Let A = {0, 1}, G = Z2, and H = Z × {0} ⊂ G. Consider the subset

X = Fix(H) ⊂ AG consisting of all the configurations x ∈ AG which arefixed by each element of H.

(a) Show that X is an irreducible subshift of finite type of AG.(b) Consider the map τ : X → X defined by τ(x)(g) = 0 for all x ∈ X and

g ∈ G. Show that τ is a cellular automaton which is pre-injective but neithersurjective nor injective.

5.47. ([Fio1, Counterexample 4.27]) Let A = {0, 1, 2} and let X ⊂ AZ be thesubshift of finite type with defining set of forbidden words {01, 02}.

(a) Show that X is not irreducible.

Exercises 149

(b) Consider the cellular automaton σ : AZ → AZ with memory set S ={0, 1} and local defining map

μ(y) =

{y(0) if y(1) �= 00 otherwise.

Show that σ(X) ⊂ X.(c) Show that the cellular automaton σ|X : X → X is surjective but not

pre-injective.

5.48. Let A, X ⊂ AZ and σ : AZ → AZ as in Exercise 5.47. Let H = Z×{0} ⊂Z

2 and let X(Z2) be the associated irreducible subshift of finite type in AZ2

defined as in Exercise 1.33 and let σZ2: AZ

2 → AZ2

be the induced cellularautomaton. Show that the cellular automaton σZ

2 |X(Z2) : X(Z2) → X(Z2) issurjective but not pre-injective.

5.49. ([Fio1, Section 3]) Let A = {0, 1} and let X ⊂ AZ be the even subshift(cf. Exercise 1.38). Let also σ : AZ → AZ be the cellular automaton withmemory set S = {0, 1, 2, 3, 4} and local defining map μ : AS → A given by

μ(y) =

{1 if y(0)y(1)y(2) ∈ {000, 111} or y(0)y(1)y(2)y(3)y(4) = 001000 otherwise.

(a) Show that σ(X) ⊂ X.(b) Set τ = σ|X : X → X. Show that τ is surjective.(c) Show that τ is not pre-injective. Hint: Show that the configurations

x1 = · · · 0000(100100)0000 · · ·

andx2 = · · · 0000(011100)0000 · · ·

satisfyτ(x1) = τ(x2) = · · · 1111(100100)1111 · · ·

Chapter 6

Finitely Generated Amenable Groups

This chapter is devoted to the growth and amenability of finitely generatedgroups. The choice of a finite symmetric generating subset for a finitely gen-erated group defines a word metric on the group and a labelled graph, whichis called a Cayley graph. The associated growth function counts the numberof group elements in a ball of radius n with respect to the word metric. Wedefine a notion of equivalence for such growth functions and observe that thegrowth functions associated with different finite symmetric generating sub-sets are in the same equivalence class (Corollary 6.4.5). This equivalence classis called the growth type of the group. The notions of polynomial, subexpo-nential and exponential growth are introduced in Sect. 6.5. In Sect. 6.7 wegive an example of a finitely generated metabelian group with exponentialgrowth. We prove that finitely generated nilpotent groups have polynomialgrowth (Theorem 6.8.1). In Sect. 6.9 we consider the Grigorchuk group. It isshown that it is an infinite periodic (Theorem 6.9.8), residually finite (Corol-lary 6.9.5) finitely generated group of intermediate growth (Theorem 6.9.17).In the subsequent section, we show that every finitely generated group ofsubexponential growth is amenable (Theorem 6.11.2). In Sect. 6.12 we provethe Kesten-Day characterization of amenability (Theorem 6.12.9) which as-serts that a group with a finite (not necessarily symmetric) generating subsetis amenable if and only if 0 is in the �2-spectrum of the associated Laplacian.Finally, in Sect. 6.13 we consider the notion of quasi-isometry for not nec-essarily countable groups and we show that amenability is a quasi-isometryinvariant (Theorem 6.13.23).

6.1 The Word Metric

Let G be a group.One says that a subset S ⊂ G generates G, or that S is a generating subset

of G, if every element g ∈ G can be expressed as a product of elements in


151

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152 6 Finitely Generated Amenable Groups

S ∪ S−1, that is, for each g ∈ G there exist n ≥ 0, s1, s2, . . . , sn ∈ S andε1, ε2, . . . , εn ∈ {1,−1} such that

g = sε11 sε2

2 · · · sεnn . (6.1)

A subset S ⊂ G is called symmetric if S = S−1, that is, s−1 ∈ S whenevers ∈ S. Observe that if S ⊂ G is a generating subset of G, then S ∪ S−1 is asymmetric generating subset of G.

Recall that G is said to be finitely generated if it admits a finite generatingsubset. If S ⊂ G is a finite generating subset of G, then S ∪ S−1 is a finitesymmetric generating subset of G. Thus any finitely generated group admitsa finite symmetric generating subset.

Let G be a finitely generated group and let S be a finite symmetric gen-erating subset of G.

The S-word-length �S(g) = �GS (g) of an element g ∈ G is the minimal

integer n ≥ 0 such that g can be expressed as a product of n elements in S,that is,

�S(g) = min{n ≥ 0 : g = s1s2 · · · sn, si ∈ S, 1 ≤ i ≤ n}. (6.2)

It immediately follows from the definition that for g ∈ G one has

�S(g) = 0 if an only if g = 1G. (6.3)

Proposition 6.1.1. One has

�S(g−1) = �S(g) (6.4)

and�S(gh) ≤ �S(g) + �S(h) (6.5)

for all g, h ∈ G.

Proof. Let g, h ∈ G. Set m = �S(g) and n = �S(h). Then there exists1, s2, . . . , sm and t1, t2, . . . , tn in S such that g = s1s2 · · · sm and h =t1t2 · · · tn. We have g−1 = s−1

m · · · s−12 s−1

1 so that �S(g−1) ≤ m = �S(g).Exchanging the roles of g and g−1 we get �S(g) ≤ �S(g−1) and therefore�S(g−1) = �S(g).

On the other hand, we have gh = s1s2 · · · smt1t2 · · · tn so that �S(gh) ≤m + n = �S(g) + �S(h). ��

Consider the map dS = dGS : G × G → N defined by

dS(g, h) = �S(g−1h) (6.6)

for all g, h ∈ G.

Proposition 6.1.2. The map dS is a metric on G.

6.2 Labeled Graphs 153

Proof. Let g, h, k ∈ G. It follows from (6.3) that dS(g, h) = 0 if and onlyif g = h. Moreover, from (6.4) we deduce that dS(g, h) = dS(h, g). Finally,using (6.5) we get

dS(g, k) + dS(k, h) = �S(g−1k) + �S(k−1h)

≥ �S((g−1k)(k−1h))

= �S(g−1h)= dS(g, h).

��

The metric dS is called the word metric on G associated with the finitesymmetric generating subset S.

Proposition 6.1.3. The metric dS is invariant by left multiplication, thatis,

dS(gg1, gg2) = dS(g1, g2)

for all g, g1, g2 ∈ G.

Proof. We have

dS(gg1, gg2) = �S((gg1)−1gg2) = �S(g−11 g−1gg2) = �S(g−1

1 g2) = dS(g1, g2).

��

We may rephrase the previous result by saying that the action of G on itselfby left multiplication is an isometric action with respect to the word metric.

For g ∈ G and n ∈ N, we denote by

BGS (g, n) = {h ∈ G : dS(g, h) ≤ n}

the ball of radius n in G centered at the element g ∈ G. When g = 1G wehave BG

S (1G, n) = {h ∈ G : �S(h) ≤ n} and we simply write BGS (n) instead

of BGS (1G, n). Also, when there is no ambiguity on the group G, we omit the

supscript “G” and we simply write BS(g, n) and BS(n) instead of BGS (g, n)

and BGS (n).

6.2 Labeled Graphs

Let S be a set. An S-labeled graph is a pair Q = (Q, E), where Q is a set,called the set of vertices, and E is a subset of Q×S×Q, called the set of edges.The projection map λ : E → S, defined by λ(e) = s for all e = (q, s, q′) ∈ E,is called the labelling map. Also consider the projection maps α, ω : E → Qdefined by α(e) = q and ω(e) = q′ for all e = (q, s, q′) ∈ E. Then, α(e) andω(e) are called the initial and terminal vertices of the edge e ∈ E.


Let Q1 = (Q1, E1) and Q2 = (Q2, E2) be two S-labeled graphs. An S-labeled graph homomorphism from Q1 to Q2 is a map φ : Q1 → Q2 such that(φ(q), s, φ(q′)) ∈ E2 for all (q, s, q′) ∈ E1. An S-labeled graph homomorphismφ : Q1 → Q2 which is bijective and such that the inverse map φ−1 : Q2 →Q1 is also an S-labeled graph homomorphism is called a S-labeled graphisomorphism. If such an S-labeled graph isomorphism exists one says thatthe S-labeled graphs Q1 and Q2 are isomorphic.

An S-labeled graph Q = (Q, E) is said to be finite if the sets Q and E areboth finite.

Let Q = (Q, E) be an S-labeled graph.An S-labeled subgraph of Q is an S-labeled graph Q′ = (Q′, E′) such that

Q′ ⊂ Q and E′ ⊂ E.Let Q′ ⊂ Q and set E′ = E ∩ (Q′ × S × Q′). Then the S-labeled graph

Q′ = (Q′, E′) is called the S-labeled subgraph of Q induced on Q′. Sometimes,by abuse of language, we shall simply denote by Q′ the S-labeled graphQ′ = (Q′, E′).

If there exist vertices q, q′ ∈ Q and labels s1 �= s2 in S such that(q, s1, q

′), (q, s2, q′) ∈ E, in other words, if there exist two edges with the

same initial and terminal vertices but with different labels, then one saysthat Q has multiple edges.

An edge of the form (q, s, q), q ∈ Q, s ∈ S, is called a loop at q.A path in Q is a finite sequence of edges π = (e1, e2, . . . , en), e1, e2, . . . , en ∈

E, such that ω(ei) = α(ei+1) for all i = 1, . . . , n − 1. The integer n is calledthe length of the path π and it is denoted by �(π). The vertices π− = α(e1)and π+ = ω(en) are called the initial and terminal vertices of π and one saysthat π connects π− to π+. The label of a path π = (e1, e2, . . . , en) is definedby λ(π) = (λ(e1), λ(e2), . . . , λ(en)) ∈ Sn. For q ∈ Q, we also admit the emptypath starting and ending at q. It has length 0 and its label is the emptyword ε. Let π1 = (e1, e2, . . . , en) and π2 = (e′1, e

′2, . . . , e

′m) be two paths with

π+1 = π−

2 . Then the path

π1π2 = (e1, e2, . . . , en, e′1, e′2, . . . , e

′m)

is called the composition of the paths π1 and π2. Note that �(π1π2) = �(π1)+�(π2) and that λ(π1π2) = λ(π1)λ(π2).

Let q, q′ ∈ Q. If there is no path connecting q to q′ we set dQ(q, q′) = ∞,otherwise, we set

dQ(q, q′) = min{�(π) : π a path connecting q to q′}. (6.7)

A path π connecting q to q′ with minimal length, that is, such that �(π) =dQ(q, q′), is called a geodesic path from q to q′.

Proposition 6.2.1. Let Q= (Q, E) be a labeled graph. Then, for all q, q′, q′′ ∈Q one has:

6.2 Labeled Graphs 155

(i) dQ(q, q′) ∈ N ∪ {∞};(ii) dQ(q, q′) = 0 if and only if q = q′;(iii) dQ(q, q′) < ∞ if and only if there exists a path which connects q to q′;(iv) dQ(q, q′′) ≤ dQ(q, q′) + dQ(q′, q′′) (triangular inequality).

Proof. The statements (i), (ii), and (iii) are trivial. Let us prove (iv). IfdQ(q, q′) = ∞ or dQ(q′, q′′) = ∞ there is nothing to prove. Otherwise, let π1

be a geodesic path connecting q to q′ and π2 a geodesic path connecting q′

to q′′. Then π1π2 connects q to q′′ and therefore

dQ(q, q′′) ≤ �(π1π2) = �(π1) + �(π2) = dQ(q, q′) + dQ(q′, q′′).

��

Note that, in general, for q, q′ ∈ Q one has dQ(q, q′) �= dQ(q′, q). For q ∈ Qand r ∈ N we denote by B(q, r) = {q′ ∈ Q : dQ(q, q′) ≤ r} the ball of radiusr centered at q.

The labeled graph Q is said to be connected if given any two vertices q, q′ ∈Q there exists a path which connects q to q′. Equivalently, Q is connected ifand only if dQ(q, q′) < ∞ for all q, q′ ∈ Q.

A path π in Q such that π− = π+ is said to be closed.The valence (or degree) δ(q) of a vertex q ∈ Q is the cardinality of the set

{e ∈ E : α(e) = q}. If δ(q) < ∞ for all q ∈ Q, one says that Q is locally finite.If all vertices have the same finite valence k then one says that Q is regularof degree k.

Suppose that S is equipped with an involution s → s. Then the labeledgraph Q is said to be edge-symmetric if for all e = (q, s, q′) ∈ E one has thatthe inverse edge e−1 = (q′, s, q) also belongs to E.

If Q is edge-symmetric and (q, s, q′) ∈ E one says that the vertices q and q′

are neighbors. Suppose that Q is edge-symmetric. If π = (e1, e2, . . . , en) is apath in Q connecting a vertex q to a vertex q′, then π−1 = (en

−1, en−1−1, . . . ,

e2−1, e1

−1) is a path in Q connecting q′ to q. The path π−1 is called theinverse of π. Note that �(π−1) = �(π).

Corollary 6.2.2. Suppose that Q is edge-symmetric and connected. Then themap dQ : Q × Q → N defined in (6.7) is a metric on the set Q of verticesof Q.

Proof. This follows immediately from Proposition 6.2.1 and the fact thatdQ(q, q′) = dQ(q′, q) for all q, q′ ∈ Q. The latter immediately follows fromthe fact that the map π → π−1 yields a length-preserving bijection from theset of paths connecting q to q′ onto the set of paths connecting q′ to q. ��

The metric dQ is called the graph metric on Q.Let π = (e1, e2, . . . , en) be a path in Q. We associate with π the sequence

πQ = (q0, q1, . . . , qn) ∈ Qn+1 defined by qi = α(ei+1) for all i = 0, 1, . . . , nand qn = ω(en). The vertices q0, q1, . . . , qn are called the vertices visited by π.


One says that π is simple if qi �= qj for all 0 ≤ i �= j ≤ n. If π is closed,one says that π is a closed simple path if qi �= qj for all 0 ≤ i, j ≤ n suchthat (i, j) �= (0, n). One says that π is proper, or with no back–tracking, ifei+1 �= e−1

i for i = 1, 2, . . . , n − 1.An edge-symmetric, connected labeled graph with no loops, no multiple

edges and with no closed proper paths is called a tree.

6.3 Cayley Graphs

Let G be a finitely generated group and S a finite symmetric generatingsubset of G. The Cayley graph of G with respect to S is the S-labeled graphCS(G) = (Q, E) whose vertices are the group elements, that is, Q = G andthe edge set is E = {(g, s, gs) : g ∈ G and s ∈ S}.

Fig. 6.1 An edge in the Cayley graph CS(G) is a triple (g, s, h), where g ∈ G, s ∈ S, and

h = gs

Note that, as S is symmetric, the inverse map s → s−1 is an involu-tion on S. Since we have (h, s−1, g) ∈ E for all (g, s, h) ∈ E, it followsthat the Cayley graph CS(G) is edge-symmetric with respect to the inversemap on S. Moreover, the Cayley graph is connected. For, given g, h ∈ G, asS generates G, we can find a nonnegative integer n and s1, s2, . . . , sn ∈ Ssuch that g−1h = s1s2 · · · sn. Then the path π = (e1, e2, . . . , en), whereei = (gs1s2 · · · si−1, si, gs1s2 · · · si−1si), i = 1, 2, . . . , n, connects g to h =gs1s2 · · · sn. Also observe that if 1G ∈ S then CS(G) has a loop at each ver-tex and that, on the contrary, CS(G) has no loops if 1G /∈ S. On the otherhand, CS(G) has no multiple edges, that is, for all g, h ∈ G, there exists atmost one s ∈ S such that (g, s, h) ∈ E, namely s = g−1h if g−1h ∈ S.

Proposition 6.3.1. Let G be a finitely generated group. Let S ⊂ G be afinite symmetric generating subset. Then, the S-distance of two group ele-ments equals the graph distance of the same elements viewed as vertices inthe associated Cayley graph CS(G). In other words,

dS(g, h) = dCS(G)(g, h) (6.8)

for all g, h ∈ G.

Proof. Let g, h ∈ G and suppose that π = (e1, e2, . . . , en) is a geodesic pathconnecting g and h. Let λ(π) = (s1, s2, . . . , sn) be the label of π. It thenfollows that dS(g, h) = �S(g−1h) = �S(s1s2 · · · sn) ≤ n = �(π) = dCS(G)(g, h).

6.3 Cayley Graphs 157

Conversely, suppose that dS(g, h) = m. Then we can find s′1, s′2, . . . , s

′m ∈ S

such that g−1h = s′1s′2 · · · s′m. Then the unique path π′ = (e′1, e

′2, . . . , e

′m)

with (π′)− = g and label λ(π′) = s′1s′2 · · · s′m clearly satisfies (π′)+ = h, that

is, it connects g to h. We deduce that dCS(G)(g, h) ≤ �(π′) = m = dS(g, h).Then (6.8) follows. ��

Note that, in particular, two distinct elements g and h in G are neighborsin the graph CS(G) if and only if dS(g, h) = 1. For all g ∈ G, the maps → gs is a bijection from S onto the set gS of all neighbors of g in CS(G).In particular, all vertices g in CS(G) have the same degree δ(g) = |S|.

Summarizing, we have that a Cayley graph is a connected, edge-symmetricand regular labeled graph.

Examples 6.3.2. We graphically represent Cayley graphs by connecting twoneighboring vertices by a single directed labeled arc e (see Fig. 6.1). Thusone should think of the inverse edge e−1 as the oppositely directed arc withlabel λ(e)−1.

(a) Let G = Z and take S = {1,−1} as a finite symmetric generatingsubset of G. Then the Cayley graph CS(Z) is represented in Fig. 6.2.

Fig. 6.2 The Cayley graph of G = Z for S = {1,−1}

(b) Let G = Z and take S = {1, 0,−1} as a finite symmetric generatingsubset of G. Then the Cayley graph CS(Z) is represented in Fig. 6.3. Notethat as 0 = 1Z ∈ S, we have a loop at each vertex in CS(Z).

Fig. 6.3 The Cayley graph of G = Z for S = {1, 0,−1}

(c) Let G = Z and S = {2,−2, 3,−3}. Then, the corresponding Cayleygraph CS(Z) is represented in Fig. 6.4.

(d) Let G = Z×(Z/2Z), where Z/2Z = {0, 1} and take S = {(1, 0), (−1, 0),(0, 1)}. Then the Cayley graph CS(G) is represented by the bi-infinite ladderas in Fig. 6.5.

(e) Let G = D∞ be the infinite dihedral group, that is, the group ofisometries of the real line R generated by the reflections r : R → R and


Fig. 6.4 The Cayley graph of G = Z for S = {2,−2, 3,−3}

Fig. 6.5 The Cayley graph of G = Z × (Z/2Z) for S = {(1, 0), (−1, 0), (0, 1)}

s : R → R defined by

r(x) = −x (symmetry with respect to 0)s(x) = 1 − x (symmetry with respect to 1/2)

for all x ∈ R. Note that r2 = s2 = 1G. Taking S = {r, s}, the Cayley graphCS(G) is as in Fig. 6.6.

Fig. 6.6 The Cayley graph of G = D∞ for S = {r, s}

(f) Let G be the infinite dihedral group, as in the previous example. Denoteby t : R → R the map defined by t(x) = x + 1 for all x ∈ R (translation).Then we have t = sr and hence s = tr. It follows that the set S = {r, t, t−1} isalso a symmetric generating subset of G and the corresponding Cayley graphCS(G) is as in Fig. 6.7.

(g) Let G = Z2 and take S = {(1, 0), (−1, 0), (0, 1), (0,−1)} as a finite and

symmetric generating subset of G. Then, the corresponding Cayley graphCS(Z2) is given in Fig. 6.8.

(h) Let G = Z2 and take

6.3 Cayley Graphs 159

Fig. 6.7 The Cayley graph of G = D∞ for S = {r, t, t−1}

Fig. 6.8 The Cayley graph of G = Z2 for S = {(1, 0), (−1, 0), (0, 1), (0,−1)}

S = {(1, 0), (−1, 0), (0, 1), (0,−1), (1, 1), (−1, 1), (1,−1), (−1,−1)}.

Then, the corresponding Cayley graph CS(Z2) is as in Fig. 6.9.(i) Let G = FX be the free group based on a nonempty finite set X and

take S = X∪X−1. Then, the Cayley graph CS(FX) is a regular tree of degree|S| = 2|X| (see Fig. 6.10). Indeed, since 1FX

/∈ S, the Cayley graph CS(FX)does not have loops. Moreover, if π is a nontrivial proper path in CS(FX)with label λ(π) = (s1, s2, . . . , sn) ∈ S∗, then, the word w = s1s2 · · · sn ∈ S∗

is nonempty and, by definition of properness, it is reduced. Therefore 1FX

and h = s1s2 · · · sn ∈ FX are distinct and we have π+ = π−h �= π−, that is,π is not closed. This shows that CS(FX) is a tree.


Fig. 6.9 The Cayley graph of G = Z2 for S = {(1, 0), (−1, 0), (0, 1), (0,−1), (1, 1), (−1, 1),

(1,−1), (−1,−1)}

6.4 Growth Functions and Growth Types

Let G be a finitely generated group and S ⊂ G a finite and symmetricgenerating subset of G. The growth function of G relative to S is the functionγG

S : N → N defined by

γGS (n) = |BG

S (n)| = |{g ∈ G : �S(g) ≤ n}| (6.9)

for all n ∈ N. When there is no ambiguity we omit the supscript “G” andsimply denote it by γS .

Note that γS(0) = |BS(0)| = |{1G}| = 1 and that γS(n) ≤ γS(n + 1) forall n ∈ N. Also, as the map (s1, s2, . . . , sn) → s1s2 · · · sn is a surjection from(S ∪ {1G})n to BS(n), one has

γS(n) ≤ |S ∪ {1G}|n (6.10)

for all n ∈ N.

Proposition 6.4.1. Let G be a finitely generated group and let S and S′ betwo finite symmetric generating subsets of G. Let c = max{�S′(s) : s ∈ S}.

6.4 Growth Functions and Growth Types 161

Fig. 6.10 The Cayley graph of G = F2 for S = {a, a−1, b, b−1}

Then the following holds.

(i) �S′(g) ≤ c�S(g) for all g ∈ G;(ii) dS′(g, h) ≤ cdS(g, h) for all g, h ∈ G;(iii) BS(n) ⊂ BS′(cn) for all n ∈ N;(iv) γS(n) ≤ γS′(cn) for all n ∈ N.

Proof. (i) Let g ∈ G. Suppose that �S(g) = n. Then there exist s1, s2, . . . , sn ∈S such that g = s1s2 · · · sn. Using (6.5) we get

�S′(g) = �S′(s1s2 · · · sn) ≤n∑

i=1

�S′(si) ≤ cn.

(ii) By applying (i) we have, for all g, h ∈ G,

dS′(g, h) = �S′(g−1h) ≤ c�S(g−1h) = cdS(g, h).


Finally, from (ii), we have that dS′(g, 1G) ≤ cn if dS(g, 1G) ≤ n for all g ∈ G.This gives (iii). Thus

γS(n) = |BS(n)| ≤ |BS′(cn)| = γS′(cn)

for all n ∈ N. ��

Two metrics d and d′ on a set X are said to be Lipschitz-equivalent if thereexist constants c1, c2 > 0 such that

c1d(x, y) ≤ d′(x, y) ≤ c2d(x, y)

for all x, y ∈ X.

Corollary 6.4.2. Let G be a finitely generated group and let S and S′ be twofinite symmetric generating subsets of G. Then, the word metrics dS and dS′

are Lipschitz-equivalent. ��

A non-decreasing function γ : N → [0,+∞) is called a growth function. Letγ, γ′ : N → [0,+∞) be two growth functions. One says that γ′ dominates γ,and one writes γ � γ′, if there exists an integer c ≥ 1 such that

γ(n) ≤ cγ′(cn) for all n ≥ 1.

One says that γ and γ′ are equivalent and one writes γ ∼ γ′ if γ � γ′ andγ′ � γ.

Proposition 6.4.3. We have the following:

(i) � is reflexive and transitive;(ii) ∼ is an equivalence relation;(iii) let γ1, γ2, γ

′1, γ

′2 : N → [0,+∞) be growth functions. Suppose that γ1 ∼

γ′1, γ2 ∼ γ′

2 and that γ1 � γ2. Then γ′1 � γ′

2.

Proof. It is clear that � is reflexive. Let γ1, γ2, γ3 : N → [0,+∞) be growthfunctions. Suppose that γ1 � γ2 and that γ2 � γ3. Let c1 and c2 be positiveintegers such that γ1(n) ≤ c1γ2(c1n) and γ2(n) ≤ c2γ3(c2n) for all n ≥ 1.Then, taking c = c1c2 one has

γ1(n) ≤ c1γ2(c1n) ≤ c1c2γ3(c2c1n) = cγ3(cn)

for all n ≥ 1. Thus � is also transitive. This shows (i).Property (ii) immediately follows from (i) and the definition of ∼.Finally, suppose that γ1, γ2, γ

′1, γ

′2 satisfy the hypotheses of (iii). Then we

have in particular, γ′1 � γ1, γ1 � γ2 and γ2 � γ′

2. From (i) we deduce thatγ′1 � γ′

2. ��

Let γ : N → [0,+∞) be a growth function. We denote by [γ] the ∼-equivalence class of γ. By abuse of notation, we shall also write [γ] ∼ γ(n).


If γ1 and γ2 are two growth functions, we write [γ1] � [γ2] if γ1 � γ2. Thisdefinition makes sense by virtue of Proposition 6.4.3(iii). Note that, this way,� becomes a partial ordering on the set of equivalence classes of growthfunctions.

Examples 6.4.4. (a) Let α and β be nonnegative real numbers. Then nα � nβ

if and only if α ≤ β, and nα ∼ nβ if and only if α = β.(b) Let γ : N → [0,+∞) be a growth function. Suppose that γ is a poly-

nomial of degree d for some d ≥ 0. Then one has γ(n) ∼ nd.(c) Let a, b ∈ (1, +∞). Then

an ∼ bn. (6.11)

Indeed, suppose for instance that a ≤ b. On the one hand we have an ≤ bn forall n ≥ 1, so that trivially an � bn. On the other, setting c = [loga b] + 1 > 1(here [ · ] denotes the integer part), one has

bn = (aloga b)n = a(loga b)n ≤ acn ≤ cacn

for all n ≥ 1, so that bn � an. We deduce that an ∼ bn. In particular, wehave an ∼ exp(n) for all a ∈ (1, +∞).

(d) Let d ≥ 0 be an integer. Then nd � exp(n) and nd �∼ exp(n). As aconsequence if γ : N → [0,+∞) is a growth function such that γ(n) � nd,then γ � exp(n) and γ �∼ exp(n). Indeed, since limn→∞

nd

exp(n) = 0, the

sequence ( nd

exp(n) )n≥1 is bounded and we can find an integer c ≥ 1 such thatnd

exp(n) ≤ c for all n ≥ 1. It follows that nd ≤ c exp(n) ≤ c exp(cn) for alln ≥ 1 and therefore nd � exp(n).

On the other hand, suppose by contradiction that exp(n) � nd. Then wecan find an integer c > 0 such that exp(n) ≤ c(cn)d for all n ≥ 1. But thenexp(n)

nd ≤ cd+1 for all n ≥ 1 contradicting the fact that limn→∞exp(n)

nd = +∞.Thus nd �∼ exp(n).

The remaining statements concerning the growth function γ immediatelyfollow from transitivity of � (cf. Proposition 6.4.3(i)) and symmetry of ∼ (cf.Proposition 6.4.3(ii)).

From Proposition 6.4.1(iii) and (6.10) we deduce the following:

Corollary 6.4.5. Let G be a finitely generated group and let S and S′ betwo finite symmetric generating subsets of G. Then, the growth functionsassociated with S and S′ are equivalent, that is, γS ∼ γS′ . Moreover, γS(n) �exp(n). ��

Let G be a finitely generated group. The equivalence class [γS ] of thegrowth functions associated with the finite symmetric generating subsets Sof G is called the growth type of G and we denote it by γ(G).


Proposition 6.4.6. Let γ : N → [0,+∞) be a growth function with γ(0) > 0.Then γ ∼ 1 if and only if γ is bounded.

Proof. Suppose first that γ is bounded. Then we can find an integer c ≥ 1such that γ(n) ≤ c for all n ≥ 1. It follows that γ(n) ≤ c1(n) ≤ c1(cn) forall n ≥ 1. Thus γ � 1. On the other hand, setting c = [ 1

γ(0) ] + 1, we have1(n) = 1 ≤ cγ(0) ≤ cγ(n) ≤ cγ(cn) for all n ≥ 1 so that 1 � γ. This showsthat γ ∼ 1.

Conversely, suppose that γ ∼ 1. Then γ � 1 and we can find an integerc ≥ 1 such that γ(n) ≤ c1(cn) = c for all n ≥ 1. This shows that γ isbounded. ��

Corollary 6.4.7. Let G be a finitely generated group. Then γ(G) ∼ 1 if andonly if G is finite. As a consequence, all finite groups have the same growthtype.

Proof. Let S be a finite and symmetric generating subset of G. Suppose thatγ(G) ∼ γS(n) ∼ 1. Then, by Proposition 6.4.6 we deduce that γS is bounded,say by an integer c ≥ 1. This shows that |G| ≤ c and therefore G is finite.Conversely, if G is finite, we have γS(n) = |BS(n)| ≤ |G| for all n ≥ 1, that is,γS is bounded. From Proposition 6.4.6 we deduce that γ(G) ∼ γS(n) ∼ 1. ��

Proposition 6.4.8. Let G be an infinite finitely generated group. Then n �γ(G).

Proof. Let S be a finite symmetric generating subset of G. Consider theinclusions

{1G} = BS(0) ⊂ BS(1) ⊂ BS(2) ⊂ · · · ⊂ BS(n) ⊂ BS(n + 1) ⊂ · · · (6.12)

Let us show that if BS(n) = BS(n+1) for some n ∈ N , then BS(n) = BS(m)for all m ≥ n. We proceed by induction on m. Suppose that BS(n) = BS(m)for some m ≥ n + 1. For all g ∈ BS(m + 1) there exist g′ ∈ BS(m) ands ∈ S such that g = g′s. By the inductive hypothesis, g′ ∈ BS(m−1) so thatg = g′s ∈ BS(m − 1)S ⊂ BS(m). Since BS(m) ⊂ BS(m + 1), it follows thatBS(m + 1) = BS(m) = BS(n).

As a consequence, if BS(n) = BS(n + 1) for some n ∈ N, then we haveG = BS(n). Since, by our assumptions, G is infinite, we deduce that allthe inclusions in (6.12) are strict. It follows that for all n ∈ N we haven ≤ |BS(n)| = γS(n). This shows that n � γS(n) and therefore n � γ(G). ��

Definition 6.4.9. Let G be a finitely generated group.One says that G has exponential (resp. subexponential) growth if γ(G) ∼

exp(n) (resp. γ(G) �∼ exp(n)).One says that G has polynomial growth if there exists an integer d ≥ 0

such that γ(G) � nd.


Proposition 6.4.10. Every finitely generated group of polynomial growth hassubexponential growth.

Proof. Let G be a finitely generated group. It follows from Example 6.4.4(d)that if γ(G) � nd for some integer d ≥ 0, then γ(G) �∼ exp(n). ��

Examples 6.4.11. (a) Let G = Z. With S = {1,−1} one has that the ball ofradius r centered at the element g ∈ G is the interval [g − r, g + r] = {n ∈Z : g− r ≤ n ≤ g + r}, see Fig. 6.11. We have γS(n) = 2n + 1. It follows thatγ(Z) ∼ γS(n) ∼ n. In particular, Z has polynomial growth.

Fig. 6.11 The ball BS(n, r) ⊂ Z with S = {1,−1}; here r = 2

(b) Let G = Z2 and S = {(1, 0), (−1, 0), (0, 1), (0,−1)}. Then the ball

of radius r centered at the element g = (n, m) ∈ G is the diagonal squarewith vertices (n+ r,m), (n− r,m), (n, m− r), (n, m+ r), see Fig. 6.12. In thelanguage of cellular automata, the ball BZ

2

S (g, 1) is commonly called the vonNeumann neighborhood of g. We have γS(n) = 1 +

∑nk=1 4k = 2n2 + 2n + 1.

It follows that γ(Z2) ∼ γS(n) ∼ n2. In particular, Z2 has polynomial growth.

Fig. 6.12 The ball BS(n, r) ⊂ Z2 with S = {(1, 0), (−1, 0), (0, 1), (0,−1)} and r = 2


(c) Let G = Z2 and

S = {(1, 0), (−1, 0), (0, 1), (0,−1), (1, 1), (−1, 1), (1,−1), (−1,−1)}.

Then the ball of radius r centered at the element g = (n, m) ∈ G is the square[n−r, n+r]×[m−r,m+r], see Fig. 6.13. In the language of cellular automata,the ball BZ

2

S (g, 1) is commonly called the Moore neighborhood of g. We haveγZ

2

S (n) = 4n2 + 4n + 1 = (2n + 1)2 = (γZ

S(n))2. Note that this yields againγ(Z2) ∼ γS(n) ∼ n2 and the polynomial growth of Z

2 (cf. Example 6.4.11(b)above).

Fig. 6.13 The ball BS(n, r) ⊂ Z2 with S = {±(1, 0),±(0, 1),±(1, 1),±(1,−1)} and r = 2

(d) Let G = Z× (Z/2Z) and S = {(1, 0), (−1, 0), (1, 1), (−1, 1)}. Then theball of radius r centered at the element (n, 0) is represented in Fig. 6.14. Wededuce that γS(n) = (2n + 1) + 2(n − 1) + 1 = 4n. Thus γ(Z × (Z/2Z)) ∼γS(n) ∼ n. In particular, Z × (Z/2Z) has polynomial growth.

(e) Let G = D∞ be the infinite dihedral group and S = {r, s}. Then theball of radius n centered at the element g ∈ G is represented in Fig. 6.15. Itfollows that γS(n) = 2n + 1. Thus γ(D∞) ∼ γS(n) ∼ n. In particular, D∞has polynomial growth.


Fig. 6.14 The ball BS((n, 0), r) ⊂ Z× (Z/2Z) with S = {(1, 0), (−1, 0), (0, 1)}; here r = 2

Fig. 6.15 The ball BS(g, n) ⊂ D∞ with S = {r, s}; here n = 3

(f) Let G = D∞ be the infinite dihedral group and S = {r, t, t−1}. Thenthe ball of radius r centered at the element g ∈ G is represented in Fig. 6.16.It follows that γS(n) = (2n + 1) + 2(n − 1) + 1 = 4n. Note that this yieldsagain γ(D∞) ∼ γS(n) ∼ n and the polynomial growth of D∞.

Fig. 6.16 The ball BS(g, n) ⊂ D∞ with S = {r, t, t−1}; here n = 2

(g) Let G = Fk be the free group of rank k ≥ 2. Let {a1, a2, . . . , ak} be afree basis and set S = {a1, a

−11 , a2, a

−12 , . . . , ak, a−1

k }. Then the ball of radiusr centered at the element g ∈ G is the finite tree rooted at g of depth r (seeFig. 6.17). We have

γFk

S (n) = 1 + 2k

n−1∑

j=0

(2k − 1)j =k(2k − 1)n − 1

k − 1.

It follows that γ(Fk) ∼ γS(n) ∼ (2k − 1)n ∼ exp(n). In particular, Fk hasexponential growth.


Fig. 6.17 The ball BS(a, 2) ⊂ F2 with S = {a, a−1, b, b−1}

6.5 The Growth Rate

Lemma 6.5.1. Let (an)n≥1 be a sequence of positive real numbers such thatan+m ≤ anam for all n, m ≥ 1. Then the limit

limn→∞

n√

an

exists and equals infn≥1n√

an.

Proof. Fix an integer t ≥ 1 and, for all n ≥ 1 write n = qt+r, with 0 ≤ r < t.Then an ≤ aqtar ≤ aq

tar and n√

an ≤ aq/nt

n√

ar. As 0 ≤ r/n < t/n andlimn→∞ t/n = 0, we have that limn→∞ r/n = 0 and limn→∞ q/n = 1/t. Inparticular limn→∞ a

q/nt a

1/nr = a

1/tt = t

√at. Thus, lim supn→∞ n

√an ≤ t

√at.

This gives lim supn→∞ n√

an ≤ inft≥1t√

at ≤ lim inft→∞ t√

at completing theproof. ��

6.5 The Growth Rate 169

Proposition 6.5.2. Let G be a finitely generated group and let S be a finitesymmetric generating subset of G. Then, the limit

λS = limn→∞

n√

γS(n) (6.13)

exists and λS ∈ [1,+∞).

Proof. From (6.5) we deduce that BS(n + m) ⊆ BS(n)BS(m) and therefore

γS(n + m) = |BS(n + m)|≤ |BS(n)BS(m)|≤ |BS(n)| · |BS(m)|= γS(n)γS(m).

Thus, the sequence {γS(n)}n≥1 satisfies the hypotheses of the previous lemmaand λS exists and is finite. As γS(n) ≥ 1 for all n ∈ N we also have λS ≥ 1. ��

Definition 6.5.3. The number λS = λGS in (6.13) is called the growth rate

of G with respect to S.

Proposition 6.5.4. Let G be a finitely generated group and let S be a fi-nite symmetric generating subset of G. Then λS > 1 if and only if G hasexponential growth.

Proof. Suppose that γ(G) ∼ exp(n). We then have exp(n) � γS so that thereexists an integer c ≥ 1 such that en ≤ cγS(cn) for all n ≥ 1. This implies

1 < c√

e= limn→∞

cn√

en ≤ limn→∞

cn√

cγS(cn) =(

limn→∞

cn√

c)·(

limn→∞

cn√

γS(cn))

= λS .

Conversely, suppose that λS > 1. By Lemma 6.5.1 we have n√

γS(n) ≥ λS sothat

γS(n) ≥ λnS .

This shows that exp(n) ∼ λnS � γS(n). By Corollary 6.4.5 we have γS(n) �

exp(n) and therefore γ(G) ∼ γS ∼ exp(n). ��

Corollary 6.5.5. Let G be a finitely generated group and let S be a finitesymmetric generating subset of G. Then λS = 1 if and only if G has subex-ponential growth. ��

As an immediate consequence of Proposition 6.5.4 and Proposition 6.4.3(ii)we deduce the following.

Corollary 6.5.6. Let G be a finitely generated group and let S and S′ be twofinite symmetric generating subsets of G. Then λS = 1 (resp. λS > 1) if andonly if λS′ = 1 (resp. λS′ > 1). ��


6.6 Growth of Subgroups and Quotients

Proposition 6.6.1. Let G be a group and let N be a normal subgroup of G.Suppose that N and the quotient group G/N are both finitely generated. ThenG is finitely generated.

Proof. Denote by π : G → G/N the quotient homomorphism. Let U ⊂ Nand T ⊂ G/N be two finite symmetric generating subsets of N and G/Nrespectively. Let S ⊂ G be a finite symmetric set such that π(S) ⊃ T andS ⊃ U . Let us prove that S generates G. Let g ∈ G. Then there exist h ≥ 0and t1, t2, . . . , th ∈ T such that π(g) = t1t2 · · · th. Let s1, s2, . . . , sh ∈ S besuch that π(si) = ti for i = 1, 2, . . . , h. Setting g′ = s1s2 · · · sh we then haveπ(g′) = π(g). It follows that n = (g′)−1g ∈ ker(π) = N . Therefore, thereexist k ≥ 0 and sh+1, sh+2, . . . , sh+k ∈ S such that n = sh+1sh+2 · · · sh+k.It follows that g = g′n = s1s2 · · · shsh+1, sh+2 · · · sh+k. This shows that Sgenerates G. ��

Proposition 6.6.2. Let G be a group and let H be a subgroup of finite indexin G. Then, G is finitely generated if and only if H is finitely generated.

Proof. Let R ⊂ G be a complete set of representatives of the right cosets ofH in G such that 1G ∈ R. Note that |R| = [G : H] < ∞. Suppose first thatH is finitely generated. Let S ⊂ H be a finite symmetric generating subsetof H. Given g ∈ G, there exist r ∈ R and h ∈ H such that g = hr. Lets1, s2, . . . , sn ∈ S be such that h = s1s2 · · · sn. Then g = s1s2 · · · snr. Thisshows that the set S ∪ R is a finite generating subset of G.

Suppose now that G is finitely generated and let S be a finite symmetricgenerating subset of G. Let us show that the finite set

S′ = RSR−1 ∩ H (6.14)

generates H. Let h ∈ H and write h = s1s2 · · · sn where si ∈ S. Then, thereexists h1 ∈ H and r1 ∈ R such that s1 = h1r1. Note that h1 = 1Gs1r

−11 ∈ S′.

By induction, for all i = 2, 3, . . . , n−1 there exist hi ∈ H and ri ∈ R such thatri−1si = hiri. Moreover, hi = ri−1sir

−1i ∈ S′. Thus, setting hn = rn−1sn we

have

h = s1s2 · · · sn

= (1Gs1r−11 )(r1s2r

−12 ) · · · (rn−2sn−1r

−1n−1)(rn−1sn)

= h1h2 · · ·hn−1hn.

Note that hn = h−1n−1 · · ·h−1

2 h−11 h ∈ H and, on the other hand, hn =

rn−1sn = rn−1sn1G ∈ RSR−1. Thus also hn ∈ S′. This shows that S′ gener-ates H. ��

6.6 Growth of Subgroups and Quotients 171

Proposition 6.6.3. Let G be a finitely generated group and let H be a finitelygenerated subgroup of G. Then γ(H) � γ(G).

Proof. Let SG (resp. SH) be a finite symmetric generating subset of G(resp. H). Then the set S = SH ∪SG is a finite symmetric generating subsetof G. As SH ⊂ S we have BH

SH(n) ⊂ BG

S (n) and therefore γHSH

(n) ≤ γGS (n)

for all n ∈ N. Thus, γ(H) � γ(G). ��

Corollary 6.6.4. Every finitely generated group which contains a finitelygenerated subgroup of exponential growth has exponential growth. ��

From Example 6.4.11(d) and Corollary 6.6.4 we deduce the following.

Corollary 6.6.5. Every finitely generated group which contains a subgroupisomorphic to the free group F2 has exponential growth. ��

In the following proposition we show that finitely generated groups andtheir finite index subgroups (which are finitely generated as well, by Propo-sition 6.6.2) have the same growth type.

Proposition 6.6.6. Let G be a finitely generated group and let H be a finiteindex subgroup of G. Then H is finitely generated and γ(H) = γ(G).

Proof. Since [G : H] < ∞, we deduce from Proposition 6.6.2 that H is finitelygenerated. It follows from Proposition 6.6.3 that γ(H) � γ(G).

Let now S be a finite symmetric generating subset of G and consider thefinite symmetric set S′ = RSR−1 ∩ H. It follows from the proof of Propo-sition 6.6.2 that S′ generates H. Let g ∈ BG

S (n) and write g = s1s2 · · · sn,where s1, s2, . . . , sn ∈ S. As in the proof of Proposition 6.6.2 we may findr0 = 1G, r1, . . . , rn ∈ R such that

g = s1s2 · · · sn

= (1Gs1r−11 )(r1s2r

−12 ) · · · (rn−2sn−1r

−1n−1)(rn−1snr−1

n )rn

= h1h2 · · ·hn−1hnrn

where hi = ri−1sir−1i ∈ S′. It follows that BG

S (n) ⊂ BHS′(n)R so that, taking

cardinalities,

γGS (n) = |BG

S (n)| ≤ |BHS′(n)||R| = [G : H]γH

S′(n) ≤ [G : H]γHS′([G : H]n).

This shows that γ(G) � γ(H). It follows that γ(G) = γ(H). ��

Two groups G1 and G2 are called commensurable if there exist finite indexsubgroups H1 ⊂ G1 and H2 ⊂ G2 such that H1 and H2 are isomorphic. Fromthe previous proposition we immediately deduce:

Corollary 6.6.7. If G1 and G2 are commensurable groups and G2 is finitelygenerated, then G1 is finitely generated and one has γ(G1) = γ(G2). ��


Proposition 6.6.8. Let G be a finitely generated group and let N be a normalsubgroup of G. Then the quotient group G/N is finitely generated and onehas γ(G/N) � γ(G). If in addition N is finite, then γ(G/N) = γ(G).

Proof. Let S be a finite symmetric generating subset of G and let π : G →G/N denote the canonical quotient homomorphism. Then S′ = π(S) is afinite symmetric generating subset of G/N . Thus, for all n ∈ N one has

BG/NS′ (n) = π(BG

S (n)) (6.15)

and therefore

γG/NS′ (n) = |BG/N

S′ (n)| ≤ |BGS (n)| = γG

S (n).

This shows that γ(G/N) � γ(G).Suppose now that N is finite. From (6.15) we deduce that BG

S (n) ⊂π−1(BG/N

S′ (n)) and therefore

γGS (n) = |BG

S (n)| ≤ |N ||BG/NS′ (n)| = |N |γG/N

S′ (n) ≤ |N |γG/NS′ (|N |n).

This shows that γ(G) � γ(G/N). It follows that γ(G) = γ(G/N). ��

Lemma 6.6.9. Let γ1, γ2, γ′1, γ

′2 : N → [0,+∞) be growth functions. Suppose

that γ1 � γ′1, γ2 � γ′

2. Then the products γ1γ2, γ′1γ

′2 : N → [0,∞) are also

growth functions and one has γ1γ2 � γ′1γ

′2.

Proof. Since the product of non-decreasing functions is also non-decreasing,it is clear that γ1γ2 and γ′

1γ′2 are also growth functions. Let c1 and c2 be

positive integers such that γ1(n) ≤ c1γ′1(c1n) and γ2(n) ≤ c2γ

′2(c2n) for all

n ≥ 1. Taking c = c1c2 we have

(γ1γ2)(n) = γ1(n)γ2(n)≤ c1c2γ

′1(c1n)γ′

2(c2n)≤ c1c2γ

′1(c1c2n)γ′

2(c1c2n)= c(γ′

1γ′2)(cn)

for all n ≥ 1. This shows γ1γ2 � γ′1γ

′2. ��

Given two growth functions γ1 and γ2 we set [γ1] · [γ2] = [γ1γ2]. Thisdefinition makes sense since if γ1 ∼ γ′

1 and γ2 ∼ γ′2 then γ1γ2 ∼ γ′

1γ′2 as it

immediately follows from Lemma 6.6.9.

Proposition 6.6.10. Let G1 and G2 be two finitely generated groups. Thenthe direct product G1 × G2 is also finitely generated and γ(G1 × G2) =γ(G1)γ(G2).

Proof. Let S1 and S2 be finite symmetric generating subsets of G1 and G2.Then the set

6.7 A Finitely Generated Metabelian Group with Exponential Growth 173

S = (S1 × {1G2}) ∪ ({1G1} × S2)

is a finite symmetric generating subset of G1×G2. Let (g1, g2) ∈ BG1×G2S (n).

Then there exist s1,1, s2,1, . . . , sk,1 ∈ S1 and s1,2, s2,2, . . . , sh,2 ∈ S2, whereh + k ≤ n, such that

(g1, g2) = (s1,1, 1G2)(s2,1, 1G2) · · · (sk,1, 1G2) · (1G1 , s1,2)(1G1 , s2,2) · · · (1G1 , sh,2)= (s1,1s2,1 · · · sk,1, s1,2s2,2 · · · sh,s).

Thus, BG1×G2S (n) ⊂ BG1

S1(n)×BG2

S2(n) and γG1×G2

S (n) ≤ γG1S1

(n)γG2S2

(n). Thisshows that γ(G1 × G2) � γ(G1)γ(G2).

On the other hand, if g1 ∈ BG1S1

(n) and g2 ∈ BG2S2

(n), then (g1, g2) ∈BG1×G2

S (2n) and one has γG1S1

(n)γG2S2

(n) ≤ γG1×G2S (2n) ≤ 2γG1×G2

S (2n).This shows that γ(G1)γ(G2) � γ(G1 × G2). It follows that γ(G1 × G2) =γ(G1)γ(G2). ��

From Example 6.4.11(a) and the previous proposition one immediatelydeduces the following.

Corollary 6.6.11. Let d be a positive integer. Then γ(Zd) ∼ nd. ��

Corollary 6.6.12. Every finitely generated abelian group has polynomialgrowth.

Proof. Let G be a finitely generated abelian group. Then there exist an in-teger d ≥ 0 and a finite group F such that G is isomorphic to the cartesianproduct Z

d × F . It follows from Proposition 6.6.10, Corollary 6.6.11 andCorollary 6.4.7 that γ(G) = γ(Zd)γ(F ) ∼ nd. This shows that G has polyno-mial growth. ��

6.7 A Finitely Generated Metabelian Groupwith Exponential Growth

We have seen in Corollary 6.6.12 that every finitely generated abelian grouphas polynomial growth. The purpose of the present section is to give anexample of a finitely generated metabelian group with exponential growth. Asevery metabelian group is solvable and hence amenable (Theorem 4.6.3), thiswill show in particular that there exist finitely generated amenable groups,and even finitely generated solvable groups, whose growth is exponential.

Proposition 6.7.1. Let G denote the subgroup of GL2(Q) generated by thetwo matrices

A =(

2 00 1

)and B =

(1 10 1

).

Then G is a finitely generated metabelian group with exponential growth.


Proof. The group G is finitely generated by definition. We have

G ={(

2k r0 1

): k ∈ Z, r ∈ Z[1/2]

}, (6.16)

where Z[1/2] is the subring of Q consisting of all dyadic rationals. Indeed, ifwe denote by H the right-hand side of (6.16), we first observe that G ⊂ Hsince H is clearly a subgroup of GL2(Q) containing A and B. On the otherhand, we also have the inclusion H ⊂ G since any r ∈ Z[1/2] can be writtenin the form r = 2mn for some m, n ∈ Z, so that

(2k r0 1

)= AmBnAk−m ∈ G.

From (6.16), we deduce that the determinant map yields a surjective ho-momorphism from G onto an infinite cyclic group whose kernel is isomorphicto the additive group Z[1/2] and is therefore abelian. Since any element in[G, G] has determinant 1, this shows that [G, G] is abelian, that is, that G ismetabelian.

It remains to show that G has exponential growth. To see this, let usestimate from below the cardinality of the ball BS(3n − 2), where S is thefinite symmetric generating subset of G defined by S = {A, B, A−1, B−1}and n ≥ 1 is a fixed integer. Consider the subset E ⊂ G consisting of allmatrices of the form

M(q) =(

1 q0 1

),

where q is an integer such that 0 ≤ q ≤ 2n − 1. Developing q in base 2, weget

q =n∑

i=1

ui2i−1,

where ui ∈ {0, 1} for 1 ≤ i ≤ n. This gives us

M(q) =(

1 10 1

)u1(

1 20 1

)u2(

1 22

0 1

)u3

· · ·(

1 2n−1

0 1

)un

.

As we have (1 2i−1

0 1

)= Ai−1BA−(i−1)

for all 1 ≤ i ≤ n, it follows that we can write M(q) in the form

M(q) = Bu1ABu2A−1A2Bu3A−2 · · ·An−2Bun−1A−(n−2)An−1BunA−(n−1)

= Bu1ABu2ABu3A · · ·ABun−1ABunA−(n−1).

6.8 Growth of Finitely Generated Nilpotent Groups 175

This shows that

�S(M(q)) ≤ (2n − 1) + (n − 1) = 3n − 2.

We deduce that |BS(3n − 2)| ≥ |E| = 2n. It follows that the growth rate ofG with respect to S satisfies

λS = limn→∞

3n−2√

|BS(3n − 2)| ≥ limn→∞

3n−2√

2n = 3√

2 > 1.

Thus G has exponential growth. ��

6.8 Growth of Finitely Generated Nilpotent Groups

In this section we prove the following

Theorem 6.8.1. Every finitely generated nilpotent group has polynomialgrowth.

In order to prove this result we need some preliminaries.

Lemma 6.8.2. Let G be any group. Let H and K be two normal subgroupsof G. Suppose that S ⊆ H and T ⊆ K generate H and K respectively. Then[H,K] is equal to the normal closure in G of the set {[s, t] : s ∈ S, t ∈ T}.

Proof. Denote by N the normal closure in G of the set {[s, t] : s ∈ S, t ∈ T}and let us show that N = [H,K]. Since {[s, t] : s ∈ S, t ∈ T} ⊂ [H,K] and[H,K] is a normal subgroup, we have

N ⊂ [H,K]. (6.17)

Consider the quotient homomorphism π : G → G/N . For all s ∈ S and t ∈ Twe have [π(s), π(t)] = π([s, t]) = 1G/N , that is, π(s) and π(t) commute. Itfollows that all elements of π(H) commute with all elements of π(K), sinceS generates H and T generates K. In other words, π([h, k]) = [π(h), π(k)] =1G/N , for all h ∈ H and k ∈ K. This gives [h, k] ∈ N for all h ∈ H and k ∈ Kand therefore [H,K] ⊂ N . From (6.17) we deduce that [H,K] = N . ��

Let G be a group. Recall that the lower central series of G is the se-quence (Ci(G))i≥0 of normal subgroups of G defined by C0(G) = G andCi+1(G) = [Ci(G), G] for all i ≥ 0. Given elements g1, g2, . . . , gi ∈ G, i ≥ 3,we inductively set

[g1, g2, . . . , gi] = [[g1, g2, . . . , gi−1], gi] ∈ Ci−1(G).

If S ⊂ G and i ≥ 2 we then denote by S(i)G the set consisting of all elements

of the form


[s1, s2, . . . , si], s1, s2, . . . , si ∈ S. (6.18)

The elements (6.18) are called simple S-commutators of weight i. Note thatS

(i)G ⊂ Ci−1(G).

Lemma 6.8.3. Let G be any group. Let SG ⊂ G be a generating subset of G.Then for all i ≥ 1, the subgroup Ci(G) is the normal closure in G of the setS

(i+1)G .

Proof. Let us prove the statement by induction on i. We have that C1(G) =[G, G] is the normal closure in G of S

(2)G = {[s1, s2] : s1, s2 ∈ SG}, as it

follows from Lemma 6.8.2 by taking H = K = G and S = T = SG. Thus, thestatement holds for i = 1. Suppose by induction that Ci−1(G) is the normalclosure in G of S

(i)G , denote by N the normal closure in G of S

(i+1)G , and

let us show that Ci(G) = N . Since S(i+1)G ⊂ Ci(G) and Ci(G) is a normal

subgroup, we deduce thatN ⊂ Ci(G). (6.19)

Denote by π : G → G/N the quotient map. Let w ∈ S(i)G and s ∈ SG.

Then, by definition, [w, s] ∈ S(i+1)G and therefore [w, s] ∈ N . It follows that

[π(w), π(s)] = π([w, s]) = 1G/N , that is, π(w) and π(s) commute. As SG gen-erates G, we have that π(SG) generates G/N and therefore π(w) ∈ Z(G/N).It follows that π(hwh−1) = π(h)π(w)π(h)−1 = π(w) for all h ∈ G, so thatπ([hwh−1, s]) = [π(hwh−1), π(s)] = [π(w), π(s)] = 1G/N . It follows that

[hwh−1, s] ∈ N. (6.20)

By the inductive hypothesis, every element in Ci−1(G) can be expressedas a product (h1w1h

−11 )(h2w2h

−12 ) · · · (hmwmh−1

m ) with wk ∈ S(i)G and hk ∈

G, k = 1, 2, . . . , m, for some m ∈ N. By taking H = Ci−1(G), K = G,S = {wh : w ∈ S

(i)G , h ∈ G} and T = SG in Lemma 6.8.2, we deduce that

Ci(G) = [Ci−1(G), G] is the normal closure in G of the set {[hwh−1, s] :w ∈ S

(i)G , h ∈ G, s ∈ SG}. Thus from (6.20) it follows that Ci(G) ⊂ N . By

using (6.19) this shows that Ci(G) = N . ��

Lemma 6.8.4. Let G be a finitely generated nilpotent group of nilpotencydegree d ≥ 1. Then the subgroups Ci(G), i = 1, 2, . . . , d − 1 are finitelygenerated.

Proof. Let SG ⊂ G be a finite generating subset of G. We first show thatthe quotient groups Ci(G)/Ci+1(G) are finitely generated, i = 0, 1, . . . , d − 1.From Lemma 6.8.3 we have that Ci(G) is the normal closure in G of thefinite set S

(i+1)G . Therefore, if π : G → G/Ci+1(G) denotes the quotient ho-

momorphism, we have that Ci(G)/Ci+1(G) = π(Ci(G)) is the normal clo-sure in G/Ci+1(G) of the set π(S(i+1)

G ). But π(S(i+1)G ) = (π(SG))(i+1) ⊂

Z(Ci(G)/Ci+1(G)) and therefore π(S(i+1)G ) generates Ci(G)/Ci+1(G).

6.8 Growth of Finitely Generated Nilpotent Groups 177

Let us now prove the statement by reverse induction starting from i = d−1.It follows from the first part of the proof that the subgroup Cd−1(G) ∼=Cd−1(G)/{1G} = Cd−1(G)/Cd(G) is finitely generated. Suppose by induc-tion that the subgroup Ci+1(G) is finitely generated for some i ≤ d − 2.From the first part of the proof we have that Ci(G)/Ci+1(G) is also finitelygenerated. Therefore from Proposition 6.6.1 we deduce that Ci(G) is finitelygenerated. ��

We are now in position to prove the main result of this section.

Proof of Theorem 6.8.1. Let G be a finitely generated nilpotent group ofnilpotency degree d. Let us prove the statement by induction on d. If d = 0then G = {1G} and therefore G has polynomial growth. Suppose now thatd ≥ 1 and that all finitely generated nilpotent groups of nilpotency degree≤ d − 1 have polynomial growth.

We first observe that the subgroup H = C1(G) is nilpotent of nilpotencydegree ≤ d − 1. Indeed, as one immediately checks by induction, we haveCi(H) ⊂ Ci+1(G) for all i = 0, 1, . . . , d − 1, so that Cd−1(H) ⊂ Cd(G) ={1G}. Moreover, by Lemma 6.8.4, we have that H is finitely generated. Thus,by the inductive hypothesis we have that H has polynomial growth. LetT ⊂ H be a finite symmetric generating subset of H. Then there exist integersc1 > 0 and q ≥ 0 such that

γHT (n) ≤ c1(c1n)q (6.21)

for all n ≥ 1.Let S = {s1, s2, . . . , sk} ⊂ G be a finite symmetric generating subset of

G. Let g ∈ G and suppose that m = �GS (g) ≤ n. Then there exist 1 ≤

i1, i2, . . . , im ≤ k such that

g = si1si2 · · · sim . (6.22)

Since g2g1 = g1g2[g−12 , g−1

1 ] and [g−12 , g−1

1 ] ∈ H for all g1, g2 ∈ G, we canpermute the generators in (6.22) and express g in the form

g = sj1sj2 · · · sjmh (6.23)

where 1 ≤ j1 ≤ j2 ≤ · · · ≤ jm ≤ k and h ∈ H. Let us estimate �HT (h). We set

L = max{�HT (w) : w ∈ S(i), 2 ≤ i ≤ d}. (6.24)

As we observed above, exchanging a pair of consecutive generators producesa simple S-commutator of weight two on their right. It is clear that oneneeds at most n such exchanges to bring sj1 , where j1 = ip1 = min{ip : p =1, 2, . . . , m}, to the leftmost place. Analogously, one needs at most n suchexchanges to bring sj2 , where j2 = min{ip : p = 1, 2, . . . ,m; p �= p1}, to theleftmost but one place (in fact on the right of sj1). And so on. Altogether,


there are at most mn ≤ n2 such exchanges. This produces at most n2 simpleS-commutators of weight two. But at each step, when moving a generator sjt

to the left, we also have to exchange it with all the simple S-commutatorson its left that were produced before (namely by sj1 , sj2 , . . . , sjt−1). As oneeasily checks by induction, that this produces, altogether, at most n3 simpleS-commutators of weight three, n4 simple S-commutators of weight four, andso on. Continuing this way, since G is nilpotent of nilpotency degree d, allsimple S-commutators of weight d+1 are equal to 1G. Thus, the total numberof simple S-commutators that are eventually produced in this process is atmost n2 + n3 + · · · + nd ≤ dnd. It follows from (6.24) that

�HT (h) ≤ Ldnd. (6.25)

On the other hand, we can bound the number of elements in G which areof the form sj1sj2 · · · sjm , where 1 ≤ j1 ≤ j2 ≤ · · · ≤ jm ≤ k, by c2n

k, wherec2 > 0 is a constant independent of n. Indeed, each such group element canbe written in the form sn1

1 sn22 · · · snk

k , where 0 ≤ ni ≤ n for all i = 1, 2, . . . , k.We deduce from (6.21) and (6.25) that

γGS (n) ≤ c2n

kc1(Ldnd)q = Cnδ ≤ C(Cn)δ

for all n ≥ 1, where δ = k + qd and C = c1c2(Ld)q. Note that C > 0 is aconstant independent of n. It follows that G has polynomial growth. ��

From Theorem 6.8.1 and Proposition 6.6.6 we deduce the following:

Corollary 6.8.5. Every finitely generated virtually nilpotent group has poly-nomial growth. ��

6.9 The Grigorchuk Group and Its Growth

In this section we present the Grigorchuk group and some of its main prop-erties, namely being infinite, periodic, residually finite and of intermediategrowth.

Let Σ = {0, 1}. We denote by Σ∗ = ∪n∈NΣn the set of all words on thealphabet Σ. Recall that Σ∗ is a monoid for the word concatenation whoseidentity element is the empty word ε (cf. Sect. D.1). Every word w ∈ Σ∗ maybe uniquely written in the form w = σ1σ2 · · ·σn, where σ1, σ2, . . . , σn ∈ Σand n = �(w) ∈ N is the length of w.

Denote by Sym(Σ∗) the symmetric group on Σ∗ (cf. Appendix C). Weintroduce a partial order � in Σ∗ by setting u � v, u, v ∈ Σ∗, if there existsw ∈ Σ∗ such that uw = v. We then set

Sym(Σ∗,�)= {g ∈ Sym(Σ∗) : g(u) � g(v) for all u, v ∈Σ∗ such that u � v}.(6.26)

6.9 The Grigorchuk Group and Its Growth 179

Note that Sym(Σ∗,�) is a subgroup of Sym(Σ∗). Moreover, for w ∈ Σ∗ thelength �(w) equals the maximum of n ∈ N such that there exists a sequence(wk)0≤k≤n of distinct words in Σ∗ such that ε = w0 � w1 � · · · � wn = w.It follows that if g ∈ Sym(Σ∗,�) then �(g(w)) = �(w) for all w ∈ Σ∗.

Consider the elements a, b, c, d ∈ Sym(Σ∗) defined as follows. We firstdefine a by setting

a(ε) = ε (6.27)

anda(0w) = 1w, a(1w) = 0w (6.28)

for all w ∈ Σ∗. Then, for all w ∈ Σ∗, we define b(w), c(w), d(w) by inductionon �(w). We start by setting

b(ε) = c(ε) = d(ε) = ε. (6.29)

Then we setb(0w) = 0a(w), b(1w) = 1c(w),c(0w) = 0a(w), c(1w) = 1d(w),d(0w) = 0w, d(1w) = 1b(w)

(6.30)

for all w ∈ Σ∗. By an obvious induction we have

a, b, c, d ∈ Sym(Σ∗,�). (6.31)

Example 6.9.1. Let w = 10110 ∈ Σ∗. Then we have

a(w) = a(10110) = 00110,

b(w) = b(10110) = 1c(0110) = 10a(110) = 10010,

c(w) = c(10110) = 1d(0110) = 10110,

d(w) = d(10110) = 1b(0110) = 10a(110) = 10010.

Definition 6.9.2. The Grigorchuk group is the subgroup G of Sym(Σ∗) gen-erated by the elements a, b, c, d.

Observe that by (6.31) we have

G ⊂ Sym(Σ∗,�). (6.32)

Proposition 6.9.3. The following relations hold in G.

a2 = b2 = c2 = d2 = 1G (6.33)

andbc = cb = d, dc = cd = b, db = bd = c. (6.34)

Proof. To prove (6.33), we have to show that

g2(w) = w (6.35)


for all w ∈ Σ∗ and g ∈ {a, b, c, d}. We have a(ε) = ε by (6.27). Moreoverfrom (6.28) we have

a2(0w) = a(1w) = 0w and a2(1w) = a(0w) = 1w, (6.36)

for all w ∈ Σ∗. This shows that a2 = 1G. To prove (6.35) for g = b, c andd, we use induction on �(w). By virtue of (6.29) this holds when �(w) = 0.Suppose that (6.35) holds when �(w) = n. From (6.30) and the fact thata2 = 1G we deduce, for all w ∈ Σ∗ with �(w) = n,

b2(0w) = b(0a(w)) = 0a2(w) = 0w, b2(1w) = b(1c(w)) = 1c2(w) = 1w,

c2(0w) = c(0a(w)) = 0a2(w) = 0w, c2(1w) = c(1d(w)) = 1d2(w) = 1w,

d2(0w) = d(0w) = 0w, d2(1w) = d(1b(w)) = 1b2(w) = 1w.

This shows that (6.35) holds for all w ∈ Σ∗ with �(w) = n + 1. This proves(6.35) for g = b, c, d and (6.33) follows.

To prove (6.34), we have to show that

ij(w) = k(w) (6.37)

for all w ∈ Σ∗ and all distinct i, j, k ∈ {b, c, d}. Again, we can use inductionon �(w). If �(w) = 0 then (6.37) follows from (6.29). Suppose by inductionthat (6.37) holds when �(w) = n. From (6.28),(6.30) and (6.33) we deduce,for all w ∈ Σ∗ with �(w) = n,

bc(0w) = b(0a(w)) = 0a2(w) = 0w = d(0w),

bc(1w) = b(1d(w)) = 1cd(w) = 1b(w) = d(1w),

cd(0w) = c(0w) = 0a(w) = b(0w),

cd(1w) = c(1b(w)) = 1db(w) = 1c(w) = b(1w),

db(0w) = d(0a(w)) = 0a(w) = c(0w),

db(1w) = d(1c(w)) = 1bc(w) = 1d(w) = c(1w).

It follows that (6.37) holds for all w ∈ Σ∗ with �(w) = n + 1 whenever(i, j, k) = (b, c, d), (c, d, b), (d, b, c). Thus, by induction bc = d, cd = b anddb = c. Finally, using (6.33) we deduce cb = cd2b = (cd)(db) = bc, dc =db2c = (db)(bc) = cd and bd = bc2d = (bc)(cd) = db. This completes theproof of (6.34). ��

It follows from (6.33) that the set S = {a, b, c, d} is a symmetric gener-ating subset of G. We denote by �S : G → N the corresponding word-lengthfunction. Every group element g ∈ G can be expressed in the form

g = s1s2 · · · sn (6.38)


with s1, s2, . . . , sn ∈ S. We say that the expression (6.38) is a reduced formof g provided that for all i, j = 1, 2, . . . , n − 1 one has that if si = a thensi+1 ∈ {b, c, d}, and if sj ∈ {b, c, d} then sj+1 = a. It immediately followsfrom (6.33) and (6.34) that every group element g ∈ G can be expressed (notnecessarily in a unique way) as in (6.38) in reduced form.

For all n ∈ N we set

Hn = {g ∈ G : g(w) = w for all w ∈ Σn}. (6.39)

Proposition 6.9.4. For all n ∈ N, the set Hn is a normal subgroup of Gand

[G : Hn] < ∞. (6.40)

Moreover,

G = H0 ⊃ H1 ⊃ H2 ⊃ · · · ⊃ Hn ⊃ Hn+1 ⊃ · · · . (6.41)

Proof. Since g(Σn) ⊂ Σn for all g ∈ G (cf. (6.32)) we may consider the(restriction) map θn : G → Sym(Σn) defined by θn(g)(w) = g(w) for allg ∈ G and w ∈ Σn. Clearly, θn is a homomorphism and ker(θn) = Hn. Thisshows that Hn is a normal subgroup and that [G : Hn] = |G/Hn| = |θn(G)| ≤| Sym(Σn)| < ∞.

Finally, let n ∈ N, u ∈ Σn and g ∈ Hn+1. Let σ ∈ Σ and set w = uσ ∈Σn+1 so that u � w. From (6.32) we deduce that g(u) � g(w) = uσ andtherefore g(u) = u. This shows that g ∈ Hn. Thus Hn ⊃ Hn+1 and (6.41)follows. ��

Corollary 6.9.5. The Grigorchuk group G is residually finite.

Proof. Let h ∈ ∩n∈NHn. We have h(w) = w for all w ∈ Σ∗ and thereforeh = 1G. This shows that ∩n∈NHn = {1G}. As [G : Hn] < ∞ for all n ∈ N wededuce from Proposition 2.1.11 that G is residually finite. ��

Proposition 6.9.6. We have:

(i) the subgroup H1 consists of all group elements g ∈ G which can be ex-pressed in the form (6.38) (not necessarily reduced) with an even numberof occurrences of the generator a;

(ii) [G : H1] = 2;(iii) the group H1 is generated by the elements b, c, d, aba, aca, ada;(iv) the group H1 equals the normal closure in G of the elements b, c and d.

Proof. It follows from (6.28) and (6.30) that a generator s ∈ S satisfiess(σ) = σ for all σ ∈ Σ if and only if s ∈ {b, c, d}. Thus, g = s1s2 · · · sn,si ∈ S, belongs to H1 if and only if |{i : si = a}| is an even number. Thisalso shows that [G : H1] = 2. Finally, let g ∈ H1. Then it can be expressedin one of the following reduced forms:


g = t0at1at2at3at4 · · · at2k−1at2k = t0(at1a)t2(at3a)t4 · · · (at2k−1a)t2k,

g = at1at2at3at4 · · · at2k−1at2k = (at1a)t2(at3a)t4 · · · (at2k−1a)t2k,

g = t0at1at2at3at4 · · · at2k−1a = t0(at1a)t2(at3a)t4 · · · (at2k−1a),g = at1at2at3at4 · · · at2k−1a = (at1a)t2(at3a)t4 · · · (at2k−1a),

where t0, t1, . . . , t2k+1 ∈ {b, c, d} and k ∈ N (observe that the occurrences ofthe a’s is 2k). This shows that the set {b, c, d, aba, aca, ada} generates H1.Now, the normal closure in G of the generators b, c, d is the subgroup H ⊂ Ggenerated by all the conjugates gtg−1 with g ∈ G and t ∈ {b, c, d}. Thustaking g = 1G, a we deduce from (iii) that H1 ⊂ H. On the other hand, sinceb, c, d ∈ H1 we also have H ⊂ H1. Thus H = H1. ��

Let h ∈ H1. For all w ∈ Σ∗ there exist w0, w1 ∈ Σ∗, �(w0) = �(w1) = �(w)such that h(0w) = 0w0 and h(1w) = 1w1. Denote by h0, h1 ∈ Sym(Σ∗,�)the maps defined by h0(w) = w0 and h1(w) = w1. We thus have

h(0w) = 0h0(w) and h(1w) = 1h1(w)

for all w ∈ Σ∗. We denote by φ0 : H1 → Sym(Σ∗,�) (resp. φ1 : H →Sym(Σ∗,�)) the map defined by φ0(h) = h0 (resp. φ1(h) = h1) and byφ : H1 → Sym(Σ∗,�) × Sym(Σ∗,�) the product map φ(h) = (h0, h1).From (6.30) we immediately deduce

φ(b) = (a, c), φ(aba) = (c, a),φ(c) = (a, d), φ(aca) = (d, a),φ(d) = (1G, b), φ(ada) = (b, 1G).

(6.42)

Proposition 6.9.7. We have:

(i) the maps φ0, φ1 : H1 → G are surjective homomorphisms;(ii) the map φ : H1 → G × G is an injective homomorphism.

Proof. Let σ ∈ {0, 1}. For all h, h′ ∈ H and w ∈ Σ∗ we have

σφσ(hh′)(w) = hh′(σw) = h(σφσ(h′)(w)) = σφσ(h)φσ(h′)(w).

This shows that φσ(hh′) = φσ(h)φσ(h′). It follows that φ0 and φ1 are ho-momorphisms. From (6.42) and Proposition 6.9.6(iii) we immediately deducethat φ0(H1) = φ1(H1) = G. This shows (i).

To show the injectivity of φ, let h ∈ H1 and suppose that φ(h) = 1G×G =(1G, 1G). We have h(ε) = ε and h(σw) = σhσ(w) = σw for all σ ∈ {0, 1} andw ∈ Σ∗. It follows that h = 1G = 1H . This shows (ii). ��

As a consequence of Proposition 6.9.7(ii), we can identify each elementh ∈ H1 with its image φ(h) ∈ G × G. Thus, we shall write h = (h0, h1) ifh ∈ H1 and φ(h) = (h0, h1).

Recall that a group is called periodic if it contains no elements of infiniteorder.


Theorem 6.9.8. The Grigorchuk group G is an infinite finitely generatedperiodic group.

Proof. Since H1 is a proper subset of G and φ0 : H1 → G is surjective (cf.Proposition 6.9.7(i)), we deduce that G is infinite.

Let now show that G is a 2–group, that is, that the order of every elementg ∈ G is a power of 2. The proof is by induction on �S(g). By (6.33) thestatement is true for �S(g) = 1.

Let us first show that every element g ∈ G\{1G} is conjugate either to anelement in S or to an element g′ which can be expressed in a reduced form

g′ = at1at2 · · · atk (6.43)

with t1, t2, . . . , tk ∈ {b, c, d}, k ≥ 1, and such that �S(g′) ≤ �S(g). Indeed,if g is not conjugate to an element in S, then any element of minimalS-length in the conjugacy class of g necessarily admits one of the follow-ing two reduced forms, besides (6.43): t1at2 · · · atka and t0at1 · · · atk witht0, t1, . . . , tk ∈ {b, c, d}, t0 �= tk, k ≥ 1. Conjugating by a and t0 respectivelyand replacing tkt0 with t ∈ {b, c, d} according to (6.34), we transform theabove two expressions into one of the form (6.43). It is clear that such aprocess does not increase the word lengths so that �S(g′) ≤ �S(g).

Suppose first that g ∈ H1 and �(g) > 1. Since the order of any elementequals the order of all its conjugates, we may suppose, up to conjugacy, thatg admits a reduced expression of the form (6.43) (with k ≥ 2 even, sinceg ∈ H1). Now, the image of each quadruple at2i−1at2i, i = 1, 2, . . . , k/2, viaφ0 or φ1 has word-length ≤ 2. Thus, �S(g0), �S(g1) ≤ �S(g)/2 < �S(g). Byinduction, g0 e g1 are 2−elements, and thus g = (g0, g1) is a 2−element aswell.

Suppose now that g /∈ H1. As before, we may suppose, up to conjugacy,that g admits a reduced expression of the form (6.43) (with k odd, sinceg �∈ H1). We distinguish three cases.

Case 1. The generator d appears in the expression of g, say d = ti forsome 1 ≤ i ≤ k. Then, up to conjugating by the element ti−1ati−2a · · · at1a,we can suppose that d = t1.

Now g2 = (adat2)(at3at4) · · · (atkad)(at2at3) · · · (atk−1atk) ∈ H. The im-age of each quadruple via φ0 (resp. φ1) has length 2, with the only exceptionfor those of the form atjad = (atja)d (resp. adatj = (ada)tj) since φ0(d) = 1G

(resp. φ1(ada) = 1G). But at least one of these quadruples occurs in g2, forinstance for j = k (resp. j = 1), so that �S(φ0(g2)) ≤ 2k − 1 < 2k = �S(g)(resp. �S(φ1(g2)) < �S(g)). By induction, φ0(g2) and φ1(g2) are 2-elementsso that g2 = (φ0(g2), φ1(g2)) and therefore g are 2-elements as well.

Case 2. Suppose now that d doesn’t occur in the expression of g but cdoes. As before, up to conjugacy, we may suppose that t1 = c. We then haveφ0(abat′) = ca, φ0(acat′) = da and φ1(abat′) = ac, φ1(acat′) = ad for allt′ ∈ {b, c}. It follows that �S(φ0(g2)) = �S(φ1(g2)) = 2k = �(g). Observe thatφ0(g2) = da · · · a (resp. φ1(g2) = a · · · ad) so that, by conjugating by a (resp.


ad) we fall into Case 1. This shows that φ0(g2) and φ1(g2) are 2-elements.Thus g2 = (φ0(g2), φ1(g2)) and therefore g are 2-elements.

Case 3. Finally, suppose that neither d nor c appear in (6.43) so thatnecessarily g = (ab)2h+1 for some h ≥ 0. We have g2 = (ab)4h+2 =((aba)b)2h+1 ∈ H, so that φ0(g2) = (ca)2h+1 and φ1(g2) = (ac)2h+1. Since�S(φ0(g2)) = �S(φ1(g2)) = 4h + 2 = �S(g), we are in Case 2 and we deducethat φ0(g2) and φ1(g2) are 2-elements. Thus g2 = (φ0(g2), φ1(g2)) and there-fore g itself are 2-elements. ��Theorem 6.9.9. The Grigorchuk group G does not have polynomial growth.

In order to prove this theorem, let us prove some preliminary results.

Lemma 6.9.10. The subgroup of G generated by a and d is dihedral of or-der 8.

Proof. The dihedral group D8 of order 8 has presentation D8 = 〈x, y :x2, y2, (xy)4〉. Since a2 = d2 = 1G by (6.33), we are only left to verify thatthe order of the element ad is 4. We have (ad)2 = (ada)d = (b, 1)(1, b) =(b, b) �= 1G, while (ad)4 = ((ad)2)2 = (b, b)2 = (b2, b2) = (1G, 1G) = 1G. ��

Recall that two groups G1 and G2 are commensurable if there exist twosubgroups of finite index K1 ⊂ G1 and K2 ⊂ G2 such that K1 and K2 areisomorphic.

Lemma 6.9.11. The Grigorchuk group G is commensurable with its ownsquare G × G.

Proof. Let us start by showing that the index of φ(H1) inside G×G is finite.Since b, c, d, aba, aca and ada generate H1 (cf. Proposition 6.9.4), we de-

duce that the elements in (6.42) generate φ(H1).Denote by B ⊂ G the normal closure in G of b. The quotient G/B is

generated by the images of the generators of G and since cd = b ∈ B, it isgenerated by the images of a and d. From Lemma 6.9.10 we deduce

[G : B] = |G/B| ≤ 8. (6.44)

Let g ∈ G and consider the element gbg−1. Since φ0 (resp. φ1) is sur-jective, there exists an element h ∈ H (resp. h′ ∈ H) such that φ0(h) = g(resp. φ1(h′) = g). It follows that (gbg−1, 1G) = φ(hadah−1) ∈ φ(H1) (resp.(1G, gbg−1) = φ(h′d(h′)−1) ∈ φ(H1)). As g varies in G the elements (gbg−1, 1)(resp. (1, gbg−1)) generate the subgroup B0 = B × {1G} ⊂ φ(H1) (resp.B1 = {1G} × B ⊂ φ(H1)). Observe that B0 � B1 � B and that B0 andB1 are normal subgroups of G × G. Moreover, B0 ∩ B1 = {1G×G}, so that(G×G)/(B0B1) � G/B×G/B. From (6.44) and the fact that B0B1 ⊂ φ(H1)we deduce that

[G × G : φ(H1)] ≤ [G × G : B0B1] = [G : B]2 = 64. (6.45)

This shows that φ(H1) has finite index in G × G.


On the other hand [G : H1] = 2 (Proposition 6.9.6(ii)) and since φ is aninjective homomorphism (Proposition 6.9.7(ii)), we have that H1 and φ(H1)are isomorphic. It follows that G and G × G are commensurable. ��

Proof of Theorem 6.9.9. Since G is infinite (Theorem 6.9.8) we have n � γ(G)(cf. Proposition 6.4.8). Since G and G×G are commensurable (Lemma 6.9.11)we deduce from Corollary 6.6.7 that γ(G) = γ(G×G). On the other hand, byProposition 6.6.10 we have γ(G×G) = γ(G)2. Using Lemma 6.6.9 it followsthat n2 � γ(G)2 = γ(G). By induction, we have that n2h � γ(G) for allh ∈ N. Thus G cannot have polynomial growth. ��

The remaining of this section is devoted to showing the following:

Theorem 6.9.12. The Grigorchuk group G has subexponential growth.

To prove this theorem we need some preliminaries. We start with a usefulcriterion for detecting that certain finitely generated groups have subexpo-nential growth.

Lemma 6.9.13. Let G be a finitely generated group. Let S ⊂ G be a finitesymmetric generating subset and suppose that there exist an integer M ≥ 2,two constants 0 < k < 1 and K ≥ 0, and an injective homomorphism

ψ : H → GM

g → (gi)Mi=1

where H ⊂ G is a finite index subgroup of G, such that

M∑

i=1

�S(gi) ≤ k�S(g) + K (6.46)

for all g ∈ H. Then G has subexponential growth.

Proof. Let us show that λS = limn→∞ γGS (n)

1n equals 1.

Fix ε > 0. Then there exists an integer n0 ≥ 1 such that γGS (n) < (λS +ε)n

for all n ≥ n0. Since λS ≥ 1, it follows that

γGS (n) ≤ γG

S (n0)(λS + ε)n (6.47)

for all n ∈ N.Let

γHS (n) = |{h ∈ H : �S(h) ≤ n}|

and fix a system T of left coset representatives of H in G. Set C =maxt∈T �S(t). Then, given g ∈ G, there exist unique h ∈ H and t ∈ Tsuch that g = th. Therefore �S(h) ≤ �S(t) + �S(g) ≤ C + �(g), so thatBG

S (n) ⊂ TBHS (n + C). We deduce that

γGS (n) ≤ [G : H]γH

S (n + C). (6.48)


On the other hand, by (6.46) we have

γHS (n) ≤

∑γG

S (n1)γGS (n2) · · · γG

S (nM )

where the sum runs over all M−tuples n1, n2, . . . , nM such that∑

ni ≤kn + K.

From (6.47) we then deduce:

γHS (n) ≤ γG

S (n0)M∑

(λS + ε)n1(λS + ε)n2 · · · (λS + ε)nM

= γGS (n0)M

∑(λS + ε)n1+n2+···+nM

≤(γG

S (n0)(kn + K))M

(λS + ε)kn+K .

Using (6.48) we then obtain

γGS (n) ≤ [G : H]γH

S (n + C) ≤ [G : H](γG

S (n0)(kn + K ′))M

(λS + ε)kn+K′

(6.49)where K ′ = K+kC. Taking the nth roots and passing to the limit for n → ∞in (6.49), the first term on the left tends to λS , while the last term on theright tends to (λS + ε)k. Thus λS ≤ (λS + ε)k. Since ε was arbitrary wededuce that λS ≤ λk

S . Since by hypothesis k < 1, we have that λS = 1.Finally, from Corollary 6.5.5 we deduce that G has subexponential growth.

��

Lemma 6.9.14. Let n ≥ 1. Then φσ(Hn+1) ⊂ Hn for σ = 0, 1.

Proof. We proceed by induction on n. We have already noticed that φσ(H1) ⊂G = H0 (Proposition 6.9.7(i)). Suppose that φσ(Hn) ⊂ Hn−1 and let us showthat φσ(Hn+1) ⊂ Hn. Let g ∈ Hn+1 and w = σu, with u ∈ Σn and σ ∈ {0, 1}.Then

σu = w = g(w) = σφσ(g)(u),

so that φσ(g)(u) = u. This shows that φσ(g) ∈ Hn. It follows thatφσ(Hn+1) ⊂ Hn. ��

As a consequence of the previous lemma, for all n ≥ 1 and w =σ1σ2 · · ·σn ∈ Σn the homomorphisms φw : Hn → G defined by

φw = φσ1 ◦ φσ2 ◦ · · · ◦ φσn

are well defined. We then define the homomorphism ψn : Hn →∏

w∈Σn Gby setting ψn(g) = (φw(g))w∈Σn for all g ∈ Hn. Note that by Proposi-tion 6.9.7(ii) ψn is injective. Thus, identifying Hn with its image ψn(Hn) ⊂∏

w∈Σn G, we simply write g = (gw)w∈Σn for all g ∈ Hn.In the following, we consider the alphabet Λ = {α, β, γ, δ} and the monoid

Λ∗. The map Λ → S given by α → a, β → b, γ → c and δ → d uniquelyextends to a surjective monoid homomorphism π : Λ∗ → G. We say that a


word w = λ1λ2 · · ·λn ∈ Λ∗ is reduced if for all i, j = 1, 2, . . . , n − 1 one hasλi = α implies λi+1 ∈ {β, γ, δ} and λj ∈ {β, γ, δ} implies λj+1 = α.

Consider the following transformation p : Λ2 → Λ∗ defined by setting

p(λλ′) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

ε if λ = λ′

δ if λ = β, λ′ = γ or λ = γ, λ′ = βγ if λ = β, λ′ = δ or λ = δ, λ′ = ββ if λ = γ, λ′ = δ or λ = δ, λ′ = γλλ′ otherwise.

(6.50)

Given a word u = λ1λ2 · · ·λn ∈ Λ∗, and 1 ≤ i ≤ n, we set

u(i) = λ1λ2 · · ·λi−1p(λiλi+1)λi+2 · · ·λn ∈ Λ∗.

Note that u is reduced if and only if u = u(i) for all i = 1, 2, . . . , n − 1.By induction we define u(i1,i2,...,ik) = (u(i1,i2,...,ik−1))ik

for 1 ≤ ik ≤�(u(i1,i2,...,ik−1)) − 1.

Let now w = λ1λ2 · · ·λn ∈ Λ∗. If w is reduced we set w = w. Otherwise,let i1 be the minimal integer such that w �= w(i1). If w(i1) is reduced we setw = w(i1). Otherwise, let i2 be the minimal integer such that w(i1) �= w(i1,i2).If w(i1,i2) is reduced we set w = w(i1,i2). And so on. Since �(w) > �(w(i1)) >�(w(i1,i2) > · · · > �(w(i1,i2,...,ik−1)) > �(w(i1,i2,...,ik)) > · · · there exists aninteger k such that w(i1,i2,...,ik) is reduced. We then set w = w(i1,i2,...,ik). Atransformation w(i1,i2,...,ij−1) → w(i1,i2,...,ij) is called a leftmost cancellation.It follows that every word in Λ∗ can be transformed into a reduced word bya finite sequence of leftmost cancellations.

We denote by Θ1 ⊂ Λ∗ the subset consisting of all reduced words won Λ which contain an even number �α(w) of occurrences of the letterα. Thus, a word w ∈ Θ1 is an alternate product of terms of the form(αλα) and λ′, with λ, λ′ ∈ {β, γ, δ}. Consider the maps Φ0 : Θ1 → Λ∗ andΦ1 : Θ1 → Λ∗ defined by setting Φσ((αλ1α)λ2 · · · ) = Φσ(αλ1α)Φσ(λ2) · · ·(resp. Φσ(λ1(αλ2α) · · · ) = Φσ(λ1)Φσ(αλ2α) · · · for all λ1, λ2, . . . ∈ {β, γ, δ}where

Φ0(β) = α, Φ0(αβα) = γ, Φ1(β) = γ, Φ1(αβα) = α,

Φ0(γ) = α, Φ0(αγα) = δ, Φ1(γ) = δ, Φ1(αβα) = α,

Φ0(δ) = ε, Φ0(αδα) = β, Φ1(β) = δ, Φ1(αβα) = ε.

For n ≥ 1 we inductively define Θn+1 ⊂ Λ∗ as the subset consisting of allreduced words w in Θn such that the reduced words Φ0(w) and Φ1(w) belongto Θn. Also, given w ∈ Θn we recursively set wσ0σ1···σn−1σ = Φσ(wσ0σ1···σn−1)for all σ, σ1, σ2, . . . , σn−1 ∈ {0, 1}.

Lemma 6.9.15. For all w ∈ Θ1 one has

�(w0) + �(w1) ≤ �(w) + 1. (6.51)


Proof. Let w ∈ Θ1. We distinguish a few cases.Case 1. Suppose that w starts and ends with α so that w = (αλ1α)λ2 · · ·

λk−1(αλkα) with λi ∈ {β, γ, δ}. Note that �(w) = 2k + 1. Then wσ = Φσ(w)and Φσ(w) = Φσ(αλ1α)Φσ(λ2) · · ·Φσ(λk−1)Φσ(αλkα) and therefore

�(wσ) = �(Φσ(w)) ≤ �(Φσ(w))

≤ �(Φσ(αλ1α)) + �(Φσ(λ2)) + · · · + �(Φσ(λk−1)) + �(Φσ(αλkα))

≤ k =(2k + 1) − 1

2

=�(w) − 1

2.

Case 2. Suppose that w starts and ends with letters in {β, γ, δ}, that is,w = λ1(αλ2α)λ3 · · · (αλk−1α)λk with λi ∈ {β, γ, δ}. Note that �(w) = 2k−1.Then wσ = Φσ(w) and Φσ(w) = Φσ(λ1)Φσ(αλ2α) · · ·Φσ(αλk−1α)Φσ(λk) andtherefore

�(wσ) = �(Φσ(w)) ≤ �(Φσ(w))

≤ �(Φσ(λ1)) + �(Φσ(αλ2α)) + · · · + �(Φσ(αλk−1α)) + �(Φσ(λk))

≤ k =(2k − 1) + 1

2

=�(w) + 1

2.

Case 3. Finally, suppose that w starts with α and ends with λ ∈ {β, γ, δ},or viceversa. To fix ideas, suppose that w = (αλ1α)λ2 · · · (αλk−1α)λk withλi ∈ {β, γ, δ}. Passing to the inverse word w−1 one handles the other possibil-ity. Note that �(w) = 2k. Then wσ = Φσ(w) and Φσ(w) = Φσ(αλ1α)Φσ(λ2) · · ·Φσ(αλk−1α)Φσ(λk) and therefore

�(wσ) = �(Φσ(w)) ≤ �(Φσ(w))

≤ �(Φσ(αλ1α)) + �(Φσ(λ2)) + · · · + �(Φσ(αλk−1α)) + �(Φσ(λk))

≤ k =2k

2

=�(w)

2.

This shows (6.51). ��

Lemma 6.9.16. For all g ∈ H3 one has

1∑

i,j,k=0

�S(gijk) ≤ 56�S(g) + 8. (6.52)


Proof. Let g ∈ H3 and let w ∈ Θ3 be such that g = π(w) and �S(g) = �(w).As a consequence of Lemma 6.9.15 we have

1∑

i,j=0

�(wij) ≤1∑

i=0

(�(wi) + 1)

=1∑

i=0

�(wi) + 2

≤ �(w) + 3

(6.53)

and therefore

1∑

i,j,k=0

�(wijk) ≤1∑

i,j=0

(�(wij) + 1)

≤1∑

i,j=0

�(wij) + 4

(by (6.53)) ≤(

1∑

i=0

�(wi) + 2

)+ 4 =

1∑

i=0

�(wi) + 6

(by (6.51)) ≤ �(w) + 1 + 6 = �(w) + 7.

(6.54)

Now we observe that by definition of the maps Φ0 and Φ1, the inequalities in(6.51), (6.53) and (6.54) can be sharpened as follows.

�(w0) + �(w1) ≤ �(w) + 1 − �δ(w), (6.55)

where �δ(w) denotes the number of occurrences of the letter δ in the wordw. Indeed, every such δ, may appear in w either isolated or in a triplet(αδα). Then, in the first case we have Φ0(δ) = ε, while in the second one,Φ1(αδα) = ε.

Similarly, since every letter γ in the word w will give rise via Φ0 or Φ1 toa letter δ in one of w0 and w1, the above argument shows that, even if somecancellation occurs,

�(w00) + �(w01) + �(w10) + �(w11) ≤ �(w) + 3 − �γ(w), (6.56)

where �γ(w) denotes the number of occurrences of the letter γ in the word w.Finally, since every letter β in the word w will give rise via Φ0 or Φ1 to a

letter γ in one of w0 and w1, and therefore to a letter δ in one of w00, w01, w10

and w11, the above arguments show that indeed, even if some cancellationoccurs,


1∑

i,j,k=0

�(wijk) ≤ �(w) + 7 − �β(w), (6.57)

where �β(w) denotes the number of occurrences of the letter β in the word w.Since �(w) = �α(w)+�β(w)+�γ(w)+�δ(w) and w is reduced, we necessarily

have �β(w) + �γ(w) + �δ(w) ≥ (w)−12 and therefore

maxλ∈{β,γ,δ}

�λ(w) >�(w)

6− 1. (6.58)

Taking into account the inequalities in (6.54) we obtain

1∑

i,j,k=0

�(wijk) ≤ min

⎧⎨

⎩

1∑

i=0

�(wi) + 6,

1∑

i,j=0

�(wij) + 4

⎫⎬

⎭ . (6.59)

Observe that π(wijk) = gijk so that �S(gijk) ≤ �(wijk) for all i, j, k = 0, 1.Thus, using first (6.59), (6.55), (6.56) and (6.57), and then (6.58) we deduce

1∑

i,j,k=0

�S(gijk) ≤1∑

i,j,k=0

�(wijk)

≤ min{�(w) + 7 − �β(w), �(w) + 7 − �γ(w), �(w) + 7 − �δ(w)}≤ �(w) + 7 − max{�β(w), �γ(w), �δ(w)}

≤ �(w) + 7 −(

�(w)6

− 1)

≤ 56�(w) + 8

=56�(g) + 8.

This shows (6.52). ��

Proof of Theorem 6.9.12. Let H = H3. By Proposition 6.9.4 we have [G :H] < ∞. Moreover, by Lemma 6.9.16 we can take ψ = ψ3, M = 8, k = 5/6and K = 8, and apply Lemma 6.9.13 to obtain that G has subexponentialgrowth. ��

A finitely generated group G is said to have intermediate growth if Ghas subexponential growth but does not have polynomial growth. Note thatevery finitely generated group of intermediate growth contains no subgroupsisomorphic to F2, by Corollary 6.6.5. From Theorem 6.9.9 and Theorem 6.9.12we then get:

Theorem 6.9.17. The Grigorchuk group G has intermediate growth. ��

6.10 The Følner Condition for Finitely Generated Groups 191

6.10 The Følner Condition for Finitely GeneratedGroups

Let G be a group. As right multiplication by elements in G is bijective wehave, for all A, B, F ⊂ G, F finite, and g ∈ G,

(A \ B)g = Ag \ Bg, (6.60)

|F \ Fg| = |Fg \ F | = |F \ Fg−1| = |Fg−1 \ F |. (6.61)

Lemma 6.10.1. Let A, B, C ⊂ G be three finite sets. Then we have:

|A \ B| ≤ |A \ C| + |C \ B|. (6.62)

Proof. Suppose that x ∈ A \ B, that is, (i) x ∈ A and (ii) x /∈ B. Wedistinguish two cases. If x /∈ C, then, by (i), x ∈ A \ C. If x ∈ C, then,by (ii), x ∈ C \ B. In both cases, x ∈ (A \ C) ∪ (C \ B). It follows thatA \ B ⊂ (A \ C) ∪ (C \ B) and (6.62) follows. ��

Let now S ⊂ G be a finite subset. The isoperimetric constant of G withrespect to S is the nonnegative number

ιS(G) = infF

|FS \ F ||F | (6.63)

where F runs over all non empty finite subsets of G. Observe that ιS(G) =ιS∪{1G}(G).

Proposition 6.10.2. Let G be a finitely generated group and let S ⊂ G be afinite generating subset. Then the following conditions are equivalent:

(a) G is amenable;(b) for all ε > 0 there is a finite subset F = F (ε) ⊂ G such that

|F\Fs| < ε|F | for all s ∈ S; (6.64)

(c) ιS(G) = 0.

Proof. Suppose (a) and fix ε > 0. By Theorem 4.9.1, G satisfies the Følnerconditions. For our convenience, we express the Følner conditions as follows.Given any finite subset K ⊂ G and ε′ > 0 there exists a finite subset F =F (K, ε′) ⊂ G such that

|F\Fk| < ε′|F | for all k ∈ K. (6.65)

Taking ε′ = ε, K = S and F = F (S, ε) in 6.65, we then immediatelyobtain (6.64), and the implication (a) ⇒ (b) follows.


Suppose (b). Taking g = s in (6.61) we immediately deduce

|F\Fs| < ε|F | for all s ∈ S ∪ S−1. (6.66)

Fix a finite set K ⊂ G and ε′ > 0. Since S generates G we can find n ∈ N

such that K ⊂ (S ∪ S−1)n. Set ε = ε′/n. Let F = F (ε) ⊂ G be a finite setsuch that (6.66) holds. Given k ∈ K we can find a1, a2, . . . , an ∈ S ∪ S−1

such that k = a1a2 · · · an. Then, recalling (6.62) and (6.60), we have

|F \ Fk| = |F \ Fa1a2 · · · an|≤ |F \ Fan| + |Fan \ Fan−1an| + |Fan−1an \ Fan−2an−1an|

+ · · · + |Fa2a3 · · · an \ Fa1a2 · · · an|= |F \ Fan| + |F \ Fan−1| + |F \ Fan−2| + · · · + |F \ Fa1|< nε|F |≤ ε′|F |

for all k ∈ K. This shows that G satisfies the Følner conditions (6.65) andtherefore G is amenable by Theorem 4.9.1. Thus (b) ⇒ (a).

Finally, the equivalence (b) ⇔ (c) immediately follows from (6.61) and theinequalities

|F \ Fs| = |Fs \ F | ≤ |FS \ F | ≤∑

s′∈S

|Fs′ \ F | =∑

s′∈S

|F \ Fs′|.

for any s ∈ S and any finite subset F ⊂ G. ��

6.11 Amenability of Groups of Subexponential Growth

In this section we show that every finitely generated group of subexponentialgrowth is amenable.

We first prove the following:

Lemma 6.11.1. Let (an)n≥1 be a sequence of positive real numbers. Then

lim infn→∞

an+1

an≤ lim inf

n→∞n√

an. (6.67)

Proof. Set α = lim infn→∞an+1an

. If α = 0 there is nothing to prove. Other-wise, let 0 < β < α. Then, there exists an integer N ≥ 1 such that an+1

an≥ β

for all n ≥ N . Thus, for all p = 1, 2 . . . one has

aN+p

aN=

aN+p

aN+p−1· aN+p−1

aN+p−2· · · aN+1

aN≥ βp.

This gives aN+p ≥ βpaN . It follows that setting n = N + p, one has, for alln ≥ N

6.12 The Theorems of Kesten and Day 193

an ≥ βn−NaN = βn(β−NaN ).

Thus, taking the nth roots we obtain n√

an ≥ β n√

β−NaN and therefore,

lim infn→∞

n√

an ≥ β. (6.68)

As (6.68) holds for all β < α, we have lim infn→∞ n√

an ≥ α = lim infn→∞an+1an

.��

We are now in position to prove the main result of this section.

Theorem 6.11.2. Every finitely generated group of subexponential growth isamenable.

Proof. Let G be a finitely generated group of subexponential growth. LetS ⊂ G be a finite symmetric generating subset of G. Then, by virtue of theprevious lemma we have 1 ≤ lim infn→∞

γS(n+1)γS(n) ≤ limn→∞

n√

γS(n) = 1, so

that lim infn→∞γS(n+1)

γS(n) = 1. Fix ε > 0 and let n0 ∈ N be such that

γS(n0 + 1)γS(n0)

< 1 + ε. (6.69)

Set F = BS(n0) and let us show that |F \Fs| < ε|F | for all s ∈ S. Let s ∈ S.Then, as BS(n0)s ⊂ BS(n0 + 1) for all s ∈ S, we have

|F \ Fs| = |Fs \ F |= |BS(n0)s \ BS(n0)|≤ |BS(n0 + 1) \ BS(n0)|= γS(n0 + 1) − γS(n0)< εγS(n0)= ε|F |,

where the last inequality follows from (6.69). From Proposition 6.10.2 wededuce that G is amenable. ��

From Theorem 6.9.12 and Theorem 6.11.2 we deduce the following:

Corollary 6.11.3. The Grigorchuk group G is amenable. ��

6.12 The Theorems of Kesten and Day

Let G be a group.Let p ∈ [1,+∞). We consider the real Banach space

�p(G) =

{x ∈ R

G :∑

g∈G

|x(g)|p < ∞}


consisting of all p-summable real functions on G. For x ∈ �p(G), the nonneg-

ative number ‖x‖p =(∑

g∈G |x(g)|p) 1

p is called the �p-norm of x.The support of a configuration x ∈ R

G is the set {g ∈ G : x(g) �= 0}. Wedenote by R[G] ⊂ R

G the vector subspace consisting of all finitely supportedconfigurations in R

G. Note that R[G] is a dense subspace in �p(G).When p = 2, it is possible to endow �2(G) with a scalar product 〈·, ·〉

defined by setting〈x, y〉 =

∑

g∈G

x(g)y(g)

for all x, y ∈ �2(G). Then, the �2-norm of an element x ∈ �2(G) is given by‖x‖2 =

√〈x, x〉. The space (�2(G), 〈·, ·〉) is a real Hilbert space.

The �p-norm of a linear map T : �p(G) → �p(G) is defined by

‖T‖p→p = supx∈p(G)‖x‖p≤1

‖Tx‖p = supx∈p(G)

x�=0

‖Tx‖p

‖x‖p.

Then, T is continuous if and only if ‖T‖p→p < ∞. We denote by L(�p(G))the space of all continuous linear maps T : �p(G) → �p(G) and by I : �p(G) →�p(G) the identity map.

Note that, by density of R[G] in �p(G), we have

‖T‖p→p = supx∈R[G]‖x‖p≤1

‖Tx‖p = supx∈R[G]

x�=0

‖Tx‖p

‖x‖p(6.70)

for all T ∈ L(�p(G)).Let p ∈ [1,+∞) and s ∈ G. For all x ∈ �p(G) and g ∈ G we set

(T (p)s x)(g) = x(gs).

We then have

‖T (p)s x‖p

p =∑

g∈G

|T (p)s x(g)|p

=∑

g∈G

|x(gs)|p

=∑

h∈G

|x(h)|p (by setting h = gs)

= ‖x‖pp

for all x ∈ �p(G). We deduce that T(p)s x ∈ �p(G) and that the linear map

T(p)s : �p(G) → �p(G) has �p-norm


‖T (p)s ‖p→p = 1. (6.71)

In particular, T(p)s ∈ L(�p(G)).

Let now S ⊂ G be a non-empty finite set. We denote by M(p)S : �p(G) →

�p(G) the map defined by M(p)S = 1

|S|∑

s∈S T(p)s . In other words,

(M (p)S x)(g) =

1|S|∑

s∈S

x(gs)

for all x ∈ �p(G) and g ∈ G. The map M(p)S is called the �p-Markov operator

on �p(G) associated with S.

Proposition 6.12.1. Let G be a group and S ⊂ G a non-empty finite set.Then the �p-Markov operator M

(p)S : �p(G) → �p(G) is linear and continuous.

Moreover,

‖M (p)S ‖p→p ≤ 1. (6.72)

Proof. Since T(p)s ∈ L(�p(G)), we deduce that M

(p)S ∈ L(�p(G)). Finally, we

have

‖M (p)S ‖p→p = ‖ 1

|S|∑

s∈S

T (p)s ‖p→p ≤ 1

|S|∑

s∈S

‖T (p)s ‖p→p = 1

where the last equality follows from (6.71). ��

Proposition 6.12.2. Let G be a group. Let S ⊂ G be a finite subset contain-ing 1G. Then, the following conditions are equivalent:

(a) ‖M (2)S ‖2→2 = 1;

(b) given ε > 0 there exists x ∈ R[G] such that ‖x‖2 = 1 and ‖M (2)S x‖2 ≥

1 − ε;(c) given ε > 0 there exists x ∈ R[G] such that x ≥ 0, ‖x‖2 = 1 and

‖M (2)S x‖2 ≥ 1 − ε;

(d) given ε > 0 there exists x ∈ R[G] such that x ≥ 0, ‖x‖2 = 1 and

‖x − T (2)s x‖2 ≤ ε for all s ∈ S; (6.73)

(e) 1 belongs to the real spectrum σ(M (2)S ) of M

(2)S .

Proof. The implication (a) ⇒ (b) follows from the definition of �2-norm andthe density of R[G] in �2(G) (cf. (6.70)).

Let now x ∈ R[G]. We have

‖M (2)S x‖2

2 ≤∥∥∥M (2)

S |x|∥∥∥

2

2. (6.74)


Indeed,

‖M (2)S x‖2

2 =

∥∥∥∥∥1|S|∑

s∈S

T (2)s x

∥∥∥∥∥

2

2

=∑

g∈G

(1|S|∑

s∈S

x(gs)

)2

≤∑

g∈G

(1|S|∑

s∈S

|x(gs)|)2

=

∥∥∥∥∥1|S|∑

s∈S

T (2)s |x|

∥∥∥∥∥

2

2

=∥∥∥M (2)

S |x|∥∥∥

2

2.

Thus, replacing x by |x| gives the implication (b) ⇒ (c).Suppose that (d) fails to hold, that is, there exists ε0 > 0 such that for all

x ∈ R[G], x ≥ 0, ‖x‖2 = 1, there exists s0 ∈ S such that ‖x − T(2)s0 x‖2 ≥ ε0.

Since �2(G) is uniformly convex (Lemma I.4.2) there exists δ0 > 0 such that∥∥∥∥∥

x + T(2)s0 x

2

∥∥∥∥∥2

≤ 1 − δ0 (6.75)

for all x ∈ R[G] such that ‖x‖2 = 1. It then follows

‖M (2)S x‖2 =

∥∥∥∥∥1|S|∑

s∈S

T (2)s x

∥∥∥∥∥2

=

∥∥∥∥∥∥2|S|

(x + T

(2)s0 x

2

)+

1|S|

∑

s∈S\{1G,s0}T (2)

s x

∥∥∥∥∥∥2

≤ 2|S|

∥∥∥∥∥x + T

(2)s0 x

2

∥∥∥∥∥2

+1|S|

∑

s∈S\{1G,s0}‖T (2)

s x‖2

(by (6.75)) ≤ 2|S| (1 − δ0) +

1|S| (|S| − 2)

= 1 − 2δ0

|S|

for all x ∈ R[G], such that ‖x‖2 = 1 and x ≥ 0. This clearly contradicts (c).We have shown (c) ⇒ (d).

Let us show (d) ⇒ (e). Suppose that (6.73) holds. Then, for every ε > 0there exists x ∈ R[G] such that


‖x − M(2)S x‖2 =

∥∥∥∥∥x − 1|S|∑

s∈S

T (2)s x

∥∥∥∥∥2

=

∥∥∥∥∥1|S|∑

s∈S

(x − T (2)s x)

∥∥∥∥∥2

≤ 1|S|∑

s∈S

‖x − T (2)s x‖2

≤ ε.

Therefore, by Corollary I.2.3, the linear map I − M(2)S ∈ L(�2(G)) is not

bijective, that is, 1 ∈ σ(M (2)S ).

Finally, the implication (e) ⇒ (a) follow from (I.14) and the fact that‖M (2)

S ‖2→2 ≤ 1 (Proposition 6.12.1). ��

Lemma 6.12.3. Let S ⊂ G be a non-empty finite subset. The following con-ditions are equivalent:

(a) 1 ∈ σ(M (2)S );

(b) 1 ∈ σ(M (2)S∪{1G}).

Proof. If S contains 1G there is nothing to prove.Suppose that 1G /∈ S and set α = |S|

|S|+1 . Note that 0 < α < 1 and that

M(2)S∪{1G} = (1 − α)I + αM

(2)S . (6.76)

We then have I − M(2)S∪{1G} = (1 − α)(I − M

(2)S ) so that I − M

(2)S∪{1G} is

bijective if and only if I −M(2)S is bijective. In other words, 1 ∈ σ(M (2)

S∪{1G})

if and only if 1 ∈ σ(M (2)S ). ��

Before stating and proving the main result of this section we need a lit-tle more work. More precisely, we want to show the equivalence betweenthe existence of an almost S-invariant positive element x ∈ R[G] of norm‖x‖2 = 1 (cf. condition (d) in Proposition 6.12.2) and the existence of analmost S-invariant non-empty finite set F ⊂ G (cf. condition (b) in Proposi-tion 6.10.2).

Let F ⊂ G be a finite set. Denote, as usual, by χF the characteristic mapof F and observe that χF ∈ R[G].

Proposition 6.12.4. Let A, B ⊂ G be two finite sets. Then

‖χA − χB‖1 = ‖χA − χB‖22 = |A \ B| + |B \ A|. (6.77)


Proof. First note that the map χA −χB takes value 1 at each point of A \B,value −1 at each point of B \A, and value 0 everywhere else, that is, outsideof the set (A \ B) ∪ (B \ A) = (G \ (A ∪ B)) ∪ (A ∩ B). We then have

‖χA − χB‖22 =∑

g∈G

|χA(g) − χB(g)|2

=∑

g∈G

|χA(g) − χB(g)|

=∑

g∈G:χA(g)−χB(g)=1

|χA(g) − χB(g)|

+∑

g∈G:χA(g)−χB(g)=−1

|χA(g) − χB(g)|

=∑

g∈G

χA\B(g) +∑

g∈G

χB\A(g)

= |A \ B| + |B \ A|.

��

Lemma 6.12.5. Let x, y ∈ �2(G) such that ‖x‖2 = ‖y‖2 = 1. Then, thefollowing holds:

(i) x2 ∈ �1(G) and ‖x2‖1 = 1;(ii) ‖x2 − y2‖1 ≤ 2‖x − y‖2;

(iii) suppose that x, y ≥ 0. Then, ‖x − y‖2 ≤(‖x2 − y2‖1

) 12 .

Proof. (i) We have ‖x2‖1 =∑

g∈G |x2(g)| = ‖x‖22 = 1.

(ii) We have

‖x2 − y2‖1 =∑

g∈G

|x2(g) − y2(g)|

=∑

g∈G

|x(g) − y(g)| · |x(g) + y(g)|

= 〈|x − y|, |x + y|〉

≤ ‖x − y‖2 · ‖x + y‖2

≤ ‖x − y‖2 · (‖x‖2 + ‖y‖2)

= 2‖x − y‖2,

where the first inequality follows from Cauchy-Schwarz.


(iii) We have

‖x − y‖22 =∑

g∈G

(x(g) − y(g))2

=∑

g∈G

|x(g) − y(g)| · |x(g) − y(g)|

≤∑

g∈G

|x(g) − y(g)| · |x(g) + y(g)|

=∑

g∈G

|x2(g) − y2(g)|

= ‖x2 − y2‖1,

where the inequality follows from the fact that x, y ≥ 0. ��

Lemma 6.12.6. Let F ⊂ G be a non-empty finite set and s ∈ G. Then thefollowing holds:

(i) T(p)s χF = χFs−1 , for p = 1, 2;

(ii) ‖χF ‖1 = ‖χF ‖22 = |F |;

(iii) ‖χF − T(1)s χF ‖1 = ‖χF − T

(2)s χF ‖2

2 = 2|F \ Fs|.

Proof. For g ∈ G and p = 1, 2, one has that (T (s)p χF )(g) = χF (gs) equals 1

if and only if gs ∈ F , that is, if and only if g ∈ Fs−1. This shows (i).Moreover, recalling that χF (g) ∈ {0, 1}, for all g ∈ G, on has

‖χF ‖22 =∑

g∈G

χF (g)2 =∑

g∈G

χF (g) = ‖χF ‖1 = |{g ∈ G : χF (g) = 1}| = |F |.

This shows (ii).Finally, (iii) follows from (i), (6.77) (with A = F and B = Fs−1)

and (6.61). ��

Lemma 6.12.7. Let x ∈ R[G] such that x ≥ 0 and ‖x‖1 = 1. Then thereexist an integer n ≥ 1, nonempty finite subsets Ai ⊂ G and real numbersλi > 0, 1 ≤ i ≤ n, satisfying A1 ⊃ A2 ⊃ · · · ⊃ An and λ1 + λ2 + · · ·+ λn = 1such that

x =n∑

i=1

λiχAi

|Ai|. (6.78)

Proof. Let 0 < α1 < α2 < · · · < αn be the values taken by x. For each1 ≤ i ≤ n, let us set

Ai = {g ∈ G : x(g) ≥ αi}.


Clearly the sets Ai are nonempty finite subsets of G such that A1 ⊃ A2 ⊃· · · ⊃ An. On the other hand, we have

x = α1χA1 + (α2 − α1)χA2 + · · · + (αn − αn−1)χAn

= λ1χA1

|A1|+ λ2

χA2

|A2|+ · · · + λn

χAn

|An|,

by setting λ1 = α1|A1| and λi = (αi − αi−1)|Ai| for 2 ≤ i ≤ n. Thus λi > 0for 1 ≤ i ≤ n and

n∑

i=1

λi = α1|A1| + (α2 − α1)|A2| + · · · + (αn − αn−1)|An|

= α1(|A1| − |A2|) + α2(|A2| − |A3|) + · · · + αn|An|

=∑

g∈G

x(g) = 1.

��

Lemma 6.12.8. With the same notation and hypotheses as in Lemma 6.12.7,we have

‖x − T (1)g x‖1 =

n∑

i=1

λi2|Ai \ Aig|

|Ai|(6.79)

for every g ∈ G.

Proof. Equality (6.78) gives us

x − T (1)g x =

n∑

i=1

λiχAi − T

(1)g χAi

|Ai|

(by Lemma 6.12.6(a)) =n∑

i=1

λiχAi − χAig−1

|Ai|.

As we observed before (cf. the proof of Proposition 6.12.4), the map χAi −χAig−1 takes value 1 at each point of Ai \ Aig

−1, value −1 at each point ofAig

−1 \ Ai, and value 0 everywhere else. Let us set

B =⋃

1≤i≤n

(Ai \ Aig−1) and C =

⋃

1≤i≤n

(Aig−1 \ Ai).

Note that the sets B and C are disjoint. Indeed, for all 1 ≤ i, j ≤ n, we have(Ai \ Aig

−1) ∩ (Aj \ Ajg−1) = ∅ since either Ai ⊂ Aj or Aj ⊂ Ai (which

implies Ajg−1 ⊂ Aig

−1).It follows that


‖x−T (1)g x‖1 =

∑

a∈G

|(x − T (1)g x)(a)|

=∑

a∈G

∣∣∣∣∣

n∑

i=1

λi(χAi − χAig−1)(a)

|Ai|

∣∣∣∣∣

=∑

a∈B

∣∣∣∣∣

n∑

i=1

λi(χAi −χAig−1)(a)

|Ai|

∣∣∣∣∣+∑

a∈C

∣∣∣∣∣

n∑

i=1

λi(χAi −χAig−1)(a)

|Ai|

∣∣∣∣∣

=n∑

i=1

λi|Ai \ Aig

−1||Ai|

+n∑

i=1

λi|Aig

−1 \ Ai||Ai|

(by (6.61)) =n∑

i=1

λi2|Ai \ Aig|

|Ai|.

��

Let G be a finitely generated group and let S be a finite (not necessarilysymmetric) generating subset of G. Consider the combinatorial LaplacianΔS : R

G → RG (cf. Example 1.4.3(b)). For all x ∈ �2(G) ⊂ R

G and g ∈ G wehave

(ΔSx)(g) = |S|x(g) −∑

s∈S

x(gs)

= |S|x(g) −∑

s∈S

(T (2)s x)(g)

= |S|(I − M(2)S )(x)(g).

This shows that for the map Δ(2)S = ΔS |2(G) one has

Δ(2)S = |S|

(I − M

(2)S

)(6.80)

and therefore Δ(2)S ∈ L(�2(G)).

Let λ ∈ R. From (6.80) we deduce that

λI − Δ(2)S = λI − |S|(I − M

(2)S ) = −|S|

((1 − λ

|S|

)I − M

(2)S

).

It follows that

λ ∈ σ(Δ(2)S ) ⇔

(1 − λ

|S|

)∈ σ(M (2)

S )

and therefore0 ∈ σ(Δ(2)

S ) ⇔ 1 ∈ σ(M (2)S ). (6.81)

Theorem 6.12.9 (Kesten-Day). Let G be a finitely generated group. LetS ⊂ G be a finite (not necessarily symmetric) generating subset of G. Then


the following conditions are equivalent:

(a) G is amenable;(b) 0 ∈ σ(Δ(2)

S ).

Proof. Suppose (a). By Theorem 4.9.1 we have that G satisfies the Følnerconditions. Fix ε > 0. By Proposition 6.10.2 there exists a finite subset F ⊂ G

such that |F \ Fs| < ε2

2 |F | for all s ∈ S. Set x = 1√|F |

χF and note that

‖x‖2 = 1. We have

‖(I − M(2)S )x‖2 = ‖x − 1

|S|∑

s∈S

T (2)s x‖2

=1|S| ‖

∑

s∈S

(x − T (2)s x)‖2

≤ 1|S|∑

s∈S

‖x − T (2)s x‖2

=1

|S| ·√|F |∑

s∈S

‖χF − T (2)s χF ‖2

(by Lemma 6.12.6(iii)) =1

|S| ·√|F |∑

s∈S

2|F \ Fs| 12

=1|S|∑

s∈S

2(|F \ Fs|

|F |

) 12

=2

|F | 12ε

12

< ε.

From Corollary I.2.3 we deduce that I − M(2)S is not bijective, that is, 1 ∈

σ(M (2)S ). From (6.81) we deduce that 0 ∈ σ(Δ(2)

S ). Thus (a) implies (b).To prove the converse implication, first observe that, by virtue of Propo-

sition 6.10.2, in order to prove (a), it suffices to show that given any ε > 0there exists a finite subset F ⊂ G such that

|F \ Fs| < ε|F | for all s ∈ S. (6.82)

Fix ε > 0. Suppose (b) and observe that by (6.81) we have 1 ∈ σ(M (2)S ).

Also note that, by Lemma 6.12.3, we can suppose that S � 1G. Thus, byvirtue of the implication (e) ⇒ (c) in Proposition 6.12.2, we can find a non-negative function xε ∈ R[G] such that ‖xε‖2 = 1 and ‖xε − T

(2)s xε‖2 ≤ ε

2|S|for all s ∈ S. Setting x = x2

ε, from Lemma 6.12.5 we deduce that ‖x‖1 = 1and

‖x − T (1)s x‖1 ≤ ε

|S| for all s ∈ S. (6.83)


By Lemma 6.12.7, there exist an integer n ≥ 1, nonempty finite subsetsAi ⊂ G and real numbers λi > 0, 1 ≤ i ≤ n, satisfying A1 ⊃ A2 ⊃ · · · ⊃ An

and λ1 + λ2 + · · · + λn = 1, such that x =∑n

i=1 λiχAi

|Ai| .Set Ω = {1, 2, . . . , n} and consider the unique probability measure μ on Ω

such that μ({i}) = λi for every i ∈ Ω. Finally, for each g ∈ G, let Ωg denotethe subset of Ω defined by

Ωg ={

i ∈ Ω :|Ai \ Aig|

|Ai|≥ ε

}.

It follows from Lemma 6.12.8 that

‖x − T (1)g x‖1 =

∑

i∈Ω

λi2|Ai \ Aig|

|Ai|

≥∑

i∈Ωg

λi2|Ai \ Aig|

|Ai|

≥ 2ε∑

i∈Ωg

λi

= 2εμ(Ωg).

Therefore, we have

μ(Ωg) ≤‖x − T

(1)g x‖1

2εfor all g ∈ G.

By using (6.83), we deduce

μ(Ωs) <1|S| for all s ∈ S,

which implies

μ

(⋃

s∈S

Ωs

)≤∑

s∈S

μ(Ωs) < 1.

Thus ⋃

s∈S

Ωs �= Ω.

This means that there is some i0 ∈ Ω such that

|Ai0 \ Ai0s||Ai0 |

< ε for all s ∈ S.

Thus, in order to satisfy (6.82), we can take F = Ai0 . ��


6.13 Quasi-Isometries

Let G and H be two groups.

Definition 6.13.1. A map ϕ : G → H is called a quasi-isometric embeddingof G into H if the following two conditions hold:

(i) for every finite subset K ⊂ G there exists a finite subset F ⊂ H suchthat

g−11 g2 ∈ K =⇒ ϕ(g1)−1ϕ(g2) ∈ F for all g1, g2 ∈ G; (6.84)

(ii) for every finite subset F ⊂ H there exists a finite subset K ⊂ G suchthat

ϕ(g1)−1ϕ(g2) ∈ F =⇒ g−11 g2 ∈ K for all g1, g2 ∈ G. (6.85)

Definition 6.13.2. A quasi-isometry from G to H is a quasi-isometric em-bedding ϕ of G into H for which there exists a finite subset C ⊂ H suchthat

ϕ(G)C = H, (6.86)

that is, for all h ∈ H there exists g ∈ G and c ∈ C such that

h = ϕ(g)c. (6.87)

If such a quasi-isometry from G to H exists then one says that G is quasi-isometric to H.

Example 6.13.3. Let G and H be two groups and let ϕ : G → H be aninjective homomorphism. Then ϕ is a quasi-isometric embedding. Indeed,given a finite subset K ⊂ G, the set F = ϕ(K) is a finite subset ofH. On the other hand, if g1, g2 ∈ G satisfy g−1

1 g2 ∈ K, then we haveϕ(g1)−1ϕ(g2) = ϕ(g−1

1 g2) ∈ F . Conversely, suppose that F ′ is a finite subsetof H and consider the set K ′ = ϕ−1(F ′) ⊂ G. Note that |K ′| ≤ |F ′| < ∞,since ϕ is injective. On the other hand, if g1, g2 ∈ G satisfy ϕ(g1)−1ϕ(g2) ∈ F ′,then we have ϕ(g−1

1 g2) = ϕ(g1)−1ϕ(g2) ∈ F ′ so that g−11 g2 ∈ K ′. This shows

that ϕ is a quasi-isometric embedding.Moreover, ϕ is a quasi-isometry if and only if the index of the image

subgroup ϕ(G) is finite in H. Indeed, suppose first that [H : ϕ(G)] < ∞ andlet R ⊂ H be a set of representatives for the right cosets of ϕ(G) in H. Notethat R is finite and

H =∐

h∈R

ϕ(G)h = ϕ(G)R. (6.88)

Thus, ϕ is a quasi-isometry. Conversely, if C ⊂ H is a finite set such thatH = ϕ(G)C one clearly has [H : ϕ(G)] = |H/ϕ(G)| ≤ |C| < ∞.

6.13 Quasi-Isometries 205

Remark 6.13.4. Let G and H be two groups. Observe that if ψ : G → H isa quasi-isometric embedding (resp. a quasi-isometry) and h0 ∈ H, then themap ϕ : G → H defined by ϕ(g) = h0ψ(g) for all g ∈ G is a quasi-isometricembedding (resp. a quasi-isometry). Indeed, ϕ(g1)−1ϕ(g2) = ψ(g1)−1ψ(g2)for all g1, g2 ∈ G and, if C ⊂ H is a finite set such that ψ(G)C = H, thenϕ(G)C = h0ψ(G)C = h0H = H. Note that taking h0 = ψ(1G)−1 we haveϕ(1G) = 1H . It follows that if there exists a quasi-isometric embedding ofG into H, then there is a quasi-isometric embedding ϕ : G → H such thatϕ(1G) = 1H . Similarly, if G is quasi-isometric to H, one can find a quasi-isometry ϕ : G → H such that ϕ(1G) = 1H .

Proposition 6.13.5. Let G, H and L be three groups and let ϕ : G → H andψ : H → L be two quasi-isometric embeddings (resp. quasi-isometries). Thenthe composition map ψ ◦ ϕ : G → L is a quasi-isometric embedding (resp.quasi-isometry).

Proof. Since ϕ is a quasi-isometric embedding, given a finite subset K ⊂ G,we can find a finite subset E ⊂ H such that if g−1

1 g2 ∈ K, g1, g2 ∈ G,then ϕ(g1)−1ϕ(g2) ∈ E. Similarly, as ψ is a quasi-isometric embedding,we can find a finite subset F ⊂ L such that if h−1

1 h2 ∈ E, h1, h2 ∈ H,then ψ(h1)−1ψ(h2) ∈ F . It follows that if g−1

1 g2 ∈ K, g1, g2 ∈ G, thenψ(ϕ(g1))−1ψ(ϕ(g2)) ∈ F . In the same way one proves that given a finite sub-set F ⊂ L one can find a finite set K ⊂ G such that if ψ(ϕ(g1))−1ψ(ϕ(g2)) ∈F , g1, g2 ∈ G, then g−1

1 g2 ∈ K. This shows that ψ ◦ ϕ is a quasi-isometricembedding.

Let now C ⊂ H and D ⊂ L be two finite sets such that ϕ(G)C = H andψ(H)D = L. As ψ is a quasi-isometric embedding there exists a finite setF ⊂ L such that h−1

1 h2 ∈ C implies ψ(h1)−1ψ(h2) ∈ F for all h1, h2 ∈ H.Let us show that

ψ(ϕ(G))FD = L. (6.89)

Given � ∈ L there exists h ∈ H such that

� ∈ ψ(h)D. (6.90)

Let also g ∈ G and c ∈ C be such that h = ϕ(g)c. Set h′ = ϕ(g) ∈ H and ob-serve that (h′)−1h = c ∈ C. It follows that ψ(ϕ(g))−1ψ(h) = ψ(h′)−1ψ(h) ∈F , that is,

ψ(h) ∈ ψ(ϕ(g))F. (6.91)

From (6.90) and (6.91) we deduce � ∈ ψ(ϕ(g))FD. This shows (6.89). Itfollows that ψ ◦ ϕ : G → L is a quasi-isometry. ��

Corollary 6.13.6. Let G and H be two groups and let ϕ : G → H be a quasi-isometric embedding. Let L ⊂ G be a subgroup of G. Then the restriction mapϕ|L : L → H is a quasi-isometric embedding.


Proof. From Example 6.13.3 we deduce that the inclusion map ι : L → G isa quasi-isometric embedding. Since ϕ|L = ϕ ◦ ι, the statement follows fromProposition 6.13.5. ��

Proposition 6.13.7. Quasi-isometry is an equivalence relation in the classof groups.

Proof. Every group is quasi-isometric to itself. Indeed, for any group G, theidentity map IdG : G → G is clearly a quasi-isometry (just take F = Kin (6.84) and (6.84), and C = {1G} in (6.86)).

Suppose that a group G is quasi-isometric to a group H. Let ϕ : G → Hbe a quasi-isometry and let C ⊂ H be a finite subset satisfying (6.86). Wedefine a map ϕ′ : H → G as follows. For every h ∈ H we can find c ∈ C andg ∈ G such that (6.87) holds. We then set ϕ′(h) = g. Let us show that ϕ′ isa quasi-isometry.

Let h1, h2 ∈ H. We set gi = ϕ′(hi) ∈ G and ci = ϕ(gi)−1hi ∈ C, i = 1, 2.Note that hi = ϕ(gi)ci, equivalently ϕ(gi) = hic

−1i , i = 1, 2.

Let now F ⊂ H be a finite subset and suppose that h−11 h2 ∈ F . Consider

the finite set F ′ = CFC−1 ⊂ H. As ϕ is a quasi-isometric embedding, we canfind a finite subset K ⊂ G such that if ϕ(g1)−1ϕ(g2) ∈ F ′ then g−1

1 g2 ∈ K.But ϕ(g1)−1ϕ(g2) = c1h

−11 h2c

−12 ∈ CFC−1 = F ′ so that ϕ′(h1)−1ϕ′(h2) =

g−11 g2 ∈ K. Conversely, let K ⊂ G be a finite subset and suppose that

ϕ′(h1)−1ϕ′(h2) = g−11 g2 belongs to K. As ϕ is a quasi-isometric embedding,

we can find a finite set F ′ ⊂ H such that if g−11 g2 ∈ K then ϕ(g1)−1ϕ(g2) ∈

F ′. Set F = C−1F ′C. It follows that

h−11 h2 = c−1

1 ϕ(g1)−1ϕ(g2)c2 ∈ c−11 F ′c2 ⊂ F.

This shows that ϕ′ is a quasi-isometric embedding of H into G.Now, as ϕ is a quasi-isometric embedding, we can find a finite set K ⊂ G

such that if ϕ(g1)−1ϕ(g2) ∈ C then g−11 g2 ∈ K. Let us show that

G = ϕ′(H)K. (6.92)

Let g ∈ G. Set h = ϕ(g) ∈ H and g′ = ϕ′(h) ∈ G. Note that by (6.86)there exists c ∈ C such that h = ϕ(g′)c. Now ϕ(g′)−1ϕ(g) = (ch−1)h =c ∈ C and therefore (g′)−1g ∈ K, that is, g ∈ g′K = ϕ′(h)K ⊂ ϕ′(H)K.This shows (6.92). We deduce that ϕ′ is a quasi-isometry. It follows that thesymmetric property of quasi-isometry holds true.

Finally, the transitivity property of quasi-isometry was proved in Propo-sition 6.13.5. ��

As a consequence, if a group G is quasi-isometric to a group H then H isquasi-isometric to G and we simply say that G and H are quasi-isometric.

Proposition 6.13.8. Let G be a group and let N be a finite normal subgroupof G. Then G and G/N are quasi-isometric.


Proof. Let ϕ : G → G/N denote the quotient homomorphism. Let us showthat ϕ is a quasi-isometry. Given a finite subset K ⊂ G, the set F = ϕ(K) isa finite subset of G/N . On the other hand, if g1, g2 ∈ G satisfy g−1

1 g2 ∈ K,then we have ϕ(g1)−1ϕ(g2) = ϕ(g−1

1 g2) ∈ F . Conversely, suppose that F ′ isa finite subset of G/N and consider the set K ′ = ϕ−1(F ′) ⊂ G. Note that|K ′| = |N | · |F ′| < ∞. On the other hand, if g1, g2 ∈ G satisfy ϕ(g1)−1ϕ(g2) ∈F ′, then we have ϕ(g−1

1 g2) = ϕ(g1)−1ϕ(g2) ∈ F ′ so that g−11 g2 ∈ K ′. This

shows that ϕ is a quasi-isometric embedding. Since ϕ is surjective, we deducethat ϕ is indeed a quasi-isometry. ��

Proposition 6.13.9. Let G be a group and let H be a subgroup of finite indexof G. Then G and H are quasi-isometric.

Proof. It follows from Example 6.13.3 that the inclusion map ι : H → G is aquasi-isometry. ��

Corollary 6.13.10. If two groups are commensurable then they are quasi-isometric. ��

Proposition 6.13.11. Let G and H be two groups. Suppose that G and Hare quasi-isometric and that G is finitely generated. Then H is finitely gen-erated.

Proof. Let ϕ : G → H by a quasi-isometry and let C ⊂ H be a finite subsetsuch that H = ϕ(G)C. Let S ⊂ G be a finite symmetric generating subset ofG. Since ϕ is a quasi-isometric embedding, we can find a finite subset F ⊂ Hsuch that ϕ(g1)−1ϕ(g2) ∈ F whenever g1, g2 ∈ G satisfy g−1

1 g2 ∈ S. Let usshow that the set T ⊂ H defined by T = {ϕ(1G)} ∪ F ∪ C is a generatingsubset for H. Let h ∈ H. As H = ϕ(G)C, we can find g ∈ G and c ∈ C suchthat h = ϕ(g)c. Since S is a symmetric generating subset of G, there existan integer n ≥ 0 and elements s1, s2, . . . , sn ∈ S such that g = s1s2 · · · sn.Consider the elements g0, g1, . . . , gn ∈ G defined by g0 = 1G and gi = gi−1si

for all 1 ≤ i ≤ n. Observe that gn = g and that ϕ(gi−1)−1ϕ(gi) ∈ F , for all1 ≤ i ≤ n, since g−1

i−1gi = si ∈ S. Writing

h = ϕ(g)c = ϕ(1G)(ϕ(g0)−1ϕ(g1)(ϕ(g1)−1ϕ(g2) · · · (ϕ(gn−1)−1ϕ(gn))c,

we deduce that T generates H. As T is finite, this shows that H is finitelygenerated. ��

In the two following propositions we characterize quasi-isometric embed-dings and quasi-isometries for finitely generated groups in terms of the wordmetrics associated with finite symmetric generating subsets of the groups.

Proposition 6.13.12. Let G and H be two finitely generated groups anddenote by dG and dH the word metric associated with two finite symmetricgenerating subsets SG and SH for G and H respectively. Let ϕ : G → H be amap. Then the following conditions are equivalent:


(a) ϕ is an isometric embedding;(b) there exist constants α ≥ 1 and β ≥ 0 such that, for all g, g′ ∈ G,

1α

dG(g, g′) − β ≤ dH(ϕ(g), ϕ(g′)) ≤ αdG(g, g′) + β. (6.93)

Proof. Suppose (a). Let K ⊂ G (resp. F ⊂ H) be a finite set such thatg−11 g2 ∈ K (resp. ϕ(g1)−1ϕ(g) ∈ F ) whenever ϕ(g1)−1ϕ(g) ∈ SH (resp.

g−11 g2 ∈ SG) for all g1, g2 ∈ G. Also let K ′ ⊂ G be a finite set such that

g−11 g2 ∈ K ′ whenever ϕ(g1)−1ϕ(g) ∈ {1H}, equivalently, ϕ(g1) = ϕ(g2), for

all g1, g2 ∈ G. We set

α = max {diamdG(K), diamdH

(F )} (6.94)

andβ =

1α

diamdG(K ′), (6.95)

where we denoted by diamdG(K) = max{dG(k1, k2) : k1, k2 ∈ K} (resp.

diamdH(F ) = max{dH(f1, f2) : f1, f2 ∈ F}) the diameter of K ⊂ G (resp.

F ⊂ H).Let now g, g′ ∈ G. Note that (6.93) trivially holds if g = g′. Suppose

first that dG(g, g′) = 1. Thus g−1g′ ∈ SG and therefore ϕ(g)−1ϕ(g′) ∈ Fso that dH(ϕ(g), ϕ(g′)) ≤ α. Suppose now that dG(g, g′) = n ≥ 1 and letg0, g1, . . . , gn ∈ G be such that g0 = g, gn = g′ and dG(gi, gi+1) = 1 for alli = 0, 1, . . . , n − 1. Using the triangular inequality we have:

dH(ϕ(g), ϕ(g′)) = dH(ϕ(g0), ϕ(gn))

≤n−1∑

i=0

dH(ϕ(gi), ϕ(gi+1))

≤ nα

= αdG(g, g′).

This shows thatdH(ϕ(g), ϕ(g′)) ≤ αdG(g, g′) (6.96)

for all g, g′ ∈ G and the second inequality in (6.93) follows.Arguing as above, we deduce that for all n ≥ 1 the condition dH(ϕ(g),

ϕ(g′)) = n implies dG(g, g′) ≤ nα. Thus, provided that dH(ϕ(g), ϕ(g′)) �= 0,we have dG(g, g′) ≤ αdH(ϕ(g), ϕ(g′)), equivalently

1α

dG(g, g′) ≤ dH(ϕ(g), ϕ(g′)). (6.97)

On the other hand, if ϕ(g) = ϕ(g′), equivalently, ϕ(g)−1ϕ(g′) ∈ {1H}, theng−1g′ ∈ K ′ and from (6.95) we deduce that 1

αdG(g, g′) ≤ β. It follows that


1α

dG(g, g′) − β ≤ 0 = dH(ϕ(g), ϕ(g′)). (6.98)

From (6.97) (for ϕ(g) �= ϕ(g′)) and (6.98) (for ϕ(g) = ϕ(g′)) we deduce that(6.98) holds for all g, g′ ∈ G. Thus also the first inequality in (6.93) is proved.This shows (a) ⇒ (b).

Conversely, suppose (b). Let K ⊂ G be a finite subset. Consider thefinite set F = BH

SH(1H , δ) ⊂ H where δ = α max{dG(1G, k) : k ∈

K} + β. Let g1, g2 ∈ G and suppose that g−11 g2 ∈ K. Using (6.93) we de-

duce that dH(1H , ϕ(g1)−1ϕ(g2)) = dH(ϕ(g1), ϕ(g2)) ≤ αdG(g1, g2) + β =αdG(1G, g−1

1 g2) + β ≤ δ, so that ϕ(g1)−1ϕ(g2) ∈ BHSH

(1H , δ) = F . On theother hand, let F ⊂ H be a finite set. Consider the finite set K = BG

SG(1G, δ′)

where δ′ = α max{dH(1H , f) : f ∈ F} + αβ. Let g1, g2 ∈ G and sup-pose that ϕ(g1)−1ϕ(g2) ∈ F . From (6.93) it follows that dG(1G, g−1

1 g2) =dG(g1, g2) ≤ αdH(ϕ(g1), ϕ(g2)) + αβ = αdH(1H , ϕ(g1)−1ϕ(g2)) + αβ ≤ δ′,so that g−1

1 g2 ∈ BGSG

(1G, δ′) = K. This shows that ϕ is a quasi-isometricembedding and the implication (b) ⇒ (a) follows. ��

Proposition 6.13.13. Let G and H be two finitely generated groups anddenote by dG and dH the word metric associated with two finite symmetricgenerating subsets SG and SH for G and H respectively. Let ϕ : G → H bean isometric embedding. Then the following conditions are equivalent:

(a) ϕ is a quasi-isometry;(b) there exists δ > 0 such that for every h ∈ H there exists g ∈ G such that

dH(ϕ(g), h) ≤ δ. (6.99)

Proof. Suppose that ϕ is a quasi-isometric embedding. Let C ⊂ H be a finitesubset such that H = ϕ(H)C and set δ = max{dH(1H , c) : c ∈ C}. Leth ∈ H, then there exist c ∈ C and g ∈ G such that h = ϕ(g)c, that is,ϕ(g)−1h = c ∈ C. We deduce that dH(ϕ(g), h) ≤ δ and (6.99) follows. Thisshows (a) ⇒ (b).

Conversely, suppose (b) and set C = BSG(1G, δ) ⊂ G. Let h ∈ H. By (6.99)

there exists g ∈ G such that dH(ϕ(g), h) ≤ δ. It follows that ϕ(g)−1h ∈ C,equivalently h ∈ ϕ(g)C. This shows that H = ϕ(G)C. Therefore ϕ is a quasi-isometry and (a) follows. ��

Proposition 6.13.14. Let G and H be two finitely generated groups. Supposethat there is a quasi-isometric embedding ϕ : G → H. Then one has γ(G) �γ(H).

Proof. By Remark 6.13.4, we can assume ϕ(1G) = 1H . Denote by dG and dH

the word metric associated with two finite symmetric generating subsets SG

and SH for G and H respectively. By Proposition 6.13.12, there exist integersα ≥ 1 and β ≥ 0 such that

1α

dG(g, g′) − β ≤ dH(ϕ(g), ϕ(g′)) ≤ αdG(g, g′) + β (6.100)


for all g, g′ ∈ G. Note that the left inequality in (6.100) implies that ϕ(g) �=ϕ(g′) whenever g, g′ ∈ G satisfy dG(g, g′) ≥ αβ+1. For n ∈ N, let BG

SG(n) ⊂ G

(resp. BHSH

(n) ⊂ H) denote the ball of radius n centered at 1G (resp. 1H).Choose a subset En ⊂ BG

SG(n) of maximal cardinality such that dG(g, g′) ≥

αβ + 1 for all g, g′ ∈ En. Setting C = |BGSG

(αβ + 1)|, we have

|BGSG

(n)| ≤ C|En| (6.101)

for all n ∈ N . Indeed, the balls of radius C centered at the elements of En

cover BGSG

(n) by the maximality of En. Now observe that the images by ϕ ofthe elements of En are all distinct and belong to BH

SH(αn + β) by the right

inequality in (6.100). This implies that

|En| ≤ |BHSH

(αn + β)|.

By using (6.101), we then deduce that

|BGSG

(n)| ≤ C|BHSH

(αn + β)| ≤ C|BHSH

(β)||BHSH

(αn)| ≤ C ′|BHSH

(C ′n)|

for all n ≥ 1, where C ′ = Cα|BHSH

(β)|. It follows that γ(G) � γ(H). ��

Corollary 6.13.15. If two finitely generated groups are quasi-isometric, thenthey have the same growth type.

Proof. If G and H are quasi-isometric finitely generated groups, then thereexist a quasi-isometric embedding from G into H and a quasi-isometric em-bedding from H into G. It follows that γ(G) � γ(H) and γ(H) � γ(G). Thuswe have γ(G) = γ(H). ��

Corollary 6.13.16. Let G and H be two quasi-isometric finitely generatedgroups. Then G has exponential (resp. subexponential, resp. polynomial, resp.intermediate) growth if and only if H has exponential (resp. subexponential,resp. polynomial, resp. intermediate) growth. ��

An important property of quasi-isometric embeddings is the fact that everyquasi-isometric embedding is uniformly finite-to-one:

Proposition 6.13.17. Let G and H be two groups. Let ϕ : G → H be aquasi-isometric embedding. Then there exists an integer M ≥ 1 such that|ϕ−1(h)| ≤ M for all h ∈ H.

Proof. Since ϕ is a quasi-isometric embedding, we can find a finite set K ⊂ Gsuch that g−1

1 g2 ∈ K whenever g1, g2 ∈ G satisfy ϕ(g1)−1ϕ(g2) ∈ {1H}.Let us set M = |K|. Let h ∈ H. Suppose that g0 ∈ ϕ−1(h). Then, everyg ∈ ϕ−1(h) satisfies ϕ(g0)−1ϕ(g) = h−1h = 1H and therefore g−1

0 g ∈ K.Thus, we have ϕ−1(h) ⊂ g0K and hence |ϕ−1(h)| ≤ |g0K| = |K| = M . ��

Corollary 6.13.18. Let G and H be two groups. Suppose that there exists aquasi-isometric embedding ϕ : G → H and that H is finite. Then G is finite.


Proof. Taking M as in the preceding proposition, we have |G| ≤ M |H|. ��

Corollary 6.13.19. Let G and H be two groups. Suppose that H is finite.Then G is quasi-isometric to H if and only if G is finite.

Proof. If G is finite then it is clear from the definition that any map fromG to H is a quasi-isometry. On the other hand, if G is quasi-isometric to Hthen G is finite by Corollary 6.13.18. ��

Proposition 6.13.20. Let G and H be two groups. Suppose that there existsa quasi-isometric embedding ϕ : G → H and that H is locally finite. Then Gis locally finite.

Proof. Let K be a finitely generated subgroup of G and let S ⊂ K be a finitesymmetric generating subset of K. Since ϕ is a quasi-isometric embedding,we can find a finite subset F ⊂ H such that ϕ(g1)−1ϕ(g2) ∈ F wheneverg1, g2 ∈ G satisfy g−1

1 g2 ∈ S. Let L denote the subgroup of H generatedby {ϕ(1G)} ∪ F . Suppose that k ∈ K. Since S is a symmetric generatingsubset of K, there exist an integer n ≥ 0 and elements s1, s2, . . . , sn ∈ Ssuch that k = s1s2 · · · sn. Consider the elements k0, k1, . . . , kn ∈ K definedby k0 = 1G and ki = ki−1si for all 1 ≤ i ≤ n. Observe that kn = k and thatϕ(ki−1)−1ϕ(ki) ∈ F , for all 1 ≤ i ≤ n, since k−1

i−1ki = si ∈ S. Thus, we have

ϕ(k) = ϕ(1G)(ϕ(k0)−1ϕ(k1))(ϕ(k1)−1ϕ(k2)) · · · (ϕ(kn−1)−1ϕ(kn)) ∈ L.

It follows that ϕ(K) ⊂ L. As H is locally finite, the subgroup L is finite.On the other hand, it follows from Proposition 6.13.17 that there exists aninteger M ≥ 1 such that |ϕ−1(h)| ≤ M for all h ∈ H. We deduce that|K| ≤ M |L| < ∞. This shows that G is locally finite. ��

Corollary 6.13.21. Let G and H be two groups. Suppose that G and H arequasi-isometric and that the group G is locally finite. Then H is locally finite.

��

Lemma 6.13.22. Let G be a group. Let E, Ω and C be three subsets of G.Then, for every c0 ∈ C one has

∂Ec0(Ω) = ∂E(Ωc−10 ). (6.102)

and∂EC(Ω) ⊃ ∂Ec0(Ω). (6.103)

Proof. By (5.3) we have Ω+Ec0 =⋃

e∈E Ω(ec0)−1 =⋃

e∈E Ωc−10 e−1 =

(Ωc−10 )+E and from (5.2) we deduce that Ω−Ec0 =

⋂e∈E Ω(ec0)−1 =⋂

e∈E Ωc−10 e−1 = (Ωc−1

0 )−E . It follows that ∂Ec0(Ω) = Ω+Ec0 \ Ω−Ec0 =(Ωc−1

0 )+E \ (Ωc−10 )−E = ∂E(Ωc−1

0 ). Similarly, we have


Ω+EC =⋃

e∈Ec∈C

Ω(ec)−1 =⋃

e∈Ec∈C

Ωc−1e−1 ⊃⋃

e∈E

Ωc−10 e−1 = (Ωc−1

0 )+E

and

Ω−EC =⋂

e∈Ec∈C

Ω(ec)−1 =⋂

e∈Ec∈C

Ωc−1e−1 ⊂⋂

e∈E

Ωc−10 e−1 = (Ωc−1

0 )+E .

We deduce that ∂EC(Ω) = Ω+EC \ Ω−EC ⊃ (Ωc−10 )+E \ (Ωc−1

0 )−E =∂E(Ωc−1

0 ) = ∂Ec0(Ω), where the last equality follows from (6.102). ��

Theorem 6.13.23. Let G and H be two quasi-isometric groups. Suppose thatH is amenable. Then G is amenable.

Proof. Let ϕ : G → H be a quasi-isometry and let C ⊂ H be a finite set suchthat

H = ϕ(G)C =⋃

c∈C

ϕ(G)c. (6.104)

Let EG ⊂ G be a finite set and ε > 0. Let us show that there exists a finiteset FG ⊂ G such that

|∂EG(FG)| < ε|FG|. (6.105)

Since ϕ is a quasi-isometric embedding, we can find a finite set E′H ⊂ H such

thatg−11 g2 ∈ EG ⇒ ϕ(g1)−1ϕ(g2) ∈ E′

H (6.106)

for all g1, g2 ∈ G. We then set EH = E′HC. Also, by Proposition 6.13.17 we

can find an integer M ≥ 1 such that

|ϕ−1(F )| ≤ M |F | (6.107)

for all finite sets F ⊂ H. Since H is amenable, it follows from Corollary 5.4.5that we can find a finite subset F ′

H ⊂ H such that

|∂EH(F ′

H)| <ε

M |C| |F′H |. (6.108)

By (6.104) we can find c0 ∈ C such that

F ′Hc−1

0 ∩ ϕ(G) �= ∅ (6.109)

and|F ′

Hc−1 ∩ ϕ(G)| ≤ |F ′Hc−1

0 ∩ ϕ(G)| (6.110)

for all c ∈ C. Set FH = F ′Hc−1

0 ⊂ H and FG = ϕ−1(FH) ⊂ G. Note thatFG �= ∅ by (6.109). Then we have

ϕ(FG) = FH ∩ ϕ(G) ⊂ FH (6.111)


andϕ(G \ FG) ⊂ H \ FH . (6.112)

Moreover,|F ′

H | ≤ |C| · |FG|. (6.113)

Indeed, from (6.104) we deduce that F ′H =

⋃c∈C(F ′

H ∩ ϕ(G)c) so that

|F ′H | ≤

∑

c∈C

|F ′H ∩ ϕ(G)c|

=∑

c∈C

|F ′Hc−1 ∩ ϕ(G)|

≤∑

c∈C

|F ′Hc−1

0 ∩ ϕ(G)| (by (6.110))

= |C| · |FH ∩ ϕ(G)|

≤ |C| · |FG| (by the equality in (6.111)).

Let us show that the set FG ⊂ G has the required property. Suppose thatg ∈ ∂EG

(FG). This means that the set gEG meets both FG and G\FG. Thus,there exist g1 ∈ FG and g2 ∈ G \ FG such that g−1g1 ∈ EG and g−1g2 ∈ EG.This implies ϕ(g)−1ϕ(g1) ∈ E′

H and ϕ(g)−1ϕ(g2) ∈ E′H by applying (6.106).

As ϕ(g1) ∈ FH (by (6.111)) and ϕ(g2) ∈ H \ FH (by (6.112)), we deducethat the set ϕ(g)E′

H meets both FH and H \ FH . In other words, we haveϕ(g) ∈ ∂E′

H(FH). This shows that

∂EG(FG) ⊂ ϕ−1(∂E′

H(FH)). (6.114)

By taking cardinalities, we finally get

|∂EG(FG)| ≤ |ϕ−1(∂E′

H(FH))| (by (6.114))

≤ M |∂E′H

(FH)| (by (6.107))

= M |∂E′Hc0(F

′H)| (by (6.102))

≤ M |∂EH(F ′

H)| (by (6.103))

≤ ε

|C| |F′H | (by (6.108))

≤ ε|FG| (by (6.113)).

This shows that FG satisfies (6.105). From Corollary 5.4.5 we deduce that Gis amenable. ��


Notes

The idea of looking at a finitely generated group with a geometer’s eye byinvestigating the properties of the family consisting of all its word metricsis due to M. Gromov ([Gro1, Gro3, Gro4]) and gave birth to the flourishingbranch of mathematics which is commonly known as geometric group theoryin the late 1970s.

In the 1950s, the notion of growth of a finitely generated group arose ingroup theory in relation to volume growth in Riemannian manifolds. Thisline of study was initiated by V.A. Efremovich [Efr] and A.S. Svarc [Sva]in the USSR and, slightly later and completely independently, by J. Milnor[Mil1] and J.A. Wolf [Wol] in the USA. In [Mil1] Milnor proved that funda-mental groups of closed Riemannian manifolds with negative sectional curva-ture have exponential growth. Wolf [Wol] proved that a polycyclic group haspolynomial growth if it contains a nilpotent subgroup of finite index and hasexponential growth otherwise. Then, Milnor [Mil2] proved that every finitelygenerated non-polycyclic solvable group has exponential growth. Finally, in1972 H. Bass [Bas] showed that the growth of a nilpotent group G with afinite symmetric generating subset S is exactly polynomial in the sense thatthere are positive constants C1 and C2 such that C1n

d ≤ γS(n) ≤ C2nd, for

all n ≥ 1, where d = d(G) ≥ 0 is an integer which can be computed explicitlyfrom the lower central series of G (see [Har1, page 201] for more informa-tion on the history and prehistory of these results). Note that the growthestimates of Milnor and Bass imply that a finitely generated solvable grouphas either polynomial or exponential growth. It was shown by J. Tits [Tits](see also [Har1]) that every finitely generated linear group either is virtuallynilpotent or contains a free subgroup of rank two. This last result, which isknown as the Tits alternative for linear groups, implies that every finitelygenerated linear group has either polynomial growth or exponential growth.

The problem of the characterization of finitely generated groups with poly-nomial growth remained open until Gromov proved in [Gro2] that a finitelygenerated group with polynomial growth contains a nilpotent subgroup offinite index. It follows from the above mentioned result of Bass and Proposi-tion 6.6.6 that a group of polynomial growth has in fact exactly polynomialgrowth. Thus, for finitely generated groups, the notions of polynomial andexactly polynomial growth coincide.

The (general) Burnside problem, posed by W. Burnside in 1902, askedwhether a finitely generated periodic group is necessarily finite. It was an-swered in the negative in 1964 by E.S. Golod and I.R. Shafarevich [GolS],who gave an example of a finitely generated infinite p-group. The Grigorchukgroup, also known as the first Grigorchuk group, was originally constructedby R. I. Grigorchuk in 1980 [Gri2] as a new example of a finitely gener-ated infinite periodic group, thus providing another counterexample to thegeneral Burnside problem. In 1984 Grigorchuk [Gri4] proved that this grouphas intermediate growth (this was announced by Grigorchuk in 1983 [Gri3]),

Notes 215

thus providing a positive answer to the Milnor problem, posed by Milnorin 1968, about the existence of finitely generated groups of intermediategrowth. More precisely, in [Gri4] Grigorchuk proved, among other things,that exp(

√n) � γ(G) � exp(ns), where s = log32(31) ≈ 0.991. The Grig-

orchuk group also provides the first example of an amenable but not ele-mentary amenable group (the class of elementary amenable groups is thesmallest class of groups containing all finite and all abelian groups that isclosed under taking subgroups, quotients, extensions, and directed unions),thus answering a question posed by Day in 1957 [Day1]. Among other inter-esting properties of the Grigorchuk group G, we mention the following (see[CMS2], [Har2], [Gri5], [GriP]): (a) G is not finitely presented (a recursiveset of defining relations for G was found by I.G. Lysenok [Lys]), (b) G is justinfinite (it is infinite but every proper quotient is finite), (c) G has solvableword problem, that is, there exists an algorithm that establishes whether,given s1, s2, . . . , sn ∈ {a, b, c, d}, one has s1s2 · · · sn = 1G or not. Originally,the Grigorchuk group was defined as a group of Lebesgue measure-preservingtransformations of the unit interval. By representing the elements of Σ∗ asthe vertices of an infinite binary rooted tree, the Grigorchik group may be alsorealized as a subgroup of the full automorphism group of the tree. Anotherdescription of this group was provided by regarding it as a group generatedby a finite automaton (see [GriNS]).

Simple random walks on groups were first considered by H. Kesten in[Kes1]. Given a finitely generated group G and a finite symmetric generatingsubset S ⊂ G, the simple random walk on G relative to S is the G-invariantMarkov chain with state space G and transition probabilities given by

p(g, h) =

{1|S| if g−1h ∈ S

0 otherwise.

This can be interpreted as follows: a “random walker” on G moves from agroup element g with equal probability to one of its |S| neighbors gs, wheres ∈ S. For g, h ∈ G, denote by p(n)(g, h) the probability of reaching h fromg after exactly n steps. We then have p(0)(g, h) = δg,h where δg,h is theKronecker symbol, p(1)(g, h) = p(g, h) and, more generally,

p(n)(g, h) =∑

k∈G

p(n−1)(g, k)p(k, h).

The quantity p(n)(g, g) does not depend on g ∈ G and is called the returnprobability after n steps. The number

ρ(G, S) = lim supn→∞

n

√p(n)(g, g) (6.115)

is called the spectral radius of the simple random walk on G relative to S.Kesten [Kes1, Kes2] proved that one always has


2√|S| − 1|S| ≤ ρ(G, S) ≤ 1

with equality on the right if and only if G is amenable. Moreover, if S containsno involutions, equality on the left holds if and only if G is a free groupand S is the symmetrization of a free base. Theorem 6.12.9 is just an �2-reformulation of the amenability criterion of Kesten’s theorem. Kesten alsoproved that if S is a finite symmetric generating subset of a group G andN ⊂ G is a normal subgroup, then, denoting by G = G/N (resp S ⊂ G) thecorresponding quotient group (resp. generating subset of G) then ρ(G, S) ≤ρ(G, S) with equality if and only if N is amenable.

Day [Day3] extended Kesten’s amenability criterion to non symmetric ran-dom walks. The associated Markov chain is then determined by a probabilitydensity whose support generates the group. In this setting, the associatedMarkov operator on �2(G) is no more self-adjoint, in general. The key ingre-dient of this new proof is the uniform convexity (uniform rotundity in Day’sterminology) of Hilbert spaces (cf. Lemma I.4.2) and more generally of �p-spaces with p > 1 (note that in fact Day considers, more generally, Markovoperators on the Banach spaces �p(G), for p > 1). For more on this we refer tothe paper [KaV] by V.A. Kaimanovich and A.M. Vershik and to W. Woess’review [Woe1] and monograph [Woe2].

Another important criterion for amenability of finitely generated groupshas been obtained by R.I. Grigorchuk [Gri1]. Let G be a group with m genera-tors. Then G is isomorphic to F/N where F is the free group on m generatorsand N ⊂ F is a normal subgroup. Let α = α(G;F,N) be defined by setting

α = lim supn→∞

n√

w(n),

where w(n) equals the number of elements in N at distance at most n fromthe identity element 1F in the free group F . The non-negative number α iscalled the cogrowth of G relative to the presentation G = 〈F ;N〉. Grigorchuk[Gri1] proved that either α = 1 (this holds if and only if N = {1F }) or

√2m − 1 ≤ α ≤ 2m − 1. (6.116)

He also showed that if ρ = ρ(G, S) is the spectral radius of the simple randomwalk on G relative to S (the symmetrization of the image of a free base ofF under the canonical quotient homomorphism F → G = F/N), then thefollowing relation holds:

ρ =

⎧⎨

⎩

√2m−1m if 1 ≤ α ≤

√2m − 1

√2m−12m

(√2m−1

α + α√2m−1

)if√

2m − 1 ≤ α ≤ 2m − 1.(6.117)

Then, from (6.117) and Kesten’s criterion he deduced that in (6.116) equalityholds on the right if and only if G is amenable. This is called the Grigorchuk

Exercises 217

criterion (or the cogrowth criterion) of amenability. Grigorchuk’s criterionwas used by Ol’shanskii [Ols] to show the existence of non-amenable groupswithout non-abelian free subgroups and by Adyan [Ady] to show that thefree Burnside groups B(m, n), with m ≥ 2 generators and exponent n ≥ 665odd, are non-amenable.

The program of the classification of all finitely generated groups up toquasi-isometries was posed and initiated by Gromov [Gro4]. The definitionof quasi-isometry presented here is modeled after Y. Shalom [Sha].

Exercises

6.1. Let S = {s1, s2, . . . , sn} be a finite subset of Z. Show that S generatesZ if and only if gcd(s1, s2, . . . , sn) = 1.

6.2. Let G be a finitely generated group and let S be a finite symmetricgenerating subset of G. Let x ∈ G and denote by Lx, Rx : G → G the mapsdefined by Lx(g) = xg and Rx(g) = gx for all g ∈ G.

(a) Show that the map g → dS(g,Rx(g)) is constant on G.(b) Show that if x is in the center of G, then Rx is an isometry of (G, dS).(c) Show that supg∈G dS(g, Lx(g)) < ∞ if and only if the conjugacy class

of x in G is finite.

6.3. Suppose that S and S′ are two finite symmetric generating subsets of agroup G with S ⊂ S′. Show that one has: (i) �S(g) ≥ �S′(g) and dS(g, h) ≥dS′(g, h) for all g, h ∈ G; (ii) BS(n) ⊂ BS′(n) and γS(n) ≤ γS′(n) for alln ∈ N; (iii) λS ≤ λS′ .

6.4. Let G1 and G2 be two finitely generated groups and let G = G1×G2. LetS1 (resp. S2) be a finite symmetric generating subset of G1 (resp. G2). Showthat S = (S1 ×{1G2})∪ ({1G1}×S2) is a finite symmetric generating subsetof G and that one has �G

S (g) = �G1S1

(g1) + �G2S2

(g2) for all g = (g1, g2) ∈ G.

6.5. Let S1 and S2 be two sets. For i = 1, 2, let Qi = (Qi, Ei) be an Si-labeled graph. We define their direct product Q1 ×Q2 as the S-labeled graphQ = (Q, E) with:

(1) S = S1

∐S2, where

∐denotes the disjoint union;

(2) Q = Q1 × Q2;(3) E = {((q1, q2), s, (q′1, q

′2)) : either q1 = q′1 and (q2, s, q

′2) ∈ E2, or q2 = q′2

and (q1, s, q′1) ∈ E1}.

Suppose that S1 (resp. S2) is endowed with an involution ι1 : S1 → S1

(resp. ι2 : S2 → S2) and that Q1 (resp. Q2) is edge-symmetric with respectto such involution. Denote by ι : S → S the map defined by ι(s) = ιi(s) ifs ∈ Si, i = 1, 2, and observe that ι is an involution. Show that Q is edge-symmetric with respect to ι.


6.6. Let G1 and G2 be two finitely generated groups and let S1 ⊂ G1 andS2 ⊂ G2 be two finite and symmetric generating subsets such that 1G1 /∈ S1

and 1G2 /∈ S2. Consider the direct product group G = G1 ×G2 together withthe (finite and symmetric) generating subset S = (S1×{1G2})∪({1G1}×S2).Denote by CS1(G1), CS2(G2) and CS(G) the corresponding Cayley graphs. Ifwe identify S1 with S1 × {1G2} (resp. S2 with {1G1} × S2), we may regardCS1(G1) (resp. CS2(G2)) as an (S1×{1G2})-labeled graph (resp. ({1G1}×S2)-labeled graph). Show that CS(G) = CS1(G1) × CS2(G2).

6.7. Let G = Z, S = {1,−1} and S′ = {2,−2, 3,−3}. Find the best possiblepositive constants C1 and C2 such that C1�S(g) ≤ �S′(g) ≤ C2�S(g) for allg ∈ G. Hint: Check that C1 = 1/3 and C2 = 2.

6.8. Let G = Zm, where m ≥ 1 is an integer. Consider the finite and sym-

metric generating subset

S = {±(1, 0, 0, . . . , 0),±(0, 1, 0, . . . , 0), . . . ,±(0, 0, . . . , 0, 1)} ⊂ Zm.

(a) Show that if g = (a1, a2, . . . , am) ∈ Zm then �S(g) = |a1|+ |a2|+ · · ·+

|am|.(b) Let n ∈ N. Set P0(n) = 1 and, for all integers t ≥ 1 denote by Pt(n)

the number of distinct t-tuples (a1, a2, . . . , at) of positive integers such thata1 + a2 + · · · + at ≤ n. Show that Pt(n) =

(nt

)for 1 ≤ t ≤ n. Hint: The map

(a1, a2, . . . , at) → {a1, a1 + a2, . . . , a1 + a2 + · · · + at} establishes a bijectionbetween the set {(a1, a2, . . . , at) ∈ N

t : ai ≥ 1 and a1 + a2 + · · · + at ≤ n}and the set of all subsets of cardinality t of the set {1, 2, . . . , n}.

(c) For n, t ∈ N and t ≥ 1 denote by Nt(n) the number of all m-tuples(a1, a2, . . . , am) ∈ Z

m with∑m

i=1 |ai| ≤ n and exactly t many of the ai’snonzero. Show that γZ

m

S (n) =∑m

t=0 Nt(n).(d) Let 0 ≤ t ≤ n and let I be a subset of {1, 2, . . . , n} such that |I| = t.

Show that there are precisely Pt(n) distinct elements g = (a1, a2, . . . , am) ∈N

m with I = {i : ai > 0} and such that �S(g) ≤ n.(e) Deduce from (d) that there are exactly

(mt

)(nt

)elements g = (a1, a2, . . . ,

am) ∈ Nm with |{i : ai > 0}| = t such that �S(g) ≤ n.

(f) Deduce from (e) that Nt(n) = 2t(mt

)(nt

).

(g) Deduce from (f) and (c) that γZm

S (n) =∑m

t=0 2t(mt

)(nt

).

6.9. Suppose that γ, γ′ : N → [0,+∞) are two growth functions such thatγ � γ′. Show that lim supn→∞

n√

γ(n) ≤ lim supn→∞n√

γ′(n).

6.10. Let α =(

2 11 1

)∈ SL2(Z). Consider the metabelian group G = Z

2�α Z,

that is, the semidirect product of Z2 by the infinite cyclic subgroup of SL2(Z)

generated by α. Recall that

G ={((

xy

), z

): x, y, z ∈ Z

}

Exercises 219

with the multiplication defined by((

x1

y1

), z1

)((x2

y2

), z2

)=((

x1

y1

)+ αz1

(x2

y2

), z1 + z2

)

for all x1, x2, y1, y2, z1, z2 ∈ Z.(a) Show that αn

(10

)=( f2n+1

f2n

)for all n ∈ N, where (fk)k∈N is the

Fibonacci sequence which is inductively defined by f0 = 0, f1 = 1, andfk = fk−2 + fk−1 for all k ≥ 2.

(b) Deduce from (a) that, for any integer n ≥ 1, the set

A(n) =

{(n∑

i=1

uiαi−1

(10

), 0

): ui ∈ {0, 1} for 1 ≤ i ≤ n

}⊂ G

has cardinality |A(n)| = 2n.(c) Consider the subset S ⊂ G defined by S = {a, b, c, a−1, b−1, c−1}, where

a =((

10

), 0)

, b =((

01

), 0)

and c =((

00

), 1)

.

Show that S is a finite symmetric generating subset of G and that one hasA(n) ⊂ BG

S (3n − 2) for all n ≥ 1.(d) Deduce from (b) and (c) that G has exponential growth.

6.11. Let n ≥ 2. Show that GLn(Z) is a finitely generated group of exponen-tial growth. Hint: Use Exercise 2.18, Lemma 2.3.2 and Corollary 6.6.5.

6.12. Let F2 denote the free group of rank two. Show that the groups GL2(Z)and F2 are commensurable. Hint: Use elementary row operations to show thatthe matrices

(1 20 1

)and(

1 02 1

)generate a finite index subgroup of GL2(Z) and

apply Lemma 2.3.2.

6.13. Growth of the Baumslag-Solitar group BS(1, m). Let m be an integersuch that |m| ≥ 2. Prove that the metabelian group G = 〈a, b : aba−1 =bm〉 studied in Exercises 2.7 and 4.21 has exponential growth. Hint: Usean argument similar to the one used for the case m = 2 in the proof ofProposition 6.7.1. More precisely, take S = {a, b, a−1, b−1} and prove thatevery element of the form g = bk, where 0 ≤ k ≤ |m|n − 1 and n ≥ 1, hasword length �S(g) ≤ |m|n + n − 2 by developing k in base |m|.

6.14. Affine representation of the Baumslag-Solitar group BS(1, m). Let mbe an integer such that |m| ≥ 2. Consider the group G given by the presen-tation G = 〈a, b : aba−1 = bm〉 (cf. Exercises 2.7, 4.21 and 6.13).

(a) Let α, β : R → R be the maps respectively defined by α(x) = mx andβ(x) = x + 1 fro all x ∈ R. Show that there is a unique homomorphismϕ : G → Sym(R) satisfying ϕ(a) = α and ϕ(b) = β.


(b) Show that ϕ is injective. Hint: Use Exercise 2.7(a).(c) Let n0 be an integer such that |m|n0 ≥ 3. Consider the elements λ, μ ∈

Sym(R) respectively defined by λ = α−n0 and μ = βλβ−1. Check that theopen intervals i = (−1/2, 1/2) and J = (1/2, 3/2) satisfy λ(J) ⊂ I andμ(I) ⊂ J .

(d) Let n ≥ 1 be an integer and let σ1, σ2, . . . , σn ∈ {λ, μ}. Prove that

σ1σ2 · · ·σn �= IdR .

Hint: Use (c) and play ping-pong as in the proof of Theorem D.5.1.(e) Use (a), (b) and (d) to get another proof of the fact that G has expo-

nential growth.

6.15. Growth of the lamplighter group. Let L = (Z/2Z) � Z denote the lamp-lighter group (cf. Exercise 4.19). Recall that L is the semidirect product ofa normal subgroup H = ⊕n∈ZAn, where each An is a subgroup of order 2,with an infinite cyclic subgroup N generated by an element t which satisfiestat−1 = (an−1)n∈Z for all a = (an)n∈Z ∈ H. Let s denote the nontrivialelement of A0.

(a) Show that S = {s, t, t−1} is a symmetric generating subset of L.(b) For n ≥ 1, let Bn = ⊕n−1

k=0Ak. Prove that Bn is a subgroup of Hgenerated by the elements s, tst−1, t2st−2, . . . , tn−1st−n+1 and that |Bn|= 2n.

(c) Deduce from (b) that 2n ≤ γLS (3n − 2) for all n ≥ 1.

(d) Deduce from (c) that L has exponential growth.

6.16. Growth of the integral Heisenberg group. Let G = HZ denote the Heisen-berg group over the ring of integers (cf. Example 4.6.5). Recall that G is thesubgroup of SL3(Z) consisting of all matrices of the form

M(x, y, z) =

⎛

⎝1 y z0 1 x0 0 1

⎞

⎠ (x, y, z ∈ Z).

Let us set A = M(1, 0, 0), B = M(0, 1, 0), C = M(0, 0, 1), and S ={A, A−1, B, B−1, C, C−1}.

(a) Verify that M(x, y, z) = AxByCz for all x, y, z ∈ Z.(b) Show that S is a finite symmetric generating subset of G.(c) Show that CxAy = AyCx, CxBy = ByCx, and BxAy = AyBxCxy for

all x, y, z ∈ Z.(d) Deduce from (c) that if P ∈ G satisfies �S(P ) ≤ n, then there exist

x, y, z ∈ Z with |x| ≤ n, |y| ≤ n, and |z| ≤ n2 + n, such that P = AxByCz.(e) Deduce from (d) that there exists a constant C1 > 0 such that γS(n) ≤

C1n4 for all n ≥ 1.

(f) Let n, x, y, z be integers such that 0 ≤ x ≤ n, 0 ≤ y ≤ n, and 0 ≤ z ≤n2. Show that there exist integers q, r with 0 ≤ q ≤ n and 0 ≤ r ≤ n−1 suchthat

Exercises 221

AxByCz = Ax−qBnAqBy−nCr.

Hint: Use (c) and Euclidean division of z by n.(g) Deduce from (f) that γS(5n) ≥ (n + 1)2(n2 + 1) for all n ≥ 0.(h) Deduce from (g) that there exists a constant C2 > 0 such that γS(n) ≥

C2n4 for all n ≥ 0.

(i) Deduce from (e) and (h) that γ(G) ∼ n4.

6.17. Let G be the Grigorchuk group.(a) Show that G acts transitively on Σn for all n ∈ N. Hint: Use induction

on n. More precisely, let w1, w2 ∈ Σn and suppose first that w1 and w2 startwith the same letter, that is, there exist x ∈ Σ and u1, u2 ∈ Σn−1 such thatw1 = xu1 and w2 = xu2. Use induction and Proposition 6.9.7(i). Otherwise,if w1 and w2 do not start with the same letter, observe that w1 and a(w2)do start with the same letter and reduce to the previous case.

(b) Use (a) to recover the fact that G is infinite (cf. Theorem 6.9.8).

6.18. Let K1 = Z/2Z = {0, 1} and, for n ≥ 2, define by induction Kn =Kn−1 �(Z/2Z). Recall that Kn−1 �(Z/2Z) = (Kn−1)Z/2Z

�(Z/2Z), so that Kn

consists of the elements (f, a) ∈ (Kn−1)Z/2Z×(Z/2Z) with the multiplicationdefined by (f1, a1)(f2, a2) = (f1f

a12 , a1 + a2), for all f1, f2 ∈ (Kn−1)Z/2Z and

a1, a2 ∈ Z/2Z, where fa(a′) = f(a + a′) ∈ Kn−1 for all f ∈ (Kn−1)Z/2Z anda, a′ ∈ Z/2Z. The group Kn is called the Kaloujnine 2-group of degree n. SetΣ = {0, 1}. For n = 1 and x ∈ Σ, we set 1(x) = 1 − x, and 0(x) = x. Forn ≥ 2 let g = (f, a) ∈ Kn and w = xu ∈ Σn, where x ∈ Σ and u ∈ Σn−1.We then set g(w) = a(x)f(x)(u) ∈ Σn (note that a(x) ∈ Σ, f(x) ∈ Kn−1,and f(x)(u) ∈ Σn−1 is defined by induction).

(a) Show that the map Kn × Σn � (g, w) → g(w) ∈ Σn defines an actionof the group Kn on Σn.

(b) Show that this action is faithful.(c) By virtue of (b), we may regard Kn as a subgroup of Sym(Σn). Show,

by simple counting arguments, that Kn is a Sylow 2-subgroup of Sym(Σn).(d) Consider the elements g1,n, g2,n, . . . , gn,n ∈ Kn defined by induction

as follows. First define g1,1 = 1 ∈ K1 = Z/2Z and then, for 1 ≤ m ≤ n,set g1,m = (f1,m, 1) ∈ Km, where f1,m : Z/2Z → Km−1 is given byf1,m(a) = 1Km−1 for all a ∈ Z/2Z. Finally, for 2 ≤ k ≤ m ≤ n, set gk,m =(fk,m, 0) ∈ Km, where fk,m : Z/2Z → Km−1 is given by fk,m(0) = gk−1,m−1

and fk,m(1) = (f1,m−1, 0). Show that g1,n, g2,n, . . . , gn,n generate Kn and ver-ify that gi,n(uxv) = u(1 − x)v ∈ Σn for all u ∈ Σi−1, x ∈ Σ and v ∈ Σn−i.

(e) Define a map πn : Kn → (Z/2Z)n by induction as follows. π1 : K1 =Z/2Z → Z/2Z is the identity map, while, for n ≥ 2 and (f, a) ∈ Kn we setπn(f, a) = (

∑b∈(Z/2Z) πn−1(f(b)), a) ∈ (Z/2Z)n−1×(Z/2Z) = (Z/2Z)n for all

f ∈ (Kn−1)Z/2Z and a ∈ Z/2Z. Show that πn is a surjective homomorphism.Hint: To prove that πn is a homomorphism use induction on n and the factthat

∑b∈(Z/2Z) πn−1(fa(b)) =

∑b∈(Z/2Z) πn−1(f(b)) for all f ∈ (Kn−1)Z/2Z


and a ∈ Z/2Z. To show surjectivity, look at the πn-images of the elementsg1,n, g2,n, . . . , gn,n ∈ Kn defined in (d).

(f) Deduce from (e) that Kn/[Kn, Kn] (the abelianization of Kn) is iso-morphic to (Z/2Z)n.

(g) Consider the map Φn : Σn → {1, 2, . . . , 2n} given by expansion inbase two, that is, Φn(i1, i2, . . . , in) = i1 + 2i2 + 4i3 + · · · + 2n−1in, for alli1, i2, . . . , in ∈ Σ. Verify that, modulo the map Φ3, one has g1,3 = (1 5)(2 6)×(3 7)(4 8), g2,3 = (1 3)(2 4), and g3,3 = (1 2).

6.19. Let G be the Grigorchuk group.(a) Consider the homomorphism Ψ3 : G → Sym(8) defined by Ψ3(g) =

g|Σ3 (observe that the map Ψ3 is well defined since �(g(w)) = �(w) for allw ∈ Σ∗ and g ∈ G). With the notation from Exercise 6.18(f), verify thatΨ3(a) = (1 5)(2 6)(3 7)(4 8), Ψ3(b) = (1 3)(2 4)(5 6), Ψ3(c) = (1 3)(2 4) andΨ3(d) = (5 6).

(b) Deduce from (a) and Exercise 6.18(e) that Ψ3(G) = K3, where K3 ⊂Sym(8) is the Kaloujnine group (cf. Exercise 6.18).

(c) By applying Proposition 6.9.3, deduce from (b) and Exercise 6.18(e)that G/[G, G] (the abelianization of G) is isomorphic to (Z/2Z) × (Z/2Z) ×(Z/2Z).

6.20. The word problem for the Grigorchuk group. Let G be the Grigorchukgroup. Describe an algorithm which, given any word w ∈ {a, b, c, d}∗, deter-mines whether w represents the identity element 1G or not. Hint: Given aword w, first count the number �a(w) of occurrences of the letter a in w.Prove that if �a(w) is odd then w does not represent 1G. If �a(w) is even, usethe maps φ0, φ1 : H1 → G and apply induction on the length of the word w.

6.21. Let us say that a net (xi)i∈I in a set X converges to infinity if, forevery finite subset F ⊂ X, there exists an element i0 ∈ I such that xi /∈ Ffor all i ≥ i0. Let ϕ : G → H be a map from a group G into a group H.Show that ϕ is a quasi-isometric embedding if and only if it satisfies thefollowing condition: for any two nets (ui)i∈I and (vi)i∈I in G having thesame index set, the net (u−1

i vi)i∈I converges to infinity in G if and only ifthe net (ϕ(ui)−1ϕ(vi))i∈I converges to infinity in H.

6.22. Let G be a group and let E(G) denote the set consisting of all quasi-isometries ϕ : G → G. Define a binary relation ∼ in E(G) by declaring thatϕ1 and ϕ2 ∈ E(G) satisfy ϕ1 ∼ ϕ2 if and only if there exists a finite subsetF ⊂ G such that ϕ1(g)−1ϕ2(g) ∈ F for all g ∈ G.

(a) Show that ∼ is an equivalence relation in E(G).(b) Show that if ϕ ∈ E(G) then there exists ψ ∈ E(G) such that ϕ ◦ ψ ∼

IdG and ψ ◦ ϕ ∼ IdG.(b) Show that the composition of maps in E(G) is compatible with ∼ and

induces a group structure on the quotient set QI(G) = E(G)/ ∼.(c) Show that if G is a finite group then the group QI(G) is trivial.

Exercises 223

(d) Show that the group QI(Z) contains a subgroup isomorphic to R×Z/2Z

and is therefore uncountable. Hint: Prove that the map ϕλ : Z → Z definedby ϕλ(x) = [λx], where λ ∈ R \ {0} and [α] denotes the integral part of α,that is, the largest integer n such that n ≤ α, is a quasi-isometry and thatone has fλ1 ∼ fλ2 if and only if λ1 = λ2.

(e) Show that if G and H are quasi-isometric groups then the groups QI(G)and QI(H) are isomorphic.

6.23. Let A be a set and let G = (Q, E) be a finite A-labeled graph. Denoteby λ : E → A the labeling map defined by λ(e) = a for every edge e =(q, a, q′) ∈ E. A bi-infinite path in G is a sequence π = (en)n∈Z of edgesen = (qn, an, q′n) ∈ E such that q′n = qn+1 for all n ∈ Z. We define the labelof a bi-infinite path π = (en)n∈Z as being the element λ(π) ∈ AZ given byλ(π)(n) = λ(en) for all n ∈ Z.

(a) Denote by XG the set of the labels λ(π) of all bi-infinite paths π inG. Show that XG is a subshift of AZ. It is called the subshift defined by theA-labeled graph G.

(b) Show that if G is connected then the subshift XG is irreducible.

6.24. Let A = {0, 1} and consider the A-labeled graphs G1 and in G2 inFig. 6.18. Check that the associated subshifts XG1 and XG2 are the evensubshift (cf. Exercise 1.38) and the golden mean subshift (cf. Exercise 1.39)respectively.

Fig. 6.18 The A-labeled graphs G1 and G2

6.25. Let A be a set and let X ⊂ AZ be a subshift of finite type. Let Mbe a positive integer such that {1, 2, . . . ,M} ⊂ Z is a memory set for X.Consider the A-labeled graph G = G(X,M) = (Q, E) defined as follows:Q = LM−1(X) is the set of all X-admissible words of length M − 1, and theedge set E consists of all triples e = (aw, a, wa′) ∈ Q×A×Q, where a, a′ ∈ Aand w ∈ AM−2 are such that awa′ ∈ LM (X).

(a) Check that XG = X.(b) Show that X is irreducible if and only if G is connected (cf. Exer-

cise 6.23).

6.26. Let A be a set and let X ⊂ AZ be a subshift of finite type. Let alsoτ : AZ → AZ be a cellular automaton. Let M be a positive integer such that{1, 2, . . . ,M} ⊂ Z is a memory set for both X and τ , and let μ : AM → Adenote the corresponding local defining map for τ . Consider the A-labeled


graph G = G(X, τ,M) = (Q, E), where Q = LM−1(X) is the set of all X-admissible words of length M − 1 and the edge set E consists of all triplese = (aw, μ(awa′), wa′) ∈ Q×A×Q, where a, a′ ∈ A and w ∈ AM−2 are suchthat awa′ ∈ LM (X). Check that XG = τ(X).

6.27. Life on Z. Let A = {0, 1} and consider the cellular automaton τ : AZ →AZ defined in Exercise 5.3. Let G be the A-labeled graph in Fig. 6.19. Checkthat XG = τ(AG).

Fig. 6.19 The A-labeled graph G = G(A, τ, 3)

6.28. Let A = {0, 1} and let τ : AZ → AZ be the majority action cellularautomaton (cf. Example 1.4.3(c)) associated with the set S = {−1, 0, 1}. LetG′ be the A-labeled graph in Fig. 6.20. Check that XG′

= τ(AG).

Fig. 6.20 The A-labeled graph G′ = G(A, τ, 3)

6.29. Let A be a set and let G = (Q, E) be a finite A-labeled graph.(a) Suppose that for each pair (a, a′) ∈ A2 there exists at most one vertex

q ∈ Q which is both the terminal vertex of an edge with label a and the

Exercises 225

initial vertex of an edge with label a′. Show that the subshift XG ⊂ AZ is offinite type. Hint: Show that in fact {0, 1} ⊂ Z is a memory set for XG .

(b) Show that the hypothesis in (a) is satisfied if the labeling map λ : E →A is injective.

6.30. Let A be a set and let G = (Q, E) be a finite A-labeled graph. We recallthat given an edge e = (q, a, q′) ∈ E we denote by α(e) = q ∈ Q (resp. ω(e) =q′ ∈ Q) the initial (resp. terminal) vertex of e. Consider the E-labeled graphG′ = (Q′, E′), where Q′ = Q and E′ = {(α(e), e, ω(e)) ∈ Q×E ×Q : e ∈ E}.Note that the labeling map λ′ : E′ → E on G′ is bijective. Identify the set{λ(e) : e ∈ E} ⊂ A formed by the labels of the edges of G with a subsetof E in an arbitrary way and consider the cellular automaton τ : EZ → EZ

with memory set S = {0} and local defining map μ : ES → E defined byμ(e) = λ(e) for all e ∈ ES = E. Observe that τ(x) ∈ AZ for all x ∈ EZ andshow that τ(XG′

) = XG .

6.31. Let A be a set and X ⊂ AZ a subshift. Show that the following condi-tions are equivalent:

(i) there exists a finite A-labeled graph G such that X = XG .(ii) there exists a set B containing A, a subshift Y ⊂ BZ of finite type and

a cellular automaton τ : BZ → BZ such that X = τ(Y ). A subshift X ⊂ AZ issaid to be sofic if it satisfies one of the two above equivalent conditions. Hint:For the implication (i) ⇒ (ii) use Exercise 6.30. For the converse implication,use Exercise 6.25.

6.32. Let A be a set.(a) Show that every subshift X ⊂ AZ of finite type is sofic.(b) Suppose that A has at least two distinct elements. Show that there

exists a subshift X ⊂ AZ which is sofic but not of finite type. Hint: The evensubshift is sofic (cf. Exercise 6.24) but not of finite type (cf. Exercise 1.38(c)).

6.33. Let A be a set. Let X ⊂ AG be a sofic subshift and let τ : AG → AG

be a cellular automaton. Show that τ(X) ⊂ AG is a sofic subshift.

6.34. A subshift which is not sofic (cf. [LiM, Example 3.1.7]). Let A ={0, 1, 2} and consider the subset X ⊂ AZ consisting of all configurationsx ∈ Z such that if x(n) = 0, x(n + 1) = x(n + 2) = · · · = x(n + h) = 1,x(n+h+1) = x(n+h+2) = · · · = x(n+h+k) = 2 and x(n+h+k +1) = 0for some n ∈ Z and h, k ∈ N, then necessarily h = k.

(a) Show that X is a subshift of AZ. It is called the context-free subshift .(b) Show that X is not sofic. Hint: Suppose by contradiction that X = XG

for some finite A-labeled graph G = (Q, E). Let r = |Q|. Observe that w =01r+12r+10 ∈ L(X) so that there exists a path π in G such that λ(π) = w.Let π′ denote the subpath of π such that λ(π′) = 1r+1. Since �(π′) = r +1 >r = |Q|, we can write π′ = π1π2π3 where π2 is a closed path of lengths = �(π2) > 0 (and π1 (resp. π3) is a possibly empty path). It follows thatπ′′ = π1π2π2π3 is a path in G and its label is λ(π′′) = 01r+1+s2r+10 /∈ L(X).


6.35. Let A be a finite set and let X ⊂ AZ be a sofic subshift. A finite A-labeled graph G = (Q, E) is said to be a minimal presentation of X if (i)X = XG , and (ii) |Q| ≤ |Q′| for all A-labeled graphs G′ = (Q′, E′) such thatX = XG′

. Suppose that X is irreducible and that G = (Q, E) is a minimalpresentation of X.

(a) Show that for every vertex q ∈ Q there exists a word wq ∈ L(X) suchthat if a path π in G satisfies λ(π) = wq, then it passes through q. Hint: Bycontradiction, if this is not the case for some q ∈ Q then the A-labeled graphG′ = (Q′, E′) where Q′ = Q\{q} and E′ = E\{e ∈ E : α(e) = q or ω(e) = q}satisfies XG′

= X and |Q′| < |Q|, contradicting the minimality of G.(b) Deduce from (a) that G is connected. Hint: Let q, q′ ∈ Q. Consider

the words wq, wq′ ∈ L(X) described in (a). Then by irreducibility of X thereexists u ∈ L(X) such that wquwq′ ∈ L(X). Let π be a path in G suchthat λ(π) = wquwq′ ; then π = π1π2π3 where λ(π1) = wq, λ(π2) = u andλ(π3) = wq′ . By definition of wq (resp. wq′) the path π1 (resp. π3) passesthrough q (resp. q′), say π1 = π′

1π′′1 with (π′

1)+ = q = (π′′

1 )− (resp. π3 = π′3π

′′3

with (π′3)

+ = q′ = (π′′3 )−). Then the path π′′

1 π2π′3 connects q to q′.

(c) Deduce from (b) that for every irreducible sofic subshift Y ⊂ AZ thereexists a connected finite A-labeled graph G such that Y = XG .

6.36. Let A be a finite set and X ⊂ AZ an irreducible sofic subshift.(a) Show that the subset Xf consisting of all configurations in X whose

Z-orbit is finite (cf. Example 1.3.1(c)) is dense in X. Hint: Let x ∈ X andlet n ∈ N. By virtue of Exercise 6.35, we can find a connected A-labeledgraph G = (Q, E) such that X = XG . Let π1 be a finite path in G suchthat λ(π1) = x(0)x(1) · · ·x(n − 1) and set q = π−

1 and q′ = π+1 . Since G is

connected, we can find a finite path π2 in G connecting q′ to q. Let m = �(π2).It follows that the path π = π1π2 is closed and t = �(π) = n+m. If w = λ(π)we deduce that wk ∈ L(X) for all k ∈ N. Since X is closed, there exists aconfiguration y ∈ X such that y(ht)y(ht + 1) · · · y((h + 1)t − 1) = w for allh ∈ Z. In particular, y(0)y(1) · · · y(n−1) = x(0)x(1) · · ·x(n−1) and y ∈ Xf .

(b) Deduce from (a) that X is surjunctive. Hint: Cf. Exercise 3.29.

6.37. Show that the Morse subshift is not sofic. Hint: An infinite minimal sub-shift is irreducible (cf. Exercise 3.35(b)) and contains no configuration whoseZ-orbit is finite (cf. Exercise 3.35(d)). On the other hand, by Exercise 6.36(a),every irreducible sofic subshift contains an abundance of configurations withfinite Z-orbit.

6.38. Let A be a finite set and let X ⊂ AZ be an infinite Toeplitz subshift.Show that X is not sofic. Hint: The same arguments as for Exercise 6.37apply.

6.39. Let A be a finite set. Show that there are at most countably manydistinct sofic subshifts X ⊂ AZ. Hint: There are at most countably manyfinite A-labeled graphs up to isomorphism.

Exercises 227

6.40. Let I be a finite set. Let B = (bij)i,j∈I be a matrix with real entriesbij ≥ 0 for all i, j ∈ I. For an integer n ≥ 1, we denote by Bn = (b(n)

ij )i,j∈I then-th power of B. One says that the matrix B is irreducible if for every i, j ∈ I

there exists an integer n = n(i, j) ≥ 1 such that b(n)ij > 0. Suppose that B is

irreducible. The period per(i) of i ∈ I is the greatest common divisor of theintegers n ≥ 1 such that b

(n)ii > 0.

(a) Show that per(i) = per(j) for all i, j ∈ I. The period per(B) of thematrix B is defined as the common value of the numbers per(i), i ∈ I. Hint:Let i, j ∈ I. We can find r, s ≥ 1 such that b

(r)ij > 0 and b

(s)ji > 0. It follows

that b(r+s)ii ≥ b

(r)ij b

(s)ji > 0 and b

(r+n+s)ii ≥ b

(r)ij b

(n)jj b

(s)ji > 0 for all n ≥ 1 such

that b(n)jj > 0. It follows from the definition that per(i) divides both r+s and

r + n + s and therefore also divides their difference n. This shows that per(i)divides per(j).

(b) Let n ≥ 1 be an integer. Show that Bn is irreducible if and only if nand per(B) are relatively prime.

6.41. Let A be a set and let G = (Q, E) be a finite A-labeled graph. Theadjacency matrix of G is the (Q × Q)-matrix BG = (bqq′)q,q′∈Q where bqq′ isthe number of edges in E with initial vertex q and terminal vertex q′.

(a) Show that G is connected if and only if the matrix BG is irreducible.(b) Suppose that G is connected. The period of G is the positive integer

defined by per(G) = per(BG). Show that per(G) is the greatest commondivisor of the lengths of all closed paths in G.

6.42. Let A be a set. Given a subshift X ⊂ AZ denote by pern(X) =|Fix(nZ) ∩ X| (cf. Example 1.3.1(c)) the number of nZ-periodic configu-rations in X. Let G be a finite connected A-labeled graph.

(a) Show that for every integer N there exists an integer n ≥ N such thatpern(XG) > 0.

(b) The period per(XG) of XG is defined as the greatest common divisorof all integers n ≥ 1 for which pern(X) > 0. Show that per(XG) = per(G).

6.43. The N th higher block subshift (cf. Exercise 1.15 and Exercise 1.34). LetA be a set and N a positive integer. Consider the map ΦN : AZ → (AN )Z

defined by ΦN (x)(n) = (x(n), x(n + 1), x(n + 2), . . . , x(n + N − 1)) for allx ∈ AZ and n ∈ Z. Similarly, consider the map ϕN : A∗ → (AN )∗ defined asfollows: ϕN (w) = ε (the empty word) if �(w) < N and

ϕ(w) = (a1, a2, . . . , aN )(a2, a3, . . . , aN+1) · · · (am+1, am+2, . . . , am+N )

if w = a1a2 · · · am+N , with m ≥ 0. Let X ⊂ AZ be a subshift and set X [N ] =ΦN (X) ⊂ (AN )Z. Let L ⊂ A∗ be a subset and set L[N ] = ϕN (L) ⊂ (AN )∗.

(a) Show that X [N ] is a subshift of (AN )Z (it is called the Nth higher blocksubshift of X) and that L(X [N ]) = (L(X))[N ].


(b) Show that X is irreducible (resp. topologically mixing, resp. stronglyirreducible) if and only if X [N ] is irreducible (resp. topologically mixing, resp.strongly irreducible).

(c) Show that if X is of finite type then X [N ] is of finite type.(d) Let Fn = {0, 1, 2, . . . , n − 1} ⊂ Z and let F = (Fn)n∈N. Show that

entF (X) = entF (X [N ]).

6.44. Let A be a set and let G = (Q, E) be an A-labeled graph. For N apositive integer, the N -higher edge graph G[N ] associated with G is the AN -labeled graph (Q[N ], E[N ]) defined as follows. For N = 1 one has G[1] = Gand, for N ≥ 2, the vertex set Q[N ] is the set of all paths of length N − 1 inG and

E[N ] = {(π,λ(e1)λ(e2) · · ·λ(eN ), π′) ∈ Q[N ] × AN × Q[N ] :π = (e1, e2, . . . , eN−1), π′ = (e2, e3, . . . , eN−1, eN )}.

(a) Show that XG[N]= (XG)[N ].

(b) Deduce from (b) that if X ⊂ AZ is a sofic subshift, then the subshiftX [N ] ⊂ (AN )Z is also sofic.

6.45. The N th higher power subshift (cf. Exercise 1.16 and Exercise 1.35).Let A be a set and N a positive integer. Consider the map ΨN : AZ → (AN )Z

defined by ΨN (x)(n) = (x(nN), x(nN +1), . . . , x((n+1)N−1)) for all x ∈ AZ

and n ∈ Z. Similarly, consider the map ψN : A∗ → (AN )∗ defined as follows:ψN (w) = ε if w ∈ A∗ \ (∪n≥1A

nN ) and

ψN (w) = (a1, a2, . . . , aN )(aN+1, aN+2, . . . , a2N )· · · (a(n−1)N+1, a(n−1)N+2, . . . , anN )

if w = a1a2 · · · anN for some n ≥ 1. Let X ⊂ AZ be a subshift and set X(N) =ΨN (X) ⊂ (AN )Z. Let L ⊂ A∗ be a subset and set L(N) = ψN (L) ⊂ (AN )∗.

(a) Show that X(N) is a subshift of (AN )Z (it is called the Nth higherpower subshift of X) and that L(X(N)) = (L(X))(N).

(b) Show that if X is of finite type then X(N) is of finite type.(c) Suppose that X is irreducible. Show that X(N) is irreducible if and

only if N and per(X) are relatively prime.(d) Let Fn = {0, 1, 2, . . . , n − 1} ⊂ Z and let F = (Fn)n∈N. Show that

entF (X) = entF (X(N)).

6.46. Let A be a set and let G = (Q, E) be an A-labeled graph. For N apositive integer, the N -higher power graph G(N) associated with G is the AN -labeled graph (Q(N), E(N)) defined as follows. The vertex set is Q(N) = Qand the edge set is

E(N) = {(π−, λ(π), π+) ∈ Q × AN × Q : π a path of length N in G},

Exercises 229

where π− ∈ Q (resp. π+ ∈ Q) is the initial (resp. terminal) vertex of thepath π. Recall that given a labeled graph H, we denote by BH its adjacencymatrix (cf. Exercise 6.40).

(a) Show that BG(N) = (BG)N .(b) Show that XG(N)

= (XG)(N).(c) Deduce from (b) that if X ⊂ AZ is a sofic subshift, then the subshift

X(N) ⊂ (AN )Z is also sofic.

6.47. (cf. [Sca, Lemma]) Let G = (Q, E) be a finite labeled graph. Supposethat G is connected and let e ∈ E. Show that there exists a positive in-teger n0 such that if π is any path in G of length n0, then there exists apath π′ = (e1, e2, . . . , en0) of the same length n0, with the same initial andterminal vertices as π, and such that ei = e for some 1 ≤ i ≤ n0. Hint:Given a path π = (e1, e2, . . . , en) in G, denote by πQ = (q0, q1, . . . , qn) theassociated sequence of visited vertices (cf. Sect. 6.2). We define a decom-position of π as follows. Let i1 be the largest index such that the verticesq0, q1, . . . , qi1−1 are all distinct. Then qi1 = qj1 for a suitable j1 < i1 and weset r1 = (e1, e2, . . . , ej1) and c1 = (ej1+1, ej1+2, . . . , ei1). Continuing this way,we obtain a decomposition of the path π = r1c1r2c2 · · · rkckrk+1 where thec1, c2, . . . , ck are closed simple paths and r1, r2, . . . , rk+1 are simple (possiblyempty) paths. With this notation, for 1 ≤ s ≤ r and a positive integer d,we say that the path πs = r1c1r2c2rsrs+1 · · · ckrk is obtained from π by col-lapsing the sth closed simple path cs. Similarly, if π′′ is a closed path suchthat (π′′)− = π− and d is a positive integer, we say that the path (π′′)dπ isobtained from π by adding d copies of π′′ at the beginning of π. Now, sinceG is connected, we can find a closed path π′′ with initial (= terminal) vertexπ− containing the edge e. If n0 is large enough then in the decompositionπ = r1c1r2c2 · · · rkckrk+1 there exists a cycle c that is repeated many times.If the length of c is �, the length of π′′ is m and the cycle c is repeated atleast m times, then we may collapse the first m copies of c and then add �copies of π′′ at the beginning of π to obtain the desired path π′.

6.48. Let G = (Q, E) be a finite labeled graph. We denote by Pn(G) the setof all paths of length n in G and we define the entropy of G as

ent(G) = limn→∞

log |Pn(G)|n

.

Show that the above limit exists and is finite. Hint: Use Lemma 6.5.1.

6.49. (cf. [Sca, Theorem]) Let G = (Q, E) be a finite connected labeled graph.Let e ∈ E and denote by H = (Q′, E′) the labeled subgraph of G were Q′ = Qand E′ = E \ {e}. Let n0 be the positive integer given by Exercise 6.47 andset α = |Pn0(G)|−1.

(a) Show that |P(k−1)n0(G)| ≥ α|Pkn0(G)| for all k = 1, 2, . . ..(b) We express a path π ∈ Pkn0(G) as the composition π = π1π2 · · ·πk,

where πi ∈ Pn0(G), 1 ≤ i ≤ k. Also, for i = 1, 2, . . . , k, we denote by ϕi the set


of all paths π ∈ Pkn0(G) such that πi contains the edge e. Show that |ϕ1| ≥|P(k−1)n0(G)| and deduce from (a) that |Pkn0(G)\ϕ1| ≤ (1−α)|Pn0(G)|. Hint:By Exercise 6.47 for any π ∈ P(k−1)n0(G) there exits a path π′π ∈ Pkn0(G)such that π′ contains e.

(c) For 1 ≤ i ≤ k we set P ikn0

(G) = Pkn0(G)\⋃i−1

t=1 ϕt, Ch = {π ∈ P ikn0

(G) :πi contains e} and denote by Dh the set of pairs (σ1, σ2) ∈ P(i−1)n0(G) ×P(k−i)n0(G) such that there exists π = σ1σσ2 ∈ P i

kn0(G). Show that |Ch| ≥

|Dh| ≥ α|P ikn0

(G)|. Hint: Use Exercise 6.47 to show that for every (σ1, σ2) ∈Di there exists π = σ1π

′σ2 ∈ Pkn0(G) such that π′ contains e.(d) Deduce from (c) that |Pkn0(G) \

⋃ki=1 ϕi| ≤ (1 − α)k|Pkn0(G)|.

(e) Observe that Pkn0(H) ⊂ Pkn0(G) \⋃k

i=1 ϕi and deduce from (d) thatent(H) ≤ log(1−α)

n0+ ent(G).

(f) Deduce from (e) that ent(H) < ent(G).

6.50. Let A be a finite set. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn)n∈N. LetX ⊂ AZ be a subshift and let w ∈ L(X). Set

Xw = {x ∈ X : x(n + 1)x(n + 2) · · ·x(n + N) �= w for all n ∈ Z},

where N = �(w).(a) Show that Xw ⊂ X is a subshift of AZ.(b) Show that (Xw)[N ] = (X [N ])ΦN (w) (cf. Exercise 6.43).(c) Suppose from now on that X is irreducible of finite type. Let M be

a positive integer such that {1, 2, . . . ,M} is a memory set for X, and let Gbe the A-labeled graph such that X = X(G, M) (cf. Exercise 6.25). Let Hdenote the labeled subgraph of G[N ] obtained by removing all edges labeledby ΦN (w). Show that (X [N ])ΦN (w) = XH.

(d) Show that entF (Xw) < entF (X). Hint: Use Exercises 6.49 and 6.43(d).

6.51. Let A be a finite set. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn)n∈N.Let X ⊂ AZ be an irreducible subshift of finite type. Show that if Y ⊂ AZ

is such that Y � X, then entF (Y ) < entF (X). Hint: Show that there existsw ∈ L(X) \ L(Y ) such that Y ⊂ Xw ⊂ X. Then apply Exercise 6.50(d) andProposition 5.7.2(ii).

6.52. (cf. [Hed-3], [CovP], and [LiM, Theorem 8.1.16]). Let A be a finite set.Let F = (Fn)n∈N, where Fn = {0, 1, . . . , n}⊂Z. Also let X ⊂AZ an irreduciblesubshift of finite type and τ : X → X a pre-injective cellular automaton.

(a) Let M be a positive integer such that {1, 2, . . . ,M} is a memory setfor X. Set Y = τ(X) and consider the labeled graph G = G(X, τ,M) (cf.Exercise 6.26). Note that for every w ∈ L(Y ) there exists a path π ∈ Gsuch that w = λ(π). Show that, by pre-injectivity of τ , for all q, q′ ∈ Q andw ∈ L(Y ) there exists at most one path π in G with initial (resp. terminal)vertex π− = q (resp. π+ = q′) such that w = λ(π).

(b) Deduce from (a) that |Ln(Y )| ≤ |Ln(X)| ≤ |Ln(Y )| · |Q|2.(c) Deduce from (b) that entF (τ(X)) = entF (X).

Exercises 231

6.53. Let A be a finite set. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn)n∈N. LetX,Y ⊂ AZ be two irreducible subshifts of finite type such that entF (X) =entF (Y ). Show that every pre-injective cellular automaton τ : X → Y issurjective. Hint: Use the results of Exercise 6.51 and Exercise 6.52.

6.54. Let A be a finite set. Let X ⊂ AG be an irreducible subshift of fi-nite type. Show that every pre-injective cellular automaton τ : X → X issurjective. Hint: Use Exercise 6.53 with Y = X.

6.55. (cf. [Hed-3], [CovP], and [LiM, Theorem 8.1.16]). Let A be a finite setand X ⊂ AZ an irreducible subshift of finite type. Let Fn = {0, 1, . . . , n} ⊂ Z

and F = (Fn)n∈N. Let τ : X → X be a cellular automaton. Suppose that τis not pre-injective.

(a) Show that there exist a positive integer N and two distinct configura-tions x1, x2 ∈ X such that x1|Z\{1,2,...,N} = x2|Z\{1,2,...,N} and τ(x1) = τ(x2).

(b) Up to enlarging N if necessary, we may suppose that {1, 2, . . . , N} isa memory set for both X and τ . Let G = G(X, τ,N) = (Q, E) (cf. Exer-cise 6.26). Show that there exist two vertices q, q′ ∈ Q and two paths π1, π2

in G with initial (resp. terminal) vertices π−1 = π−

2 = q (resp. π+1 = π+

2 = q′)such that λ(π1) = x1|{1,2,...,N} and λ(π2) = x2|{1,2,...,N}. One says that thetwo paths π1, π2 constitute a diamond in G (see Fig. 6.21).

Fig. 6.21 A diamond in an A-labeled graph G

(c) Up to further enlarging N if necessary, we may suppose that N isrelatively prime with per(X) (cf. Exercise 6.40 and Exercise 6.42). By Exer-cise 6.45(c) the Nth power graph G(N) is also irreducible and π1 and π2 areedges in G(N) with the same initial vertex and same terminal vertex. Let Hbe the A-labeled subgraph of G(N) obtained by removing the edge π1. Deducefrom Exercise 6.49 that entF (XH) < entF (XG(N)).

(d) Show that XH = Y (N), where Y = τ(X).(e) Deduce from (c) and (d) and from Exercise 6.45(d) that entF (τ(X)) <

entF (X).

6.56. Let A be a finite set. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn)n∈N.Let X,Y ⊂ AZ be two subshifts such that X is irreducible of finite type andentF (X) = entF (Y ). Show that every surjective cellular automaton τ : X →Y is pre-injective. Hint: Use the result of Exercise 6.55.

6.57. The Garden of Eden theorem for irreducible subshifts of finite typeover Z [Fio1, Corollary 2.19]. Let A be a finite set and X ⊂ AZ an irreducible


subshift of finite type. Let τ : X → X be a cellular automaton. Show thatτ is surjective if and only if it is pre-injective. Hint: Combine together theresults from Exercise 6.56 with Y = X and Exercise 6.54.

6.58. Let A = {0, 1}. Consider the subset X ⊂ AZ consisting of all con-figurations x ∈ AZ such that {n ∈ Z : x(n) = 1} is an interval of Z. Letσ : AZ → AZ be the cellular automaton with memory set S = {0, 1} andlocal defining map μ : AS → A given by

μ(y) =

{1 if (y(0), y(1)) = (0, 1),y(0) otherwise.

(a) Show that X is a sofic subshift.(b) Show that X is neither of finite type nor irreducible.(c) Check that τ(X) ⊂ X(d) Show that the cellular automaton τ = σ|X : X → X is injective (and

therefore pre-injective) but not surjective.(e) Deduce from (d) that X is not surjunctive.

Chapter 7

Local Embeddability and Sofic Groups

In this chapter we study the notions of local embeddability and soficity forgroups. Roughly speaking, a group is locally embeddable into a given classof groups provided that the multiplicative table of any finite subset of thegroup is the same as the multiplicative table of a subset of some group in theclass (cf. Definition 7.1.3). In Sect. 7.1 we discuss several stability propertiesof local embeddability. Subgroups of locally embeddable groups are locallyembeddable (Proposition 7.1.7). Moreover, as the name suggests, local em-beddability is a local property, that is, a group is locally embeddable into aclass of groups if and only if all its finitely generated subgroups are locallyembeddable into the class (Proposition 7.1.8). When the class is closed underfinite direct products, the class of groups which are locally embeddable in theclass is closed under (possibly infinite) direct products (Proposition 7.1.10).We also show that a marked group which is a limit of groups belonging toa given class is locally embeddable into this class (Theorem 7.1.16). Con-versely, we prove that if the given class is closed under taking subgroups andthe marking group is free, then any marked group which is locally embed-dable is a limit of marked groups which are in the class (Theorem 7.1.19). IfC is a class of groups which is closed under finite direct products, then everygroup which is residually C is locally embeddable into C (Corollary 7.1.14).Conversely, under the hypothesis that C is closed under taking subgroups, ev-ery finitely presented group which is locally embeddable into C is residually C(Corollary 7.1.21).

In Sect. 7.2 we present a characterization of local embeddability in termsof ultraproducts: a group is locally embeddable into C if and only if it can beembedded into an ultraproduct of a family of groups in C.

Section 7.3 is devoted to LEF and LEA-groups. A group is called LEF(resp. LEA) if it is locally embeddable into the class of finite (resp. amenable)groups. As the class of finite (resp. amenable) groups is closed under takingsubgroups and taking finite direct products, all the results obtained in theprevious section can be applied. This implies in particular that every locallyresidually finite (resp. locally residually amenable) group is LEF (resp. LEA).


233

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234 7 Local Embeddability and Sofic Groups

We give an example of a finitely generated amenable group which is LEF butnot residually finite (Proposition 7.3.9). This group is not finitely presentablesince every finitely presented LEF-group is residually finite.

The Hamming metric, which is a bi-invariant metric on the symmetricgroup of a finite set, is introduced in Sect. 7.4.

In Sect. 7.5 we define the class of sofic groups. These groups are the groupsadmitting local approximations by finite symmetric groups equipped withtheir Hamming metric. Subgroups and direct products of sofic groups aresofic. Every LEA-group is sofic (Corollary 7.5.11). In particular, residuallyamenable groups, and therefore amenable groups and residually finite groups,are sofic. A group is sofic if and only if it can be embedded into an ultra-product of a family of finite symmetric groups equipped with their Hammingmetrics (Theorem 7.6.6). In Sect. 7.7 we give a characterization of finitelygenerated sofic groups in terms of their Cayley graphs. More precisely, weshow that a finitely generated group G with a finite symmetric generatingsubset S ⊂ G is sofic if and only if, for every integer r ≥ 0 and every ε > 0,there exists a finite S-labeled graph Q such that there is a proportion of atleast 1− ε of vertices q ∈ Q such that the ball of radius r centered at q in Qis isomorphic, as a labeled graph, to a ball of radius r in the Cayley graphof G associated with S (Theorem 7.7.1). The last section of this chapter isdevoted to the proof of the surjunctivity of sofic groups (Theorem 7.8.1).

7.1 Local Embeddability

Definition 7.1.1. Let G and C be two groups. Given a finite subset K ⊂ G,a map ϕ : G → C is called a K-almost-homomorphism of G into C if itsatisfies the following conditions:

(K-AH-1) ϕ(k1k2) = ϕ(k1)ϕ(k2) for all k1, k2 ∈ K;(K-AH-2) the restriction of ϕ to K is injective.

Note that in the preceding definition, the map ϕ is not required to be ahomomorphism nor to be globally injective.

Remark 7.1.2. If ϕ : G → C is a K-almost-homomorphism and ϕ′ : G → Cis a map which coincides with ϕ on K ∪ K2, then ϕ′ is also a K-almost-homomorphism.

Let now C be a class of groups, that is, a collection of groups satisfyingthe following condition: if C ∈ C and C ′ is a group which is isomorphic to C,then C ′ ∈ C. For example, C might be the class of finite groups, the class ofnilpotent groups, the class of solvable groups, the class of amenable groups,or the class of free groups.

7.1 Local Embeddability 235

Definition 7.1.3. Let C be a class of groups. One says that a group G islocally embeddable into the class C if, for every finite subset K ⊂ G, thereexist a group C ∈ C and a K-almost-homomorphism ϕ : G → C.

Examples 7.1.4. (a) Let C be a class of groups and let G be a group in C.Then G is locally embeddable into C. Indeed, the identity map IdG : G → Gis a K-almost-homomorphism of G into itself for every finite subset K ⊂ G.

(b) The group Z is locally embeddable into the class of finite groups.Indeed, let K be a finite subset of Z. Choose an integer n ≥ 0 such thatK ⊂ [−n, n]. Then, the quotient homomorphism ϕ : Z → Z/(2n + 1)Z is aK-almost-homomorphism.

Remark 7.1.5. Let G be a group and let C be a class of groups which is closedunder taking subgroups. Suppose that G is finite and it is locally embeddableinto C. Then G ∈ C. Indeed, any G-almost homomorphism ϕ : G → C fromG into a group C ∈ C is an injective homomorphism.

The following characterization of local embeddability will not be used inthe sequel but it presents some interest on its own.

Proposition 7.1.6. Let G be a group and let C be a class of groups which isclosed under taking subgroups. Then the following conditions are equivalent:

(a) G is locally embeddable into C;(b) for every finite subset K ⊂ G, there exist a set L with K ⊂ L ⊂ G and

a binary operation � : L × L → L such that (L,�) is a group in C andk1k2 = k1 � k2 for all k1, k2 ∈ K.

Proof. Suppose (a). Fix a finite subset K ⊂ G. If G is finite, then G ∈ Cby Remark 7.1.5. Therefore, condition (b) is satisfied in this case since wecan take as (L,�) the group G itself. So let us assume that G is infinite.Consider the finite subset K ′ ⊂ G defined by K ′ = K ∪ K2. Since G islocally embeddable into C, we can find a group C ′ ∈ C and a K ′-almost-homomorphism ϕ′ : G → C ′. Note that, as K ′ is finite, the subgroup C ⊂ C ′

generated by ϕ′(K ′) is countable. Since G is infinite, it follows that thereexists a surjective map σ : G → C which coincides with ϕ′ on K ′. Let usdefine a map ψ : C → G as follows. If c ∈ σ(K ′), we set ψ(c) = k′, where k′

is the unique element in K ′ such that σ(k′) = c. For c ∈ C \ σ(K ′), we takeas ψ(c) an arbitrary element in σ−1(c). Denote by L = ψ(C) ⊂ G the imageof ψ. Observe that K ⊂ K ′ ⊂ L and that ψ induces a bijection from C ontoL. Let us use this bijection to transport the group structure from C to L. Ifwe denote by � the corresponding group operation on L, this means that ψinduces a group isomorphism from C onto (L,�). It follows that we have

�1 � �2 = ψ(σ(�1)σ(�2))

for all �1, �2 ∈ L. As C is closed under taking subgroups, C and hence (L,�)belong to C. Moreover, for all k1, k2 ∈ K, we have


k1 � k2 = ψ(σ(k1)σ(k2)) = ψ(ϕ′(k1)ϕ′(k2)) = ψ(ϕ′(k1k2)) = ψ(σ(k1k2)).

Since K2 ⊂ K ′, we deduce that k1 �k2 = k1k2 for all k1, k2 ∈ K. This showsthat condition (b) is satisfied.

Conversely, suppose (b). Fix a finite subset K ⊂ G. Set C = (L,�),where L is as in (b). Let ϕ : G → C be any map extending the identity mapι : L → L. Then ϕ|K = ι|K is injective. On the other hand, for all k1, k2 ∈ K,we have k1k2 = k1 � k2 ∈ L and therefore

ϕ(k1k2) = ι(k1k2)= k1k2

= k1 � k2

= ϕ(k1) � ϕ(k2)

Thus, ϕ is a K-almost-homomorphism of G into the group C in C. This showsthat G is locally embeddable into C. �

Proposition 7.1.7. Let C be a class of groups. Every subgroup of a groupwhich is locally embeddable into C is locally embeddable into C.

Proof. Let G be a group which is locally embeddable into C and let H bea subgroup of G. Given a finite subset K ⊂ H, let ϕ : G → C be a K-almost-homomorphism of G into a group C ∈ C. Then the restriction mapϕ|H : H → C is a K-almost-homomorphism of H into C. This shows that His locally embeddable into C. �

As the name suggests, local embeddability is a local property for groups:

Proposition 7.1.8. Let G be a group and let C be a class of groups. Then Gis locally embeddable into C if and only if every finitely generated subgroup ofG is locally embeddable into C.

Proof. The necessity of the condition follows from Proposition 7.1.7. Con-versely, suppose that every finitely generated subgroup of G is locally em-beddable into C. Let K ⊂ G be a finite subset and denote by H ⊂ G thesubgroup generated by K. Then, as H is locally embeddable into C, there ex-ists a K-almost-homomorphism ϕ : H → C of H into a group C ∈ C. Extendarbitrarily ϕ to G, for example by setting ϕ(g) = 1C for all g ∈ G \H. Thenϕ : G → C is a K-almost-homomorphism of G into C. This shows that G islocally embeddable into C. �

Let C be a class of groups. Denote by C the class consisting of all groupswhich are locally embeddable into C.

Proposition 7.1.9. Let C be a class of groups. Then C = C.


Proof. Let G be a group in C. Then, given any finite subset K ⊂ G thereexist a group C ′ ∈ C and a K-almost-homomorphism ϕ : G → C ′. Considerthe finite subset K ′ = ϕ(K) ⊂ C ′ and let ϕ′ : C ′ → C be a K ′-almost-homomorphism of C ′ into a group C ∈ C. Then the composition map Φ =ϕ′ ◦ϕ : G → C is a K-almost-homomorphism. Indeed, for k1, k2 ∈ K one has

Φ(k1k2) = ϕ′(ϕ(k1k2))= ϕ′(ϕ(k1)ϕ(k2))= ϕ′(ϕ(k1))ϕ′(ϕ(k2))= Φ(k1)Φ(k2).

Moreover, as ϕ|K and ϕ′|K′ are injective, and K ′ = ϕ(K), one has that Φ|K isalso injective. It follows that G is locally embeddable into C. This shows thatC ⊂ C. The inclusion C ⊂ C follows from the observation in Example 7.1.4(a).This shows that C = C. �

A class of groups C is said to be closed under finite direct products if onehas G1 × G2 ∈ C whenever G1 ∈ C and G2 ∈ C.

Proposition 7.1.10. Let C be a class of groups which is closed under finitedirect products. Let (Gi)i∈I be a family of groups which are locally embeddableinto C. Then, their direct product G =

∏i∈I Gi is locally embeddable into C.

Proof. For each i ∈ I, let πi : G → Gi denote the projection homomorphism.Fix a finite subset K ⊂ G. Then there exists a finite subset J ⊂ I suchthat the projection homomorphism πJ =

∏j∈J πj : G → GJ =

∏j∈J Gj is

injective on K. Since the group Gj is locally embeddable into C, we can find,for each j ∈ J , a πj(K)-almost homomorphism ϕj : Gj → Cj of Gj into agroup Cj ∈ C. As the class C is closed under finite direct products, the groupC =

∏j∈J Cj is also in C. Consider the map ϕ : G → C defined by ϕ = ϕJ◦πJ ,

where ϕJ =∏

j∈J ϕj : GJ → C. For all k = (ki)i∈I , k′ = (k′

i)i∈I ∈ K, wehave

ϕ(kk′) = ϕJ(πJ (kk′))= ϕJ((kjk

′j)j∈J )

= (ϕj(kjk′j))j∈J

= (ϕj(kj)ϕj(k′j))j∈J

= (ϕj(kj))j∈J (ϕj(k′j))j∈J

= ϕ(k)ϕ(k′).

Moreover, ϕ|K is injective. Indeed, given k = (ki)i∈I , k′ = (k′

i)i∈I ∈ K, ifϕ(k) = ϕ(k′) then ϕj(kj) = ϕj(k′

j) for all j ∈ J . By injectivity of ϕj |πj(K),we deduce that kj = k′

j for all j ∈ J . This implies that k = k′ since πJ |Kis injective. This shows that ϕ is a K-almost-homomorphism of G into C. Itfollows that G is locally embeddable into C. �


Corollary 7.1.11. Let C be a class of groups which is closed under finitedirect products. Let (Gi)i∈I be a family of groups which are locally embeddableinto C. Then their direct sum G =

⊕i∈I Gi is locally embeddable into C.



g = (gi)i∈I ∈ P for which gi = 1Gi for all but finitely many i ∈ I. �

Corollary 7.1.12. Let C be a class of groups which is closed under finitedirect products. If a group G is the limit of a projective system of groupswhich are locally embeddable into C, then G is locally embeddable into C.

Proof. Let (Gi)i∈I be a projective system of groups which are locally embed-dable into C such that G = lim←−Gi. By construction of a projective limit (seeAppendix E), G is a subgroup of the group

∏i∈I Gi. We deduce that G is

locally embeddable into C by using Proposition 7.1.10 and Proposition 7.1.7.�

Recall that if C is a class of groups, then a group G is called residuallyC if for each element g ∈ G with g �= 1G, there exist a group C ∈ C and asurjective homomorphism φ : G → C such that φ(g) �= 1C .

Proposition 7.1.13. Let C be a class of groups which is closed under finitedirect products. Let G be a group which is residually C. Then, for every finitesubset K ⊂ G, there exist a group C ∈ C and a homomorphism ϕ : G → Cwhose restriction to K is injective.

Proof. Fix a finite subset K ⊂ G and consider the set

L = {hk−1 : h, k ∈ K and h �= k}.

Since G is residually C, we can find, for each g ∈ L, a group Cg ∈ C and ahomomorphism φg : G → Cg such that φg(g) �= 1Cg . The group C =

∏g∈L Cg

is in the class C since L is finite and C is closed under finite direct products.Consider the group homomorphism ϕ =

∏g∈L φg : G → C. If h and k are

distinct elements in K, then g = hk−1 ∈ L and φg(hk−1) = φg(g) �= 1Cg . Itfollows that ϕ(hk−1) �= 1C and hence ϕ(h) �= ϕ(k). Thus, the restriction ofϕ to K is injective. �

Corollary 7.1.14. Let C be a class of groups which is closed under finitedirect products. Then every group which is residually C is locally embeddableinto C.

Proof. Let G be a group which is residually C and let K ⊂ G be a finite subset.By Proposition 7.1.13, there exist a group C ∈ C and a homomorphismϕ : G → C whose restriction to K is injective. Such a ϕ is a K-almost-homomorphism. Consequently, G is locally embeddable into C. �


Corollary 7.1.15. Let C be a class of groups which is closed under finitedirect products. Then every group which is locally residually C is locally em-beddable into C.

Proof. This immediately follows from Corollary 7.1.14 and Proposition 7.1.8.�

Given a group Γ , let N (Γ ) denote the space of all Γ -marked groups (cf.Sect. 3.4). Recall that N (Γ ) may be identified with the set consisting of allnormal subgroups of Γ and that N (Γ ) is a compact Hausdorff space for thetopology induced by the prodiscrete topology on P(Γ ) = {0, 1}Γ .

Theorem 7.1.16. Let Γ be a group and let C be a class of groups. Let N ∈N (Γ ). Suppose that there exists a net (Ni)i∈I which converges to N in N (Γ )such that Γ/Ni ∈ C for all i ∈ I. Then the group Γ/N is locally embeddableinto C.

Proof. Denote by ρ : Γ → Γ/N and ρi : Γ → Γ/Ni the quotient homomor-phisms. Fix a finite subset K ⊂ Γ/N and let F ⊂ Γ be a finite symmetricsubset such that 1Γ ∈ F and ρ(F ) = K∪K−1∪{1Γ/N}. Since the net (Ni)i∈I

converges to N , we can find i0 ∈ I such that

N ∩ F 4 = Ni0 ∩ F 4. (7.1)

Set C = Γ/Ni0 and define a map ϕ : Γ/N → C by setting

ϕ(g) =

{ρi0(f

′) if g = ρ(f ′) for some f ′ ∈ F 2

1C otherwise.(7.2)

Note that ϕ is well defined. Indeed, suppose that g = ρ(f ′1) = ρ(f ′

2), forsome f ′

1, f′2 ∈ F 2. Then, from 1Γ/N = ρ(f ′

1)−1ρ(f ′

2) = ρ((f ′1)

−1f ′2), we deduce

that(f ′

1)−1f ′

2 ∈ ker(ρ) = N. (7.3)

As (f ′1)

−1f ′2 ∈ F 4, we deduce from (7.3) and (7.1) that (f ′

1)−1f ′

2 ∈ Ni0 =ker(ρi0). Therefore, we have ρi0(f

′1) = ρi0(f

′2).

Let us check now that ϕ is a K-almost-homomorphism. Let k1, k2 ∈ K.Then, there exist f1, f2 ∈ F such that ρ(f1) = k1 and ρ(f2) = k2. Observethat f1, f2 ∈ F 2 since F ⊂ F 2. Thus, we get ϕ(k1) = ρi0(f1) and ϕ(k2) =ρi0(f2) by applying (7.2). On the other hand, we also have f1f2 ∈ F 2 andk1k2 = ρ(f1)ρ(f2) = ρ(f1f2). Therefore, by applying again (7.2), we obtain

ϕ(k1k2) = ρi0(f1f2)= ρi0(f1)ρi0(f2)= ϕ(k1)ϕ(k2).

Moreover, if ϕ(k1) = ϕ(k2), then ρi0(f1) = ρi0(f2) and hence f−11 f2 ∈

ker(ρi0) = Ni0 . As f−11 f2 ∈ F 2 ⊂ F 4, we deduce from (7.1) that f−1

1 f2 ∈


N = ker(ρ). Therefore k1 = ρ(f1) = ρ(f2) = k2. This shows that ϕ is injec-tive on K. We have shown that ϕ is a K-almost-homomorphism of Γ/N intothe group C ∈ C. Therefore, Γ/N is locally embeddable into C. �Corollary 7.1.17. Let Γ be a group and let C be a class of groups. Then theset of all N ∈ N (Γ ) such that Γ/N is locally embeddable into C is closed(and hence compact) in N (Γ ).

Proof. Let (Ni)i∈I be a convergent net in N (Γ ) and suppose that the groupsΓ/Ni are locally embeddable into C for all i ∈ I. Let N = limi∈I Ni. By ap-plying Theorem 7.1.16, we deduce that Γ/N is locally embeddable into theclass of groups which are locally embeddable into C. It follows from Proposi-tion 7.1.9 that Γ/N is locally embeddable into C. �

For the next theorem we need some (slightly technical) preliminaries.Let n ≥ 2 be an integer and G, C two groups. Given a finite subset K ⊂ G,

we say that a map ϕ : G → C is an n-K-almost-homomorphism if it satisfiesthe following conditions:

(n-K-AH-1) ϕ(kε11 kε2

2 · · · kεtt ) = ϕ(k1)ε1ϕ(k2)ε2 · · ·ϕ(kt)εt

for all ki ∈ K ∪ {1G}, εi ∈ {−1, 1}, 1 ≤ i ≤ t ≤ n;(n-K-AH-2) ϕ(1G) = 1C ;(n-K-AH-3) the restriction of ϕ to (K ∪ K−1 ∪ {1G})n is injective.

Lemma 7.1.18. Let G be a group and let C be a class of groups. Let n ≥ 2be an integer. Then, the following conditions are equivalent:

(a) G is locally embeddable into C;(b) for every finite subset K ⊂ G, there exist a group C ∈ C and an n-K-

almost-homomorphism ϕ : G → C.

Proof. It is clear that any n-K-almost-homomorphism is also a K-almost-homomorphism (take ε1 = ε2 = 1 and t = 2 in (n-K-AH-1), and observethat K ⊂ (K ∪ K−1 ∪ {1G})n). Thus, (b) implies (a).

Conversely, suppose (a). Given a finite subset K ⊂ G, set K ′ = (K ∪K−1 ∪ {1G})n. As G is locally embeddable into C, there exists a K ′-almost-homomorphism ϕ : G → C of G into a group C ∈ C. Let us show thatϕ is an n-K-almost-homomorphism. As the restriction of ϕ to K ′ is injec-tive, Property (n-K-AH-3) is trivially satisfied. On the other hand, we haveϕ(k1k2) = ϕ(k1)ϕ(k2) for all k1, k2 ∈ K ′. For k1 = k2 = 1G, this gives usϕ(1G) = ϕ(1G)2, so that Property (n-K-AH-2) also holds. By taking k1 ∈ K ′

and k2 = k−11 , we get 1C = ϕ(1G) = ϕ(k1k

−11 ) = ϕ(k1)ϕ(k−1

1 ). This im-plies that ϕ(k−1

1 ) = ϕ(k1)−1 for all k1 ∈ K ′. Then, Property (n-K-AH-1)immediately follows by induction on t. This shows that ϕ is an n-K-almost-homomorphism of G into C. �Theorem 7.1.19. Let F be a free group and let C be a class of groups whichis closed under taking subgroups. Let N ∈ N (F ) and suppose that the groupF/N is locally embeddable into C. Then there exists a net (Ni)i∈I whichconverges to N in N (F ) such that F/Ni ∈ C for all i ∈ I.


Proof. Let X ⊂ F be a free base of F and denote by ρ : F → F/N thequotient homomorphism. Let I denote the set of all finite subsets E of Fpartially ordered by inclusion. Given E ∈ I, we denote by UE ⊂ X thesubset of X consisting of all elements which appear in the reduced form ofsome element in E. Since E is finite, UE is also finite and there exists nE ∈ N

such that each w ∈ E may be written in the form

w = xε11 xε2

2 · · ·xεtt (7.4)

for suitable xi ∈ UE , εi ∈ {−1, 1}, i = 1, 2, . . . , t, and 1 ≤ t ≤ nE . SetKE = ρ(BnE

) ⊂ F/N , where BnEdenotes the ball of radius nE centered at

the identity element 1F in the Cayley graph of the subgroup of F generatedby UE . Since F/N is locally embeddable into C it follows from Lemma 7.1.18that there exists a group CE ∈ C and an nE-KE-almost-homomorphismϕE : F/N → CE . As F is a free group with base X, there exists a uniquehomomorphism ρE : F → CE such that ρE(x) = ϕE(ρ(x)) for all x ∈ X.Setting NE = ker(ρE) we have NE ∈ N (F ). Moreover, since C is closed undertaking subgroups, we have that the quotient group F/NE , being isomorphicto ρE(F ), which is a subgroup of CE , also belongs to C.

Let us show that the net (NE)E∈I converges to N . Fix a finite subsetE0 ⊂ F . Let E ∈ I be such that E0 ⊂ E. Let us show that

N ∩ E0 = NE ∩ E0. (7.5)

Given w ∈ E0, we can write w as in (7.4). Using the fact that ϕE is annE-KE-almost-homomorphism, we get

ρE(w) = ρE(xε11 xε2

2 · · ·xεtt )

= ρE(x1)ε1ρE(x2)ε2 · · · ρE(xt)εt

= ϕE(ρ(x1))ε1ϕE(ρ(x2))ε2 · · ·ϕE(ρ(xt))εt

= ϕE(ρ(x1)ε1ρ(x2)ε2 · · · ρ(xt)εt)= ϕE(ρ(xε1

1 xε22 · · ·xεt

t ))= ϕE(ρ(w)).

As ϕE is injective on KE , and ρ(w) ∈ ρ(BE) = KE , we deduce that ρE(w) =1F/NE

if and only if ρ(w) = 1F/N , equivalently, w ∈ NE if and only if w ∈ N .Thus (7.5) follows. This shows that N = limE NE . �

From Theorem 7.1.16 (with Γ a free group F ) and Theorem 7.1.19 weimmediately deduce the following.

Corollary 7.1.20. Let F be a free group and let C be a class of groups whichis closed under taking subgroups. Let N ∈ N (F ). Then the following condi-tions are equivalent.

(a) the group F/N is locally embeddable into C;


(b) there exists a net (Ni)i∈I which converges to N in N (F ) such that F/Ni ∈C for all i ∈ I. �

Corollary 7.1.21. Let C be a class of groups which is closed under takingsubgroups. Then every finitely presented group which is locally embeddableinto C is residually C.

Proof. Let G be a finitely presented group which is locally embeddable into C.Fix g ∈ G \ {1G}. Since G is finitely presented, there exists a free group F offinite rank and a finite subset R ⊂ F such that G = F/N , where N ∈ N (F )denotes the normal closure of R in F . By Theorem 7.1.19, there exists a net(Ni)i∈I in N (F ) converging to N such that F/Ni ∈ C for all i ∈ I. Letρ : F → G denote the quotient homomorphism and choose an element f ∈ Fsuch that ρ(f) = g. Since the net (Ni)i∈I converges to N , we can find i0 ∈ Isuch that

N ∩ (R ∪ {f}) = Ni0 ∩ (R ∪ {f}). (7.6)

As R ⊂ N , this implies R ⊂ Ni0 and hence N ⊂ Ni0 . Thus, there is acanonical epimorphism φ : G = F/N → C = F/Ni0 . Observe now that f /∈ Nand therefore f /∈ Ni0 by (7.6). It follows that φ(g) �= 1C . As C ∈ C, thisshows that G is residually C. �

Corollary 7.1.22. Let C be a class of groups which is closed under takingsubgroups and under finite direct products. Then a finitely presented group islocally embeddable into C if and only if it is residually C.

Proof. The fact that every finitely presented group which is locally embed-dable into C is residually C follows from the previous corollary. The converseimplication follows from Corollary 7.1.14. �

We end this section by producing an example showing that Theorem 7.1.19becomes false if we omit the hypothesis that F is free. Let us first establishthe following:

Proposition 7.1.23. Let C be a class of groups. Let G be a group. Supposethat there exists a net (Ni)i∈I which converges to {1G} in N (G) such thatG/Ni ∈ C for all i ∈ I. Then G is residually C.

Proof. Let g ∈ G \ {1G}. Let us set F = {1G, g}. Since the net (Ni)i∈I

converges to {1G} in N (G), there exists i0 ∈ I such that Ni0 ∩F = {1G}∩F .As F ∩{1G} = {1G}, this implies g /∈ Ni0 . Thus, the quotient homomorphismφ : G → G/Ni0 satisfies φ(g) �= 1G/Ni0

. As G/Ni0 ∈ C, this shows that G isresidually C. �

Now, to exhibit the promised example showing the necessity of the freenesshypothesis on F in Theorem 7.1.19, we consider the additive group G = Q

of rational numbers and take as C the class of finite groups. It follows fromExample 2.1.9 that Q is not residually finite. Thus, by the preceding propo-

7.2 Local Embeddability and Ultraproducts 243

sition, there is no net (Ni)i∈I converging to {1G} in N (G) such that G/Ni

is finite for all i ∈ I. On the other hand, G is locally residually finite sinceevery finitely generated abelian group is residually finite by Corollary 2.2.4.Therefore, G is locally embeddable into C by Corollary 7.1.15.

7.2 Local Embeddability and Ultraproducts

Suppose that we are given a family of groups (Gi)i∈I and a filter ω (cf.Sect. J.1) on the index set I. Consider the group

P =∏

i∈I

Gi.

Let α = (αi)i∈I and β = (βi)i∈I be elements of P . We write α ∼ω β if{i ∈ I : αi = βi} ∈ ω.

Proposition 7.2.1. One has:

(i) α ∼ω α;(ii) α ∼ω β if and only if β ∼ω α;(iii) if α ∼ω β and β ∼ω γ, then α ∼ω γ;(iv) if α ∼ω β and γ ∼ω δ, then αγ ∼ω βδ;(v) α ∼ω β if and only if α−1 ∼ω β−1;

for all α, β, γ, δ ∈ P .

Proof. Let α = (αi)i∈I , β = (βi)i∈I , γ = (γi)i∈I and δ = (δi)i∈I ∈ P . Wehave α ∼ω α since {i ∈ I : αi = αi} = I belongs to ω (cf. (F-6) in Sect. J.1).This shows (i). Also, since {i ∈ I : αi = βi} = {i ∈ I : βi = αi} we deduce(ii). Suppose now that α ∼ω β and β ∼ω γ. We have {i ∈ I : αi = γi} ⊃{i ∈ I : αi = βi} ∩ {i ∈ I : βi = γi}. Thus (iii) follows from the fact thatω, being a filter, is closed under finite intersections and taking supersets (cf.(F-2) and (F-3) in Sect. J.1). Suppose now that α ∼ω β and γ ∼ω δ, that is,{i ∈ I : αi = βi} and {i ∈ I : γi = δi} both belong to ω. We have

{i ∈ I : αiγi = βiδi} ⊃ {i ∈ I : αi = βi} ∩ {i ∈ I : γi = δi}.

As ω is closed under finite intersections and taking supersets, we deduce that{i ∈ I : αiγi = βiδi} ∈ ω, that is, αγ ∼ω βδ. This shows (iv). Finally, from{i ∈ I : αi = βi} = {i ∈ I : α−1

i = β−1i } we deduce (v). �

Note that it follows from (i), (ii) and (iii) in Proposition 7.2.1 that ∼ω isan equivalence relation in P . Consider now the subset Nω ⊂ P defined by

Nω = {α ∈ P : α ∼ω 1P = (1Gi)i∈I}.


Proposition 7.2.2. The set Nω is a normal subgroup of P .

Proof. This is an easy consequence of the properties of ∼ω stated in Propo-sition 7.2.1. Indeed, we immediately deduce from Proposition 7.2.1(i) that1P ∈ Nω. Suppose now that α, β ∈ Nω, that is, α ∼ω 1P and β ∼ω 1P . Itfollows from Proposition 7.2.1(iv) that αβ ∼ω 1P 1P = 1P , that is, αβ ∈ Nω.Similarly, from Proposition 7.2.1(v) we deduce that α−1 ∼ω 1−1

P = 1P ,that is, α−1 ∈ Nω. Thus, Nω is a subgroup of P . Finally, let γ ∈ P .Proposition 7.2.1(iv) implies that γαγ−1 ∼ω γ1P γ−1 = 1P . It follows thatγαγ−1 ∈ Nω for all α ∈ Nω and γ ∈ P . This shows that Nω is a normalsubgroup of P . �

Observe that, given α and β in P , one has

αNω = βNω ⇐⇒ α ∼ω β. (7.7)

Indeed, one has αNω = βNω if and only if αβ−1 ∈ Nω, that is, if and only ifαβ−1 ∼ω 1P . This is equivalent to α ∼ω β by Proposition 7.2.1(iv).

The quotient group Pω = P/Nω is called the reduced product of the familyof groups (Gi)i∈I with respect to the filter ω. In the particular case whenω is an ultrafilter, one also says that Pω is the ultraproduct of the family ofgroups (Gi)i∈I with respect to the ultrafilter ω.

Theorem 7.2.3. Let C be a class of groups, (Gi)i∈I a family of groups suchthat Gi ∈ C for all i ∈ I, and ω an ultrafilter on the index set I. Then theultraproduct Pω of the family of groups (Gi)i∈I with respect to the ultrafilterω is locally embeddable into C.

Proof. Fix a finite subset K ⊂ Pω. We want to show that there exist a groupC ∈ C and a K-almost-homomorphism ϕ : Pω → C.

Choose a representative of each element g ∈ Pω, that is, an element g =(gi)i∈I ∈ P such that g = gNω. If h and k are arbitrary elements of K, wehave hkNω = hkNω and therefore hk ∼ω hk by (7.7). It follows that the setIh,k = {i ∈ I : hki = hiki} belongs to ω. Thus, since ω is closed under finiteintersections (cf.(F-5) in Sect. J.1), we have that

IK =⋂

h,k∈K

Ih,k

also belongs to ω. On the other hand, if h and k are distinct elements of K,the subset {i ∈ I : hi = ki} does not belongs to ω. As ω is an ultrafilter,this implies that I ′h,k = I \ {i ∈ I : hi = ki} = {i ∈ I : hi �= ki} belongs toω (cf. (UF) in Sect. J.1). Using again the fact that ω is closed under finiteintersections, we deduce that

I ′K =⋂

h,k∈Kh�=k

I ′h,k

7.2 Local Embeddability and Ultraproducts 245

also belongs to ω. Since ω is closed under finite intersections and ∅ /∈ ω (cf.(F-1) in Sect. J.1), we can find an index i0 ∈ I such that

i0 ∈ IK ∩ I ′K .

Consider the map ϕ : Pω → Gi0 defined by ϕ(g) = gi0 for all g ∈ Pω. Themap ϕ satisfies the following properties:

(1) ϕ(hk) = hki0 = hi0 ki0 = ϕ(h)ϕ(k) for all h, k ∈ K (because i0 ∈ IK);(2) ϕ(h) = hi0 �= ki0 = ϕ(k) for all h, k ∈ K such that h �= k (because

i0 ∈ I ′K).

This shows that ϕ is a K-almost homomorphism. It follows that Pω is locallyembeddable into the class C. �

Remark 7.2.4. When the ultrafilter ω is principal, then the group Pω itselfbelongs to C. Indeed, in this case, there is an element i0 ∈ I such that ωconsists of all subsets of I containing i0. This implies that α ∼ω β if and onlyif αi0 = βi0 for all α = (αi)i∈I , β = (βi)i∈I ∈ P . Therefore, Nω consists ofall α ∈ P such that αi0 = 1Gi0

. It follows that the group Pω is isomorphic tothe group Gi0 and therefore that Pω ∈ C.

Theorem 7.2.5. Let C be a class of groups and let G be a group. The fol-lowing conditions are equivalent:

(a) G is locally embeddable into C;(b) there exists a family of groups (Gi)i∈I such that Gi ∈ C for all i ∈ I

and an ultrafilter ω on I such that G is isomorphic to a subgroup of theultraproduct Pω of the family (Gi)i∈I with respect to the ultrafilter ω.

Proof. If G satisfies (b), then G is locally embeddable into C since Pω islocally embeddable into C by Theorem 7.2.3 and every subgroup of a groupwhich is locally embeddable into C is itself locally embeddable into C byProposition 7.1.7.

Conversely, suppose that G is locally embeddable into C. Consider the setI consisting of all finite subsets of G.

For each K ∈ I we define the set

IK = {K ′ ∈ I : K ⊂ K ′}.

Observe that IK �= ∅ as K ∈ IK . Moreover, the family of nonempty subsets(IK)K∈I is closed under finite intersections, since

IK1 ∩ IK2 = IK1∪K2

for all K1, K2 ∈ I. It follows from Proposition J.1.3 and Theorem J.1.6 thatthere exists an ultrafilter ω on I such that IK ∈ ω for all K ∈ I.

As G is locally embeddable into C, we can find, for each K ∈ I, a groupGK in C and a K-almost-homomorphism ϕK : G → GK .


Consider the ultraproduct Pω of the family (GK)K∈I with respect to ω. ByTheorem 7.2.3, Pω is locally embeddable into C. Let ϕ : G → P =

∏K∈I GK

denote the product map ϕ =∏

K∈I ϕK . Thus, we have

ϕ(g) = (ϕK(g))K∈I

for all g ∈ G. Let ρ : P → Pω = P/Nω denote the canonical epimorphism.Let us show that the composite map Φ = ρ ◦ ϕ : G → Pω is an injectivehomomorphism. This will prove that G is isomorphic to a subgroup of Pω.

Let g, h ∈ G. Consider the element K0 = {g, h} ∈ I. If K ∈ I satisfiesK0 ⊂ K, then

ϕK(gh) = ϕK(g)ϕK(h)

since ϕK is a K-almost-homomorphism. Thus, the set

{K ∈ I : ϕK(gh) = ϕK(g)ϕK(h)}

contains IK0 and therefore belongs to ω. This implies that

ϕ(gh) ∼ω ϕ(g)ϕ(h).

Therefore, we have Φ(gh) = Φ(g)Φ(h) by (7.7). This shows that Φ is a homo-morphism.

On the other hand, if g �= h, we have

ϕK(g) �= ϕK(h)

for all K ∈ I such that K0 ⊂ K. This implies that {K ∈ I : ϕK(g) �=ϕK(h)} ∈ ω. As ω is a filter, it follows that {K ∈ I : ϕK(g) = ϕK(h)} /∈ ω,that is,

ϕ(g) �∼ω ϕ(h).

Therefore we have Φ(g) �= Φ(h), again by (7.7).Consequently, Φ is injective. This shows that (a) implies (b). �

Remark 7.2.6. Suppose that G is a group and that C is a class of groups whichis closed under taking subgroups. If G is locally embeddable into C and G /∈ Cthen for any family of groups (Gi)i∈I and any ultrafilter ω on the index set Isatisfying condition (b) in Theorem 7.2.5, we deduce from Remark 7.2.4 thatthe ultrafilter ω is necessarily non-principal.

7.3 LEF-Groups and LEA-Groups

Let us rewrite the results obtained in Sect. 7.1 in the particular case whenC is either the class of finite groups or the class of amenable groups. Notethat these classes are closed under taking subgroups and taking finite direct

7.3 LEF-Groups and LEA-Groups 247

products (this is trivial for finite groups and follows from Proposition 4.5.1and Corollary 4.5.6 for amenable groups), so that all the results establishedin Sect. 7.1 apply to these two classes.

We shall use the following terminology, which is very popular in the field.A group which is locally embeddable into the class of finite groups is brieflycalled an LEF -group. Similarly, a group which is locally embeddable into theclass of amenable groups is called an LEA-group. Note that every LEF-groupis LEA since the class of finite groups is contained in the class of amenablegroups by Proposition 4.4.6.

We deduce from Proposition 7.1.6 the following characterization of LEF-groups.


(a) G is an LEF-group;(b) for every finite subset K ⊂ G, there exist a finite set L with K ⊂ L ⊂ G

and a binary operation � : L × L → L such that (L,�) is a group andk1k2 = k1 � k2 for all k1, k2 ∈ K.

�

Analogously, we deduce from Proposition 7.1.6 the following characteriza-tion of LEA-groups.


(a) G is an LEA-group;(b) for every finite subset K ⊂ G, there exist a set L with K ⊂ L ⊂ G and

a binary operation � : L × L → L such that (L,�) is an amenable groupand k1k2 = k1 � k2 for all k1, k2 ∈ K.

�

From Theorem 7.2.5 we deduce the following characterization of LEF (resp.LEA) groups in terms of ultraproducts.

Corollary 7.3.3. Let G be a group. The following conditions are equivalent:

(a) G is an LEF (resp. LEA) group;(b) there exists a family of finite (resp. amenable) groups (Gi)i∈I and an ul-

trafilter ω on I such that G is isomorphic to a subgroup of the ultraproductPω of the family (Gi)i∈I with respect to the ultrafilter ω.

�

From Corollary 7.1.15, we get:

Proposition 7.3.4. Every locally residually finite group is LEF and thereforeLEA. �


Corollary 7.3.5. All finite groups, all residually finite groups, all profinitegroups, all free groups, all abelian groups, and all locally finite groups areLEF and therefore LEA.

Proof. In order to complete the proof, it suffices to recall that free groups(resp. profinite groups) are residually finite by Theorem 2.3.1 (resp. Corol-lary 2.2.8) and that abelian groups are locally residually finite by Corol-lary 2.2.4. �

Similarly, by applying Corollary 7.1.15 to the class of amenable groups,we get:

Proposition 7.3.6. Every locally residually amenable group is LEA. �

From Proposition 7.1.7, Proposition 7.1.8, Proposition 7.1.10, Corol-lary 7.1.11, Corollary 7.1.12, Corollary 7.1.17, and Theorem 7.1.19, we get:

Proposition 7.3.7. The following assertions hold:

(i) every subgroup of an LEF-group (resp. LEA-group) is an LEF-group(resp. LEA-group);

(ii) a group is LEF (resp. LEA) if and only if all its finitely generated sub-groups are LEF (resp. LEA);

(iii) let (Gi)i∈I be a family of LEF-groups (resp. LEA-groups). Then theirdirect product

∏i∈I Gi is an LEF-group (resp. LEA-group);

(iv) let (Gi)i∈I be a family of LEF-groups (resp. LEA-groups). Then theirdirect sum

⊕i∈I Gi is an LEF-group (resp. LEA-group);

(v) let (Gi)i∈I be a projective system of LEF-groups (resp. LEA-groups).Then their projective limit G = lim←−Gi is an LEF-group (resp. LEA-group);

(vi) Let Γ be a group. Then the set of all N ∈ N (Γ ) such that Γ/N is LEF(resp. LEA) is closed (and hence compact) in N (Γ ).

(vii) let F be a free group and let N ∈ N (F ). Then F/N is an LEF-group(resp. LEA-group) if and only if there exists a net (Ni)i∈I in N (F ) withF/Ni finite (resp. amenable) for all i ∈ I such that N = limi Ni.

�

Finally, we deduce from Corollary 7.1.21 the following:

Proposition 7.3.8. A finitely presented group is LEF (resp. LEA) if andonly if it is residually finite (resp. residually amenable). �

As we have observed at the end of the previous section, the additive groupQ is LEF but not residually finite. On the other hand, it follows from Propo-sition 7.3.8 that every finitely presented LEF-group is residually finite. Thegroup Q is not finitely generated. However, there exist finitely generatedLEF-groups which are not residually finite. An example of such a group isprovided by the group G1 introduced in Sect. 2.6:

7.3 LEF-Groups and LEA-Groups 249

Proposition 7.3.9. The group G1 of Sect. 2.6 is a finitely generated amenableLEF-group which is not residually finite.

Proof. By construction, the group G1 is finitely generated since it is definedas being a subgroup of Sym(Z) generated by two elements, namely the trans-position (0 1) and the translation n �→ n + 1.

The group G1 is not residually finite by Proposition 2.6.1.Recall that, denoting by Sym0(Z) the normal subgroup of Sym(Z) con-

sisting of all permutations of Z with finite support, the group G1 is thesemidirect product of Sym0(Z) with the infinite cyclic group generated bythe translation T : n �→ n + 1 (see Lemma 2.6.4). Therefore, each ele-ment g ∈ G1 can be uniquely written in the form g = T i(g)σ(g), wherei(g) ∈ Z and σ(g) ∈ Sym0(Z). Note that we have i(gh) = i(g) + i(h) andσ(gh) = T−i(h)σ(g)T i(h)σ(h) for all g, h ∈ G1.

To prove that G1 is an LEF-group, consider a finite subset K ⊂ G1. Letus show that there exist a finite group F and a K-almost-homomorphism ofG1 into F .

Let � = maxk∈K |i(k)| and choose an integer r ≥ � such that the supportsof the elements T−jσ(k)T j ∈ Sym0(Z) are contained in the interval [−r, r]for all −� ≤ j ≤ � and k ∈ K. Let us set R = 4r,

X = {−R,−R + 1, . . . ,−1, 0, 1, . . . , R − 1, R},

and F = Sym(X). Consider the (2R + 1)-cycle γ ∈ F given by

γ = (−R − R + 1 − R + 2 · · · R − 1 R).

For all σ ∈ Sym0(Z) whose support is contained in X, let σ ∈ F denote theelement defined by σ(x) = σ(x) for all x ∈ X. Observe that if σ, σ′ ∈ Sym0(Z)have both their support contained in X, then so does σσ′ and that we have

σσ′ = σσ′. (7.8)

Moreover, for all k ∈ K and j ∈ Z such that −� ≤ j ≤ �, the supports ofσ(k) and T−jσ(k)T j are contained in X and we have

T−jσ(k)T j = γ−jσ(k)γj . (7.9)

Indeed, suppose first that x ∈ [−2r, 2r]. Then x + j ∈ [−3r, 3r] ⊂ X and wehave

[T−jσ(k)T j ](x) = [T−jσ(k)](x + j)

= T−j (σ(k)(x + j))= σ(k)(x + j) − j,

and


[γ−jσ(k)γj ](x) = [γ−jσ(k)](x + j)

= γ−j(σ(k)(x + j)

)

= γ−j (σ(k)(x + j))= σ(k)(x + j) − j.

Suppose now that x ∈ X \ [−2r, 2r]. Then [T−jσ(k)T j ](x) = x since thesupport of T−jσ(k)T j is contained in [−�+ r, r + �] ⊂ [−2r, 2r]. On the otherhand, if we denote, for y ∈ Z, by y the unique element in {y + (2R + 1)n :n ∈ Z} ∩ [−R, R], we have

x + j ∈ X \ [−r, r] (7.10)

and

[γ−jσ(k)γj ](x) = [γ−jσ(k)](x + j)

= γ−j(σ(k)(x + j)

)

by (7.10) = γ−j(x + j)= x.

This shows (7.9).Define a map ϕ : G1 → F by setting ϕ(g) = γi(g)σ(g) if the support of

σ(g) is contained in X, and ϕ(g) = 1F otherwise. Let us show that ϕ is aK-almost-homomorphism.

Let k1, k2 ∈ K. We have

ϕ(k1k2) = ϕ[T i(k1)+i(k2)(T−i(k2)σ(k1)T i(k2)σ(k2))]

= γi(k1)+i(k2)T−i(k2)σ(k1)T i(k2)σ(k2)

(by (7.8)) = γi(k1)+i(k2) · T−i(k2)σ(k1)T i(k2)σ(k2)

(by (7.9)) = γi(k1)+i(k2) · γ−i(k2)σ(k1)γi(k2) · σ(k2)

= γi(k1)σ(k1)γi(k2)σ(k2)

= ϕ[T i(k1)σ(k1)]ϕ[T i(k2)σ(k2)]= ϕ(k1)ϕ(k2).

Let us show that ϕ|K is injective. Let k1, k2 ∈ K and suppose that ϕ(k1) =ϕ(k2). This implies that γi(k1)σ(k1) = γi(k2)σ(k2), that is, γi(k1)−i(k2) =σ(k2) · σ(k1)

−1. But the support of σ(k2) · σ(k1)

−1is contained in [−r, r]

while the support of γi, i ∈ Z, is the whole set X if i is not a multiple of2R + 1. As |i(k1) − i(k2)| ≤ 2� < 2R + 1, we deduce that i(k1) − i(k2) = 0and σ(k2) · σ(k1)

−1= 1F . This implies i(k1) = i(k2) and σ(k1) = σ(k2), and

therefore k1 = k2. This shows that the restriction of ϕ to K is injective. It

7.4 The Hamming Metric 251

follows that ϕ is a K-almost homomorphism and therefore that G1 is an LEFgroup.

Finally, observe that the group Sym0(Z) is locally finite and thereforeamenable by Corollary 4.5.12. Now, G1 is the semidirect product of theamenable group Sym0(Z) with an infinite cyclic (and therefore amenable)group so that, by Proposition 4.5.5, it is an amenable group as well. �Remarks 7.3.10. (a) From Proposition 7.3.4 and Proposition 7.3.9, we deducethat the class of locally residually finite groups is strictly contained in theclass of LEF-groups.

(b) Proposition 7.3.8 and Proposition 7.3.9 imply that the group G1 is notfinitely presentable.

7.4 The Hamming Metric

Let G be a group.A metric d on G is called left-invariant (resp. right-invariant) if d(hg1,

hg2) = d(g1, g2) (resp. d(g1h, g2h) = d(g1, g2)) for all g1, g2, h ∈ G. A metricon G which is both left and right-invariant is called bi-invariant.

Note that a left-invariant (resp. right-invariant) metric d on G is entirelydetermined by the map g �→ d(1G, g), g ∈ G, since d(h, k) = d(1G, h−1k)(resp. d(h, k) = d(1G, kh−1)) for all h, k ∈ G.

Let now F be a nonempty finite set and consider the symmetric groupSym(F ). For α ∈ Sym(F ), we denote by Fix(α) the set {x ∈ F : α(x) = x} offixed points of α. The support of α is the set {x ∈ F : α(x) �= x} = F \Fix(α),so that we have

|{x ∈ F : α(x) �= x}| = |F | − |Fix(α)|. (7.11)

Consider the map dF : Sym(F ) × Sym(F ) → R defined by

dF (α1, α2) =|{x ∈ F : α1(x) �= α2(x)}|

|F | (7.12)

for all α1, α2 ∈ Sym(F ). Observe that the set {x ∈ F : α1(x) �= α2(x)} ={x ∈ F : x �= α−1

1 α2(x)} is the support of α−11 α2, so that (7.11) gives us

dF (α1, α2) = 1 − |Fix(α−11 α2)|

|F | . (7.13)

Proposition 7.4.1. Let F be a nonempty finite set. Then dF is a bi-invariantmetric on Sym(F ).

Proof. It is immediate from the definition that dF (α1, α2)≥ 0 and dF (α1, α2)=dF (α2, α1) for all α1, α2 ∈ Sym(F ). Moreover, the equality dF (α1, α2) = 0holds if and only if α1(x) = α2(x) for all x ∈ F , that is, if and only if α1 = α2.


Let now α1, α2, α3 ∈ Sym(F ). If x ∈ F satisfies α1(x) �= α2(x), thenα1(x) �= α3(x) or α2(x) �= α3(x). Thus, we have the inclusion

{x∈F : α1(x) �= α2(x)} ⊂ {x∈F : α1(x) �= α3(x)} ∪ {x∈F : α2(x) �= α3(x)}.

This implies

dF (α1, α2) =1|F | |{x ∈ F : α1(x) �= α2(x)}|

≤ 1|F | |{x ∈ F : α1(x) �= α3(x)} ∪ {x ∈ F : α2(x) �= α3(x)}|

≤ 1|F | (|{x ∈ F : α1(x) �= α3(x)}| + |{x ∈ F : α2(x) �= α3(x)}|)

= dF (α1, α3) + dF (α2, α3).

This shows that dF also satisfies the triangle inequality. Therefore, dF is ametric on Sym(F ).

It remains to show that dF is bi-invariant. Let α1, α2, β ∈ Sym(F ). Sinceβ is bijective, we have

{x ∈ F : βα1(x) �= βα2(x)} = {x ∈ F : α1(x) �= α2(x)}.

This implies that

dF (βα1, βα2) =1|F | |{x ∈ F : βα1(x) �= βα2(x)}|

=1|F | |{x ∈ F : α1(x) �= α2(x)}|

= dF (α1, α2).

Thus, dF is left-invariant. On the other hand, we have

{x ∈ F : α1β(x) �= α2β(x)} = β−1({x ∈ F : α1(x) �= α2(x)}),

which implies

dF (α1β, α2β) =1|F | |{x ∈ F : α1β(x) �= α2β(x)}|

=1|F | |β

−1({x ∈ F : α1(x) �= α2(x)})|

=1|F | |{x ∈ F : α1(x) �= α2(x)}|

= dF (α1, α2).

Consequently, dF is also right-invariant. �Definition 7.4.2. Let F be a nonempty finite set. The bi-invariant metricdF is called the (normalized) Hamming metric on Sym(F ).

Suppose now that m is a positive integer and that F1, F2, . . . , Fm arenonempty finite sets. Consider the Cartesian product F = F1×F2×· · ·×Fm

7.4 The Hamming Metric 253

and the natural group homomorphism Φ :∏m

i=1 Sym(Fi) → Sym(F ) definedby

Φ(α)(x) = (α1(x1), α2(x2), . . . , αm(xm))

for all α = (α1, α2, . . . , αm) ∈∏m

i=1 Sym(Fi) and x = (x1, x2, . . . , xm) ∈ F .

Proposition 7.4.3. With the above notation, one has

dF (Φ(α), Φ(β)) = 1 −m∏

i=1

(1 − dFi(αi, βi)) (7.14)

for all α = (αi)1≤i≤m and β = (βi)1≤i≤m in∏m

i=1 Sym(Fi).

Proof. First observe that if α = (αi)1≤i≤m ∈∏m

i=1 Sym(Fi), then we haveFix(Φ(α)) =

∏mi=1 Fix(αi), and therefore

dF (IdF , Φ(α)) = 1 − |Fix(α)||F |

= 1 −∏m

i=1 |Fix(αi)|∏mi=1 |Fi|

= 1 −m∏

i=1

|Fix(αi)||Fi|

= 1 −m∏

i=1

(1 − dFi(IdFi , αi)).

We deduce that, for all α = (αi)1≤i≤m, β = (βi)1≤i≤m ∈∏m

i=1 Sym(Fi), wehave

dF (Φ(α), Φ(β)) = d(IdF , Φ(α)−1Φ(β)) (by left-invariance of dF )

= dF (IdF , Φ(α−1β))

= 1 −m∏

i=1

(1 − dFi(IdFi , α−1i βi)

= 1 −m∏

i=1

(1 − dFi(αi, βi))

where the last equality follows from the left-invariance of dFi . �Corollary 7.4.4. Let m be a positive integer and let F be a nonempty finiteset. Consider the homomorphism

Ψ : Sym(F ) → Sym(Fm) (7.15)

defined by

Ψ(α)(x1, x2, . . . , xm) = (α(x1), α(x2), . . . , α(xm))


for all α ∈ Sym(F ) and x1, x2, . . . , xm ∈ F . Then one has

dF m(Ψ(α), Ψ(β)) = 1 − (1 − dF (α, β))m (7.16)

for all α, β ∈ Sym(F ). �

7.5 Sofic Groups

Definition 7.5.1. Let G be a group, K ⊂ G a finite subset, and ε > 0. LetF be a nonempty finite set. A map ϕ : G → Sym(F ) is called a (K, ε)-almost-homomorphism if it satisfies the following conditions:

((K, ε)-AH-1) for all k1, k2 ∈ K, one has dF (ϕ(k1k2), ϕ(k1)ϕ(k2)) ≤ ε;((K, ε)-AH-2) for all k1, k2 ∈ K, k1 �= k2, one has dF (ϕ(k1), ϕ(k2)) ≥ 1 − ε,

where dF denotes the normalized Hamming metric on Sym(F ).

Definition 7.5.2. A group G is called sofic if it satisfies the following condi-tion: for every finite subset K ⊂ G and every ε > 0, there exist a nonemptyfinite set F and a (K, ε)-almost-homomorphism ϕ : G → Sym(F ).

Proposition 7.5.3. Every finite group is sofic.

Proof. Let G be a finite group. Consider the map L : G → Sym(G) definedby L(g)(h) = gh for all g, h ∈ G. As L is a homomorphism (cf. the proof ofCayley’s theorem (Theorem C.1.2)), we have

dG(L(g1g2), L(g1)L(g2)) = 0 (7.17)

for all g1, g2 ∈ G. Moreover, for all distinct g1, g2 ∈ G we have L(g1)(h) =g1h �= g2h = L(g2)(h) for all h ∈ G so that

dG(L(g1), L(g2)) = 1. (7.18)

This shows that L is a (K, ε)-almost-homomorphism for all K ⊂ G and ε > 0.It follows that G is sofic. �

Proposition 7.5.4. Every subgroup of a sofic group is sofic.

Proof. Let G be a sofic group and let H be a subgroup of G. Fix a finitesubset K ⊂ H and ε > 0. As G is sofic, there exists a nonempty finite set Fand a (K, ε)-almost-homomorphism ϕ : G → Sym(F ). Then the restrictionmap ϕ|H : H → Sym(F ) is a (K, ε)-almost-homomorphism. This shows thatH is sofic. �

Proposition 7.5.5. Every locally sofic group is sofic.

7.5 Sofic Groups 255

Proof. Let G be a locally sofic group. Let K ⊂ G be a finite subset and ε > 0.Denote by H the subgroup of G generated by K. Then, as H is sofic, thereexist a nonempty finite set F and a (K, ε)-almost-homomorphism ψ : H →Sym(F ). Extend arbitrarily ψ to a map ϕ : G → Sym(F ), for example bysetting ϕ(g) = IdF for all g ∈ G \ H. It is clear that ϕ is a (K, ε)-almost-homomorphism. This shows that G is sofic. �

From Proposition 7.5.3 and Proposition 7.5.5, we immediately deduce thatevery locally finite group is sofic. As any locally finite group is amenable byCorollary 4.5.12, this is actually covered by the following:

Proposition 7.5.6. Every amenable group is sofic.

Proof. Suppose that G is an amenable group. Let K ⊂ G be a finite subsetand ε > 0. Set S = ({1G} ∪K ∪K−1)2. Since G is amenable, it follows fromTheorem 4.9.1 and Proposition 4.7.1(a) that there exists a nonempty finitesubset F ⊂ G such that

|F \ sF | ≤ ε

|S| |F | (7.19)

for all s ∈ S. Consider the set E =⋂

s∈S sF . Observe that E ⊂ F since1G ∈ S. In fact, as S = S−1, we get

sE ⊂ F (7.20)

for all s ∈ S. Moreover, we have

|F \ E| = |F \⋂

s∈S

sF |

= |⋃

s∈S

(F \ sF )|

≤∑

s∈S

|F \ sF |

≤ ε|F | by (7.19).

(7.21)

This implies|E| ≥ (1 − ε)|F |. (7.22)

For each g ∈ G, we have |F | = |gF | and hence |F \gF | = |gF \F |. Therefore,we can find a bijective map αg : gF \F → F \ gF . Consider the map ϕ : G →Sym(F ) defined by setting

ϕ(g)(f) =

{gf if gf ∈ F

αg(gf) otherwise

for all g ∈ G and f ∈ F (see Fig. 7.1).


Fig. 7.1 The maps αg : gF \ F → F \ gF and ϕ(g) ∈ Sym(F ). We have ϕ(g)(f1) = gf1,ϕ(g)(f2) = αg(gf2) and ϕ(g)(f3) = αg(gf3)

Now, suppose that k1, k2 ∈ K and f ∈ E. Then we have k2, k1k2 ∈ S,so that k2f, k1k2f ∈ F by (7.20). This implies ϕ(k1k2)(f) = k1k2f and(ϕ(k1)ϕ(k2))(f) = ϕ(k1)(ϕ(k2)(f)) = ϕ(k1)(k2f) = k1k2f . Therefore, thepermutations ϕ(k1k2) and ϕ(k1)ϕ(k2) coincide on E. From (7.21), we deducethat

dF (ϕ(k1k2), ϕ(k1)ϕ(k2)) ≤|F \ E||F | ≤ ε

for all k1, k2 ∈ K.On the other hand, if k1, k2 ∈ K, k1 �= k2, and f ∈ E, then we have

k1f, k2f ∈ F so that ϕ(k1)(f) = k1f �= k2f = ϕ(k2)(f). By using (7.22), wededuce that

dF (ϕ(k1), ϕ(k2)) ≥|E||F | ≥ 1 − ε

for all k1, k2 ∈ K with k1 �= k2. Thus, the map ϕ : G → Sym(F ) is a (K, ε)-almost-homomorphism. This shows that G is a sofic group. �

Observe that, when G is finite, the proof of Proposition 7.5.6 reduces tothat of Proposition 7.5.3 by taking F = G.

Proposition 7.5.7. Let (Gi)i∈I be a family of sofic groups. Then, their directproduct G =

∏i∈I Gi is sofic.

Proof. For each i ∈ I, let πi : G → Gi denote the projection homomorphism.Fix a finite subset K ⊂ G and ε > 0. Then there exists a finite subset J ⊂ I


such that the projection πJ =∏

j∈J πj : G → GJ =∏

j∈J Gj is injective onK. Choose a constant 0 < η < 1 small enough so that

1 − (1 − η)|J| ≤ ε (7.23)

andη ≤ ε. (7.24)

Since the group Gj is sofic for each j ∈ J , we can find a nonempty finite setFj and a (πj(K), η)-almost homomorphism ϕj : Gj → Sym(Fj). Consider thenonempty finite set F =

∏j∈J Fj and the map ϕ : G → Sym(F ) defined by

ϕ(g)(f) = (ϕj(gj)(fj))j∈J

for all g = (gi)i∈I ∈ G, and f = (fj)j∈J ∈ F . Then, for all k, k′ ∈ K, wehave, by applying (7.14),

dF (ϕ(kk′), ϕ(k)ϕ(k′)) = 1 −∏

j∈J

(1 − dFj (ϕj(kjk

′j), ϕj(kj)ϕj(k′

j))

≤ 1 − (1 − η)|J|

≤ ε (by (7.23)).

On the other hand, if k and k′ are distinct elements in K, then there existsj0 ∈ J such that kj0 �= k′

j0. This implies, again by using (7.14),

dF (ϕ(k), ϕ(k′)) = 1 −∏

j∈J

(1 − dFj (ϕj(kj), ϕj(k′j))

≥ 1 − (1 − dFj0(ϕj0(kj0), ϕj0(k

′j0))

≥ 1 − η

≥ 1 − ε (by (7.24)).

This shows that ϕ is a (K, ε)-almost-homomorphism of G. It follows that Gis sofic. �

Corollary 7.5.8. Let (Gi)i∈I be a family of sofic groups. Then their directsum G =

⊕i∈I Gi is sofic.



g = (gi) ∈ P for which gi = 1Gi for all but finitely many i ∈ I. �

Corollary 7.5.9. The limit of a projective system of sofic groups is sofic.

Proof. Let (Gi)i∈I be a projective system of sofic groups such that G =lim←−Gi. By construction of a projective limit (see Appendix E), G is a sub-


group of the group∏

i∈I Gi. We deduce that G is sofic by using Proposi-tion 7.5.7 and Proposition 7.5.4. �

Proposition 7.5.10. Every group which is locally embeddable into the classof sofic groups is sofic.

Proof. Let G be a group which is locally embeddable into the class of soficgroups. Let K ⊂ G be a finite subset and ε > 0. By definition of localembeddability, there exists a sofic group G′ and a K-almost-homomorphismϕ : G → G′. Set K ′ = ϕ(K). By soficity of G′, there exists a nonemptyfinite set F and a (K ′, ε)-almost-homomorphism ϕ′ : G′ → Sym(F ). Let usprove that the composite map Φ = ϕ′ ◦ ϕ : G → Sym(F ) is a (K, ε)-almost-homomorphism. Let k1, k2 ∈ K. Then we have

dF (Φ(k1k2), Φ(k1)Φ(k2)) = dF (ϕ′(ϕ(k1k2)), ϕ′(ϕ(k1))ϕ′(ϕ(k2)))= dF (ϕ′(ϕ(k1)ϕ(k2)), ϕ′(ϕ(k1))ϕ′(ϕ(k2)))≤ ε (as ϕ(k1), ϕ(k2) ∈ K ′).

Finally, let k1, k2 ∈ K be such that k1 �= k2. Since ϕ|K is injective, we haveϕ(k1) �= ϕ(k2) and therefore

dF (Φ(k1), Φ(k2)) = dF (ϕ′(ϕ(k1)), ϕ′(ϕ(k2))) ≥ 1 − ε.

Thus, Φ is a (K, ε)-almost-homomorphism. It follows that G is sofic. �

Since every amenable group is sofic by Proposition 7.5.6, an immediateconsequence of Proposition 7.5.10 is the following:

Corollary 7.5.11. Every LEA-group is sofic. In particular, every LEF-group, every locally residually amenable group, every locally residually finitegroup, every residually amenable group, and every residually finite group issofic. �

As the class of sofic groups is closed under direct products by Proposi-tion 7.5.7, it follows from Corollary 7.1.15 that every locally residually soficgroup is locally embeddable into the class of sofic groups. By applying Propo-sition 7.5.10, we get:

Corollary 7.5.12. Every locally residually sofic group is sofic. �

It follows from Proposition 7.5.10 that the class of groups which are locallyembeddable into the class of sofic groups coincide with the class of soficgroups. By applying Corollary 7.1.17, we then deduce the following:

Corollary 7.5.13. Let Γ be a group. Then the set of Γ -marked groups N ∈N (Γ ) such that Γ/N is sofic is closed (and hence compact) for the prodiscretetopology on N (Γ ) ⊂ P(Γ ) = {0, 1}Γ . �


We end this section by showing that extensions of sofic groups by amenablegroups are sofic. This is a generalization of Proposition 7.5.6 since everyamenable group is an extension of the trivial group by itself.

Proposition 7.5.14. Let G be a group. Suppose that G contains a normalsubgroup N such that N is sofic and G/N is amenable. Then G is sofic.

Proof. Let K ⊂ G be a finite subset and 0 < ε < 1. Denote by g the image ofan element g ∈ G under the canonical epimorphism of G onto G/N . Fix a setT ⊂ G of representatives for the cosets of N in G and denote by σ : G/N → Tthe map which associates with each element in G/N its representative in T .Note that σ(g)−1g ∈ N for all g ∈ G. Also set ε′ = 1 −

√1 − ε, so that

0 < ε′ < ε and (1 − ε′)2 = 1 − ε.Since G/N is amenable and hence sofic by Proposition 7.5.6, there exist

a nonempty finite set F1 and a (K, ε′)-almost-homomorphism ϕ1 : G/N →Sym(F1). In fact, in the proof of Proposition 7.5.6 it is shown that we can takeF1 ⊂ G/N such that there exists a subset E1 ⊂ F1 with |E1| ≥ (1 − ε′)|F1|satisfying ϕ1(k)(f1) = kf1 ∈ F1 and ϕ1(hk)(f1) = hkf1 ∈ F1 for all h, k ∈ Kand f1 ∈ E1.

Set M = N ∩ (σ(F1)−1 ·K ·σ(F1)) ⊂ N . As N is sofic, we can find a finiteset F2 and an (M, ε′)-almost-homomorphism ϕ2 : N → Sym(F2). Thus, forall m, m′ ∈ M we can find a set E2 ⊂ F2 such that |E2| ≥ (1 − ε)|F2| and

ϕ2(mm′)(f2) = ϕ2(m)(ϕ2(m′)(f2)) for all f2 ∈ E2. (7.25)

Set F = F1 × F2 and E = E1 × E2 and observe that

|E| = |E1| · |E2| ≥ (1 − ε′)2|F1| · |F2| = (1 − ε)|F |. (7.26)

Consider the map Φ : G → Sym(F ) defined by setting

Φ(g)(f1, f2) = (ϕ1(g)(f1), ϕ2(σ(gf1)−1gσ(f1))(f2))

for all g ∈ G and (f1, f2) ∈ F . Let us show that Φ is a (K, ε)-almost-homomorphism.

Let h, k ∈ K and (f1, f2) ∈ E. Recall that the elements kf1 = ϕ1(k)(f1)and hkf1 = ϕ1(hk)(f1) = ϕ1(h)(ϕ1(k)(f1)) both belong to F1 for all f1 ∈ E1.It follows that

Φ(h)Φ(k)(f1, f2) = Φ(h)(ϕ1(k)(f1), ϕ2(σ(kf1)−1kσ(f1))(f2))

= (ϕ1(h)(ϕ1(k)(f1)), ϕ2(σ(hϕ1(k)(f1))−1hσ(ϕ1(k)(f1)))

(ϕ2(σ(kf1)−1kσ(f1))(f2)))

= (ϕ1(hk)(f1), ϕ2(σ(hkf1)−1hσ(kf1))

(ϕ2(σ(kf1)−1kσ(f1))(f2)))

=∗ (ϕ1(hk)f1, ϕ2(σ(hkf1)−1hkσ(f1))(f2))= Φ(hk)(f1, f2)


where =∗ follows from (7.25) since m = σ(hkf1)−1hσ(kf1) and m′ =σ(kf1)−1kσ(f1) both belong to (σ(F1)−1Kσ(F1)) ∩ N = M . It follows from(7.26) that dF (Φ(hk), Φ(h)Φ(k)) ≤ ε.

Let now h, k ∈ K be such that h �= k. We distinguish two cases.If h �= k then, as ϕ1 is a (K, ε)-almost-homomorphism, we have dF1(ϕ1(h),

ϕ1(k)) ≥ 1 − ε. It follows that there exists a subset B ⊂ F1 such that |B| ≥(1 − ε)|F1| such that ϕ1(h)(b) �= ϕ1(k)(b) for all b ∈ B. Setting B′ = B × F2

we have that

|B′| = |B| · |F2| ≥ (1 − ε)|F1| · |F2| = (1 − ε)|F | (7.27)

andΦ(h)(b′) �= Φ(k)(b′) for all b′ ∈ B′. (7.28)

Suppose now that h = k. As ϕ2 is an (M, ε)-almost-homomorphism onehas dF2(ϕ2(m), ϕ2(m′)) > 1 − ε for all m, m′ ∈ M such that m �= m′. Itfollows that for all distinct m, m′ ∈ M there exists a subset D ⊂ F2 suchthat |D| ≥ (1 − ε)|F2| and ϕ2(m)(d) �= ϕ2(m′)(d) for all d ∈ D.

From h = k we deduce that for all f1 ∈ F1 one has that if m = σ(hf1)−1 ×hσ(f1) and m′ =σ(kf1)−1kσ(f1), then m, m′ ∈M and m= σ(hf1)−1hσ(f1) =σ(kf1)−1hσ(f1) �= σ(kf1)−1kσ(f1) = m′. Thus ϕ2(σ(hf1)−1hσ(f1))(d) �=ϕ2(σ(kf1)−1kσ(f1))(d) for all d ∈ D. Set D′ = D × F2 so that

|D′| = |D| · |F2| ≥ (1 − ε)|F1| · |F2| = (1 − ε)|F |. (7.29)

It follows that for all d′ ∈ D′ one has

Φ(h)(d′) �= Φ(k)(d′) for all d′ ∈ D′. (7.30)

From (7.28) and (7.30), and taking into account (7.27) and (7.29) respectively,we deduce that in either cases dF (Φ(h), Φ(k)) ≥ 1 − ε.

This shows that Φ : G → Sym(F ) is a (K, ε)-almost-homomorphism. ThusG is sofic. �

7.6 Sofic Groups and Metric Ultraproducts of FiniteSymmetric Groups

Suppose that we are given a triple T = (I, ω,F) consisting of the followingdata: a set I, an ultrafilter ω on I, and a family F = (Fi)i∈I of nonemptyfinite sets indexed by I. Our first goal in this section is to associate with sucha triple T a sofic group GT . We start by forming the direct product group

PT =∏

i∈I

Sym(Fi).

7.6 Sofic Groups and Metric Ultraproducts of Finite Symmetric Groups 261

Let α = (αi)i∈I and β = (βi)i∈I be elements of PT . Since 0 ≤ dFi(αi, βi) ≤ 1for all i ∈ I, it follows from Corollary J.2.6 that the Hamming distancesdFi(αi, βi) have a limit

δω(α, β) = limi→ω

dFi(αi, βi) ∈ [0, 1]

along the ultrafilter ω.


(i) δω(α, α) = 0;(ii) δω(β, α) = δω(α, β);(iii) δω(α, β) ≤ δω(α, γ) + δω(γ, β);(iv) δω(γα, γβ) = δω(α, β);(v) δω(αγ, βγ) = δω(α, β).

for all α, β, γ ∈ PT .

Proof. Let α = (αi)i∈I , β = (βi)i∈I , γ = (γi)i∈I ∈ PT . For each i ∈ I, wehave dFi(αi, αi) = 0, dFi(βi, αi) = dFi(αi, βi), dFi(αi, βi) ≤ dFi(αi, γi) +dFi(γi, βi), dFi(γiαi, γiβi) = dFi(αi, βi), and dFi(αiγi, βiγi) = dFi(αi, βi)since dFi is a bi-invariant metric on Sym(Fi). This gives us properties (i),(ii), (iii), (iv), and (v) for δω by taking limits along ω (cf. Corollary J.2.10).

�

Consider now the subset NT ⊂ PT defined by

NT = {α ∈ PT : δω(1PT, α) = 0}.

Proposition 7.6.2. The set NT is a normal subgroup of PT .

Proof. This is an easy consequence of the properties of δω stated in Proposi-tion 7.6.1. Indeed, we deduce from Property (i) that δω(1PT

, 1PT) = 0, that

is, 1PT∈ NT . On the other hand, by using successively (iv), (iii), and (ii), we

get

δω(1PT, α−1β) = δω(α, β) ≤ δω(α, 1PT

)+δω(1PT, β) = δω(1PT

, α)+δω(1PT, β)

for all α, β ∈ PT . This implies that α−1β ∈ NT if α, β ∈ NT . Thus, NT is asubgroup of PT . Finally, Properties (iv) and (v) imply that

δω(1PT, γαγ−1) = δω(γ−1, αγ−1) = δω(γ−1γ, α) = δω(1PT

, α)

for all α, γ ∈ PT . It follows that γαγ−1 ∈ NT for all α ∈ NT and γ ∈ PT .This shows that NT is a normal subgroup of PT . �

Observe that, given α = (αi)i∈I and β = (βi)i∈I in PT , one has

αNT = βNT ⇐⇒ δω(α, β) = 0. (7.31)


Indeed, one has αNT = βNT if and only if α−1β ∈ NT , that is, if and only ifδω(1PT

, α−1β) = 0. This is equivalent to δω(α, β) = 0 by the left-invarianceof δω.

Theorem 7.6.3. The group GT = PT /NT is sofic.

Proof. Fix a finite subset K ⊂ GT and ε > 0. We want to show that thereexist a nonempty finite set F and a (K, ε)-almost-homomorphism ϕ : GT →Sym(F ).

Choose a representative of each element g ∈ GT , that is, an element g =(gi)i∈I ∈ PT such that g = gNT .

We first introduce some constants that will be used in the proof. If h andk are distinct elements in K, then δω(h, k) > 0 by (7.31). Let us set

η = minh,k∈K

h�=k

δω(h, k)2

. (7.32)

Note that 0 < η ≤ 1/2.Now choose an integer m ≥ log ε/ log(1 − η), so that

1 − (1 − η)m ≥ 1 − ε. (7.33)

Finally, choose a real number ξ with 0 < ξ < 1 sufficiently small to make

1 − (1 − ξ)m ≤ ε. (7.34)

If h and k are arbitrary elements of K, we have hkNT = hkNT andtherefore δω(hk, hk) = 0 by (7.31). It follows that the set

A(h, k) = {i ∈ I : dFi(hki, hiki) ≤ ξ} (7.35)

belongs to ω. On the other hand, if h and k are distinct elements of K, wehave δω(h, k) ≥ 2η by (7.32). As η > 0, this implies that the set

C(h, k) = {i ∈ I : dFi(hi, ki) ≥ η} (7.36)

belongs to ω. As any finite intersection of elements of ω is in ω and thereforenonempty, we deduce that there exists an index j ∈ I such that

j ∈

⎛

⎝⋂

h,k∈K

A(h, k)

⎞

⎠⋂

⎛

⎜⎜⎝⋂

h,k∈K

h�=k

C(h, k)

⎞

⎟⎟⎠ .

Consider the map ψ : GT → Sym(Fj) defined by ψ(g) = gj for all g ∈ GT .It immediately follows from (7.35) and (7.36) that the map ψ satisfies thefollowing properties:

7.6 Sofic Groups and Metric Ultraproducts of Finite Symmetric Groups 263

(1) dFj (ψ(hk), ψ(h)ψ(k)) ≤ ξ for all h, k ∈ K;(2) dFj (ψ(h), ψ(k)) ≥ η for all h, k ∈ K such that h �= k.

Consider now the Cartesian product

F = Fj × Fj × · · · × Fj︸︷︷︸m times

and the homomorphism

Ψ : Sym(Fj) → Sym(F )

defined by

Ψ(σ)(x1, x2, . . . , xm) = (σ(x1), σ(x2), . . . , σ(xm))

for all σ ∈ Sym(Fj) and (x1, x2, . . . , xm) ∈ F . By Corollary 7.4.4 we have

dF (Ψ(σ), Ψ(σ′)) = 1 − (1 − dFj (σ, σ′))m

for all σ, σ′ ∈ Sym(Fj). It follows that the composite map

ϕ = Ψ ◦ ψ : G → Sym(F )

satisfies the following properties:

(1’) dF (ϕ(hk), ϕ(h)ϕ(k)) ≤ 1 − (1 − ξ)m for all h, k ∈ K;(2’) dF (ϕ(h), ϕ(k)) ≥ 1 − (1 − η)m for all h, k ∈ K such that h �= k.

As 1−(1−ξ)m ≤ ε by (7.34) and 1−(1−η)m ≥ 1−ε by (7.33), we deduce thatϕ is a (K, ε)-almost-homomorphism. This shows that GT is a sofic group. �

Remark 7.6.4. When the ultrafilter ω is principal, then the group GT is finite.Indeed, in this case, there is an element i0 ∈ I such that ω consists of allsubsets of I containing i0. This implies that δω(α, β) = dFi0

(αi0 , βi0) for allα = (αi)i∈I , β = (βi)i∈I ∈ PT . Therefore, NT consists of all α ∈ PT suchthat αi0 = 1Sym(Fi0 ). It follows that the group GT is isomorphic to the groupSym(Fi0).

Remark 7.6.5. Let α, α′, β, β′ ∈ PT such that αNT = α′NT and βNT =β′NT . By applying Proposition 7.6.1(iii) and (7.31), we get

dω(α, β) ≤ dω(α, α′) + dω(α′, β′) + dω(β′, β) = dω(α′, β′).

By exchanging the roles of α and α′ and of β and β′, we obtain dω(α′, β′) ≤dω(α, β). It follows that dω(α, β) = dω(α′, β′). Therefore, if g, h ∈ GT andα, β ∈ PT are such that g = αNT and h = βNT , the quantity

Δω(g, h) = δω(α, β) (7.37)


is well defined. Moreover, the map Δω : GT ×GT → [0, 1] given by (7.37) is abi-invariant metric on GT . This follows immediately from Proposition 7.6.1taking into account that NT is a normal subgroup.

Theorem 7.6.6. Let G be a group. The following conditions are equivalent:

(a) G is sofic;(b) there exists a triple T = (I, ω,F), where I is a set, ω is an ultrafilter on

I, and F = (Fi)i∈I is a family of nonempty finite sets indexed by I, suchthat G is isomorphic to a subgroup of the group GT = PT /NT .

Proof. If G satisfies (b), then G is sofic since GT is sofic by Theorem 7.6.3and every subgroup of a sofic group is itself sofic by Proposition 7.5.4.

Conversely, suppose that G is sofic. Consider the set I consisting of allpairs (K, ε), where K is a finite subset of G and ε > 0. We partially orderthe set I by setting (K, ε) � (K ′, ε′) if K ⊂ K ′ and ε′ ≤ ε.

For each i = (K, ε) ∈ I we define the set

Ii = {j ∈ I : i � j} ⊂ I.

Observe that Ii �= ∅ as i ∈ Ii. Moreover, the family of nonempty subsets{Ii}i∈I is closed under finite intersections, since

I(K1,ε1) ∩ I(K2,ε2) = I(K1∪K2,min{ε1,ε2})

for all (K1, ε1), (K2, ε2) ∈ I. It follows from Proposition J.1.3 and Theo-rem J.1.6 that there exists an ultrafilter ω on I such that I(K,ε) ∈ ω for all(K, ε) ∈ I.

As G is sofic, we can find, for each i = (K, ε) ∈ I, a nonempty finiteset Fi and a (K, ε)-almost-homomorphism ϕi : G → Sym(Fi). Consider thetriple T = (I, ω,F), where F = (Fi)i∈I , and the associated sofic groupGT = PT /NT . Let ϕ : G → PT denote the product map ϕ =

∏i∈I ϕi. Thus,

we haveϕ(g) = (ϕi(g))i∈I

for all g ∈ G. Let ρ : PT → GT = PT /NT denote the canonical epimorphism.Let us show that the composite map Φ = ρ ◦ ϕ : G → GT is an injectivehomomorphism. This will prove that G is isomorphic to a subgroup of GT .

Let g, h ∈ G and let η > 0. Consider the element i0 = ({g, h}, η) ∈ I. Ifi = (K, ε) ∈ I satisfies i0 � i, then

dFi(ϕi(gh), ϕi(g)ϕi(h)) ≤ ε ≤ η

since ϕi is a (K, ε)-almost-homomorphism. Thus, the set

{i ∈ I : dFi(ϕi(gh), ϕi(g)ϕi(h)) ≤ η}

contains Ii0 and therefore belongs to ω. This implies that

7.7 A Characterization of Finitely Generated Sofic Groups 265

δω(ϕ(gh), ϕ(g)ϕ(h)) = limi→ω

dFi(ϕi(gh), ϕi(g)ϕi(h)) = 0.

Therefore, we have Φ(gh) = Φ(g)Φ(h) by (7.31). This shows that Φ is ahomomorphism.

On the other hand, if g �= h, we have

dFi(ϕi(g), ϕi(h)) ≥ 1 − ε ≥ 1 − η

for all i ∈ I such that i0 � i. This implies that

δω(ϕ(g), ϕ(h)) = limi→ω

dFi(ϕi(g), ϕi(h)) = 1. (7.38)

Therefore, we have Φ(g) �= Φ(h) by (7.31). Consequently, Φ is injective. Thisshows that (a) implies (b). �

Remarks 7.6.7. (a) If G is an infinite sofic group and T = (I, ω,F) is a triplesatisfying condition (b) in Theorem 7.6.6, then the ultrafilter ω is necessarilynon-principal by Remark 7.6.4.

(b) Let G be a sofic group and let Φ : G → GT be as in the proof ofTheorem 7.6.6. Using the notation from Remark 7.6.5, we deduce from (7.38)that Δω(Φ(g), Φ(h)) = 1 for all g, h ∈ G such that g �= h. It follows that therestriction of the bi-invariant metric Δω to the subgroup Φ(G) ⊂ GT is thediscrete metric.

7.7 A Characterization of Finitely Generated SoficGroups

In this section we give a geometric characterization of finitely generated soficgroups in terms of a finiteness condition on their Cayley graphs.

Let G be a finitely generated group and let S be a finite symmetric gen-erating subset of G. Given r ∈ N, we denote by BS(r) the ball of radius rcentered at the vertex corresponding to the identity element 1G of G in theCayley graph CS(G) of G with respect to S, with the induced S-labeled graphstructure (cf. Sects. 6.1, 6.2 and 6.3).

Let also Q = (Q, E) be an S-labeled graph. Given q ∈ Q and r ∈ N, wedenote by B(q, r) the ball of radius r centered at q with the induced S-labeledgraph structure.

Given r ∈ N we denote by Q(r) the set of all q ∈ Q such that there existsan S-labeled graph isomorphism

ψq,r : BS(r) → B(q, r) (7.39)

satisfyingψq,r(1G) = q. (7.40)


Observe that if such a map ψq,r exists it is unique. We have the inclusions

Q = Q(0) ⊃ Q(1) ⊃ Q(2) ⊃ · · · ⊃ Q(r) ⊃ Q(r + 1) ⊃ · · · (7.41)

Fig. 7.2 The inclusions Q ⊃ Q(1) ⊃ Q(2) ⊃ Q(3)

Note also that since CS(G) (and therefore the induced S-labeled subgraphBS(r)) is edge-symmetric (with respect to the involution s �→ s−1 on S(cf. Sect. 6.3)) and ψq,r is an S-labeled graph isomorphism, then B(q, r) =ψq,r(BS(r)) is edge-symmetric as well.

Theorem 7.7.1. Let G be a finitely generated group and let S be a finitesymmetric generating subset of G. The following conditions are equivalent:

(a) the group G is sofic;(b) for all ε > 0 and r ∈ N, there exists a finite S-labeled graph Q = (Q, E)

such that|Q(r)| ≥ (1 − ε)|Q|, (7.42)

where Q(r) ⊂ Q denotes the set consisting of all vertices q ∈ Q for whichthere exists an S-labeled graph isomorphism ψq,r : BS(r) → B(q, r) fromthe ball BS(r) in the Cayley graph CS(G) of G with respect to S onto theball B(q, r) in Q satisfying ψq,r(1G) = q.

Before starting the proof of the theorem we present some preliminaryresults.

Lemma 7.7.2. Let Q = (Q, E) be an S-labeled graph and r0, i ∈ N. Supposethat q0 ∈ Q((i + 1)r0). Then B(q0, r0) ⊂ Q(ir0).

Proof. Let q′ ∈ B(q, r0) and let us show that q′ ∈ Q(ir0). It follows fromthe triangle inequality that the ball B(q′, ir0) is entirely contained in theball B(q0, (i + 1)r0). Moreover, since ψq0,(i+1)r0 is isometric, setting g =


ψ−1q0,(i+1)r0

(q′), we have g ∈ BS(r0) so that gh ∈ B((i+1)r0) for all h ∈ B(ir0).It follows that the map

ψq′,ir0 : BS(ir0) → B(q′, ir0) (7.43)

defined by ψq′,ir0(h) = ψq0,(i+1)r0(gh) for all h ∈ BS(ir0) yields an S-labeledgraph isomorphism satisfying ψq′,ir0(1G) = ψq0,(i+1)r0(g) = q′. This showsthat q′ ∈ Q(ir0). We deduce that B(q0, r0) ⊂ Q(ir0). �

Lemma 7.7.3. Let Q = (Q, E) be an S-labeled graph and r0 ∈ N. Let q1, q2 ∈Q(2r0) such that q1 �= q2 and g ∈ BS(r0). Then we have

ψq1,2r0(g) �= ψq2,2r0(g). (7.44)

Proof. If g = 1G we have ψq1,2r0(g) = ψq1,2r0(1G) = q1 �= q2 = ψq2,2r0(1G) =ψq2,2r0(g). Suppose now that g �= 1G. Suppose by contradiction thatψq1,2r0(g) = ψq2,2r0(g) = q0. Since ψq1,2r0 is isometric we have q0 ∈ B(q1, r0).It follows from Lemma 7.7.2 that q0 ∈ Q(r0).

As g ∈ BS(r0), we can find 1 ≤ r′ ≤ r0 and s1, s2, . . . , sr′ ∈ S such thatg = s1s2 · · · sr′ . Consider the path

π = ((1G, s1, s1), (s1, s2, s1s2), . . . , (s1s2 · · · sr′−1, sr′ , g))

and observe that it is contained in BS(r0). Now, π is mapped by ψq1,2r0 andψq2,2r0 into two paths π1 and π2 in Q with initial vertices π−

1 = q1, π−2 = q2

and same terminal vertex π+1 = ψq1,2r0(g) = q0 = ψq2,2r0(g) = π+

2 . Notethat since ψq1,2r0 and ψq2,2r0 are isometric π1 and π2 are both contained inB(q0, r0). Moreover, since ψq1,2r0 and ψq2,2r0 are label-preserving, π1 and π2

have the same label s1s2 · · · sr′ . The inverse images of π1 and π2 under theS-label preserving graph isomorphism ψq0,r0 : BS(r0) → B(q0, r0) are bothequal to π. Indeed in a Cayley graph there exists a unique path which ends ata given vertex and with a given label. It follows that π1 = π2 and thereforeq1 = π−

1 = π−2 = q2. This contradicts our assumptions. We deduce that

ψq1,2r0(g) �= ψq2,2r0(g). �

Lemma 7.7.4. Let Q = (Q, E) be an S-labeled graph and r0 ∈ N. Let h, k ∈BS(r0) and q0 ∈ Q(2r0). We have

ψq0,2r0(h) ∈ Q(r0) (7.45)

andψq0,2r0(hk) = ψψq0,2r0 (h),r0(k), (7.46)

where ψq,r, q ∈ Q(r), r ∈ N, is as in (7.39) and (7.40).

Proof. Since ψq0,2r0 is isometric we have ψq0,2r0(h) ∈ B(q0, r0). Then (7.45)follows from Lemma 7.7.2. Let us show (7.46). First note that (7.46) makessense by virtue of (7.45). If k = 1G (7.46) follows from ψq0,2r0(hk) =


ψq0,2r0(h) = ψψq0,2r0 (h),r0(1G) = ψψq0,2r0 (h),r0(k). Now suppose that k �= 1G.Then we can find 1 ≤ r′ ≤ r0 and s1, s2, . . . , sr′ ∈ S such that k = s1s2 · · · sr′ .Consider the path

π1 = ((h, s1, hs1), (hs1, s2, hs1s2), . . . , (hs1s2 · · · sr′−1, sr′ , hk))

and observe that it is contained in BS(2r0), since h, k ∈ BS(r0). The path π1

is mapped by ψq0,2r0 into the path

π1 = ((ψq0,2r0(h),s1, ψq0,2r0(hs1)), (ψq0,2r0(hs1), s2, ψq0,2r0(hs1s2)), . . .. . . , (ψq0,2r0(hs1s2 · · · sr′−1), sr′ , ψq0,2r0(hk))).

As ψq0,2r0 is isometric, we have that q′ = ψq0,2r0(h) belongs to B(q0, r0) andsince q0 ∈ Q(2r0) we deduce from Lemma 7.7.2 that q′ ∈ Q(r0). Consider theinverse image of the path π1 under the S-labeled graph isomorphism ψq′,r0 .Since ψ−1

q′,r0((π1)−) = ψ−1

q′,r0(ψq0,2r0(h)) = ψ−1

q′,r0(q′) = 1G and ψq′,r0 preserves

the label, this inverse image is necessarily equal to the path

π2 = ((1G, s1, s1), (s1, s2, s1s2), . . . , (s1s2 · · · sr′−1, sr′ , k)),

since in a Cayley graph there exists a unique path which starts at a given ver-tex and with a given label. It follows that ψ−1

q′,r0(ψq0,2r0(hk)) = ψ−1

q′,r0(π+

1 ) =π+

2 = k, that is, ψq0,2r0(hk) = ψq′,r0(k). Since q′ = ψq0,2r0(h), we de-duce (7.46). �

We are now in position to prove Theorem 7.7.1.

Proof of Theorem 7.7.1. Suppose that G is sofic. Fix ε > 0 and r ∈ N. SetK = BS(2r + 1) and

ε′ = ε(1 + |BS(r)| · |S| + |BS(r)|2)−1. (7.47)

Since G is sofic, there exists a nonempty finite set F and a (K, ε′)-almost-homomorphism ϕ : G → Sym(F ). We construct an S-labeled graph Q =(Q, E) as follows. We take as vertex set Q = F . Then, as set of edges wetake the set E ⊂ Q×S×Q consisting of all the triples (q, s, ϕ(s−1)(q)), wereq ∈ Q and s ∈ S. Note that Q may have loops and multiple edges and thatQ is not necessarily edge-symmetric with respect to the involution s �→ s−1

on S. Observe however that, if q ∈ Q and s ∈ S are fixed, then there existsa unique edge in Q with initial vertex q and label s.

For each q ∈ Q denote by ψq : G → Q the map defined by settingψq(g) = ϕ(g−1)(q) for all g ∈ G. Denote by Q0 the subset of Q consist-ing of all q ∈ Q satisfying the following conditions:

(∗) ψq(1G) = q,(∗∗) ψq(gs) = ψψq(g)(s) for all g ∈ BS(r) and s ∈ S,(∗ ∗ ∗) ψq(g) �= ψq(h) for all g, h ∈ BS(r) with g �= h.


Suppose that q ∈ Q0. Let g ∈ BS(r). If g = 1G we have ψq(g) = ψq(1G) =q ∈ B(q, r), by (∗). If g �= 1G, then there exist 1 ≤ r′ ≤ r and s1, s2, . . . , sr′ ∈S such that g = s1s2 · · · sr′ . Consider the sequence of edges

e1 = (q, s1, ϕ(s−11 )(q)) = (q, s1, ψq(s1)),

e2 = (ψq(s1), s2, ϕ(s−12 )ψq(s1))

= (ψq(s1), s2, ψψq(s1)(s2))

= (ψq(s1), s2, ψq(s1s2)) (by (∗∗)),

· · · · · ·

er′ = (ψq(s1s2 · · · sr′−1), sr′ , ϕ(s−1r′ )(ψq(s1s2 · · · sr′−1)))

= (ψq(s1s2 · · · sr′−1), sr′ , ψψq(s1s2···sr′−1)(sr′))

= (ψq(s1s2 · · · sr′−1), sr′ , ψq(s1s2 · · · sr′−1sr′)) (by (∗∗))

= (ψq(s1s2 · · · sr′−1), sr′ , ψq(g)).

The path π = (e1, e2, . . . , er′) connects q to ψq(g) and has length �(π) ≤ r.This shows that the graph distance from q to ψq(g) in Q does not exceed r sothat ψq(BS(r)) ⊂ B(q, r). Conversely, let q′ ∈ B(q, r). If q′ = q then by (∗)we have q′ = q = ψq(1G) ∈ ψq(BS(r)). If q �= q′ then there exist 1 ≤ r′ ≤ rand a sequence of edges (q, s1, q1), (q1, s2, q2), . . . , (qr′−1, sr′ , q′) ∈ E. Using(∗∗) as above, we get q′ = ψq(g), where g = s1s2 · · · sr′ ∈ BS(r). This showsthat B(q, r) ⊂ ψq(BS(r)). Thus

B(q, r) = ψq(BS(r)).

Moreover, by condition (∗ ∗ ∗), the map ψq|BS(r) is injective. Finally, from(∗∗) we deduce that if g ∈ BS(r) and s ∈ S then we have

(ψq(g), s, ψq(gs)) = (ψq(g), s, ψψq(g)(s)) = (ψq(g), s, ϕ(s−1)ψq(g)) ∈ E.

Thus, the map ψq,r = ψq|BS(r) : BS(r) → B(q, r) is an S-labeled graph iso-morphism such that ψq,r(1G) = ψq(1G) = q, where the last equality followsfrom (∗). We deduce that Q0 ⊂ Q(r).

Let’s now estimate from below the cardinality of Q0. We denote by dQ thenormalized Hamming metric on Sym(Q).

In order to estimate the cardinality of the set of q ∈ Q for which condition(∗) is satisfied, let us first observe that

dQ(IdQ, ϕ(1G)) ≤ ε′. (7.48)

Indeed, since ϕ is a (K, ε′)-almost-homomorphism, taking k1 = k2 = 1G inproperty (i) of Definition 7.5.1, we deduce that dQ(ϕ(1G), ϕ(1G)ϕ(1G)) ≤ ε′

which, by left-invariance of dQ, implies (7.48). From (7.48) we deduce the


existence of a subset Q′ ⊂ Q of cardinality |Q′| ≤ ε′|Q| such that

ϕ(1G)(q) = q (7.49)

for all q ∈ Q \Q′. It follows that ψq(1G) = ϕ(1−1G )(q) = ϕ(1G)(q) = q, for all

q ∈ Q \ Q′ so that condition (∗) is satisfied in Q \ Q′.Let now estimate the cardinality of the set of q ∈ Q for which condi-

tion (∗∗) is satisfied. Let g ∈ BS(r) and s ∈ S. As ϕ is a (K, ε′)-almost-homomorphism we have dQ(ϕ(s−1g−1), ϕ(s−1)ϕ(g−1)) ≤ ε′ so that thereexists a subset Q′′(g, s) ⊂ Q with |Q′′(g, s)| ≤ ε′|Q| such that

ψq(gs) = ϕ((gs)−1)(q)

= ϕ(s−1g−1)(q)

= ϕ(s−1)ϕ(g−1)(q)

= ϕ(s−1)(ψq(g))

= ψψq(g)(s)

for all q ∈ Q \ Q′′(g, s). Setting

Q′′ =⋃

g∈BS(r)s∈S

Q′′(g, s)

we have |Q′′| ≤ |BS(r)| · |S|ε′|Q| and condition (∗∗) holds for all q ∈ Q \Q′′.Fix now two distinct elements g, h ∈ BS(r). Since ϕ is a (K, ε′)-almost-

homomorphism and g−1, h−1 ∈ K, by property (ii) of Definition 7.5.1we deduce that dQ(ϕ(g−1), ϕ(h−1)) ≥ 1 − ε′. Thus we can find a subsetQ′′′(g, h) ⊂ Q of cardinality |Q′′′(g, h)| ≤ ε′|Q| such that

ψq(g) = ϕ(g−1)(q) �= ϕ(h−1)(q) = ψq(h) (7.50)

for all q ∈ Q \ Q′′′(g, h).Let now g �= h vary in BS(r) and set

Q′′′ =⋃

g,h∈BS(r)g �=h

Q′′′(g, h).

Observe that |Q′′′| ≤ |BS(r)|2ε′|Q|. It follows from (7.50) that condition(∗ ∗ ∗) holds for all q ∈ Q \ Q′′′.

In conclusion, conditions (∗), (∗∗) and (∗ ∗ ∗) are satisfied for all q ∈ Qoutside of Q′ ∪ Q′′ ∪ Q′′′. We have

|Q′ ∪ Q′′ ∪ Q′′′| ≤ (1 + |BS(r)| · |S| + |BS(r)|2)ε′|Q| = ε|Q|,


where the equality follows from (7.47). We deduce that

|Q(r)| ≥ |Q0| ≥ (1 − ε)|Q|.

Thus G satisfies condition (b). This shows (a) ⇒ (b).Conversely, suppose (b). Fix a finite set K ⊂ G and ε > 0. Let r0 ∈ N be

such that K ∪ K2 ⊂ BS(r0). Let Q = (Q, E) be the finite S-labeled graphgiven by condition (b) corresponding to r = 2r0 and ε.

Let g ∈ BS(r0). Since the map from Q(2r0) into Q defined by q �→ ψq,2r0(g)is injective (by Lemma 7.44), we have that

|{ψq,2r0(g) : q ∈ Q(2r0)}| = |Q(2r0)| (7.51)

and therefore

|Q \ {ψq,2r0(g) : q ∈ Q(2r0)}| = |Q \ Q(2r0)|. (7.52)

As a consequence, there exists a bijection αg : Q \ Q(2r0) → Q \ {ψq,2r0(g) :q ∈ Q(2r0)}. Since ψq,2r0(1G) = q for all q ∈ Q(2r0), we have {ψq,2r0(1G) :q ∈ Q(2r0)} = Q(2r0) and therefore we can take

α1G= IdQ\Q(2r0) . (7.53)

Consider now the map ϕ : G → Sym(Q) defined by

ϕ(g)(q) =

⎧⎪⎨

⎪⎩

ψq,2r0(g−1) if g ∈ BS(r0) and q ∈ Q(2r0);

αg−1(q) if g ∈ BS(r0) and q ∈ Q \ Q(2r0);

q otherwise.

(7.54)

Note that ϕ(g) ∈ Sym(Q) for all g ∈ G, by construction.Let us show that the map ϕ : G → Sym(Q) is a (K, ε)-almost-homomor-

phism. Let k1, k2 ∈ K ⊂ BS(r0) and q ∈ Q(2r0). We have

ϕ(k1k2)(q) = ψq,2r0(k−12 k−1

1 )

= ψψq,2r0 (k−12 ),r0

(k−11 ) (by (7.46))

= ϕ(k1)(ψq,2r0(k−12 ))

= [ϕ(k1)ϕ(k2)](q).

This shows that on Q(2r0) we have ϕ(k1k2) = ϕ(k1)ϕ(k2). As

|Q(2r0)| = |Q(r)| ≥ (1 − ε)|Q| (7.55)

we deduce that dQ(ϕ(k1k2), ϕ(k1)ϕ(k2)) ≤ ε.Finally, suppose that k1 �= k2. We have ϕ(k1)(q) = ψq,2r0(k

−11 ) �=

ψq,2r0(k−12 ) = ϕ(k2)(q), since k−1

1 , k−12 ∈ BS(2r0) and ψq,2r0 is injective.


From (7.55) we deduce that dQ(ϕ(k1), ϕ(k2)) ≥ 1 − ε. It follows that ϕis a (K, ε)-almost-homomorphism. Therefore G is sofic. This shows that(b) ⇒ (a). �

7.8 Surjunctivity of Sofic Groups

In this section we prove that sofic groups are surjunctive. Note that thisresult covers the fact that locally residually amenable groups are surjunctive,which had been previously established in Corollary 5.9.3. Indeed, all locallyresidually amenable groups are sofic by Corollary 7.5.11.

Theorem 7.8.1 (Gromov-Weiss). Every sofic group is surjunctive.


Lemma 7.8.2. Let G be a group, A a finite set, and equip AG with the prodis-crete topology. Let X ⊂ AG be a closed G-invariant subset and let f : X → AG

be a continuous G-equivariant map. Then there exists a cellular automatonτ : AG → AG such that f = τ |X .

Proof. From the continuity of f we deduce the existence of a finite set S ⊂ Gsuch that if two configurations y, z ∈ X coincide on S then f(y)(1G) =f(z)(1G). Let a0 ∈ A and consider the map μ : AS → A defined by setting

μ(u) =

{f(x)(1G) if there exists x ∈ X such that x|S = u

a0 otherwise

for all u ∈ AS . Then μ is well defined and if we denote by τ : AG → AG

the cellular automaton with memory set S and local defining map μ, by G-equivariance of f we clearly have τ |X = f . �

Proof of Theorem 7.8.1. Let G be a sofic group. Let A be a finite set ofcardinality |A| ≥ 2 and let τ : AG → AG be an injective cellular automaton.We want to show that τ is surjective. Every subgroup of a sofic group is soficby Proposition 7.5.4. On the other hand it follows from Proposition 3.2.2 thata group is surjunctive if all its finitely generated subgroups are surjunctive.Thus we can assume that G is finitely generated.

Let then S ⊂ G be a finite symmetric generating subset of G. As usual,for r ∈ N, we denote by BS(r) ⊂ G the ball of radius r centered at 1G in theCayley graph of G with respect to S. We set Y = τ(AG). Observe that Y isG-invariant and, by Lemma 3.3.2, it is closed in AG.

The inverse map τ−1 : Y → AG is G-equivariant and, by compactness ofAG, it is also continuous. By Lemma 7.8.2, there exists a cellular automatonσ : AG → AG such that σ|Y = τ−1 : Y → AG. Choose r0 large enough so

7.8 Surjunctivity of Sofic Groups 273

that the ball BS(r0) is a memory set for both τ and σ. Let μ : ABS(r0) → Aand ν : ABS(r0) → A denote the corresponding local defining maps for τ andσ respectively.

We proceed by contradiction. Suppose that τ is not surjective, that is,Y � AG. Then, since Y is closed in AG, there exists a finite subset Ω ⊂ G suchthat πΩ(Y ) � AΩ , where, for a subset E ⊂ G, we denote by πE : AG → AE

the projection map. It is not restrictive, up to taking a larger r0, again ifnecessary, to suppose that Ω ⊂ BS(r0). Thus, πBS(r0)(Y ) � ABS(r0).

Fix ε > 0 such that

ε < 1 − |B(2r0)| · log |A||B(2r0)| · log |A| − log(1 − |A|−|BS(r0)|)

. (7.56)

Note that log(1 − |A|−|BS(r0)|) < 0, so that the right hand side of (7.56) iswell defined and positive.

Since G is sofic, it follows from Theorem 7.7.1 that we can find a finiteS-labeled graph Q such that

|Q(3r0)| ≥ (1 − ε)|Q|, (7.57)

where we recall that Q(r), r ∈ N, denotes the set of all q ∈ Q such thatthere exists an S-labeled graph isomorphism ψq,r : BS(r) → B(q, r) satisfyingψq,r(1G) = q (cf. Theorem 7.7.1).

We have the inclusions

Q(r0) ⊃ Q(2r0) ⊃ · · · ⊃ Q(ir0) ⊃ Q((i + 1)r0) ⊃ · · · .

(cf. (7.41); see also Fig. 7.2). Also recall from Lemma 7.7.2 that B(q, r0) ⊂Q(ir0) for all Q((i + 1)r0) and i ∈ N.

For each integer i ≥ 1, we define the map μi : AQ(ir0) → AQ((i+1)r0) bysetting, for all u ∈ AQ(ir0) and q ∈ Q((i + 1)r0),

μi(u)(q) = μ(u|B(q,r0) ◦ ψq,r0

)(1G),

where ψq,r0 is the unique isomorphism of S-labeled graphs from BS(r0) ⊂ Gto B(q, r0) ⊂ Q sending 1G to q.

Similarly, we define the map νi : AQ(ir0) → AQ((i+1)r0) by setting, for allu ∈ AQ(ir0) and q ∈ Q((i + 1)r0),

νi(u)(q) = ν(u|B(q,r0) ◦ ψq,r0

)(1G).

From the fact that σ ◦ τ = σ|Y ◦ τ = τ−1 ◦ τ is the identity map onAG, we deduce that the composite νi+1 ◦ μi : AQ(ir0) → AQ((i+2)r0) satisfies(νi+1◦μi)(u)(q) = u(q) for all u ∈ AQ(ir0) and q ∈ Q((i+2)r0). In other words,denoting by ρi : AQ(ir0) → AQ((i+2)r0) the restriction map, we have thatνi+1 ◦ μi = ρi for all i ≥ 1. In particular, we have ν2 ◦ μ1 = ρ1. Thus, setting


Z = μ1(AQ(r0)) ⊂ AQ(2r0), we deduce that ν2(Z) = ρ1(AQ(r0)) = AQ(3r0). Itfollows that |Z| ≥ |A||Q(3r0)|, so that, by taking logarithms, we get

log |Z| ≥ |Q(3r0)| · log |A|. (7.58)

In order to estimate the cardinality of Z from above, we first show theexistence of a subset Q′ ⊂ Q(3r0) satisfying

|Q′| ≥ |Q(3r0)||B(2r0)|

(7.59)

and such that the balls B(q′, r0) with q′ ∈ Q′ are all disjoint.Indeed, let Q′ be a maximal subset of Q(3r0) such that the balls B(q′, r0)

with q′ ∈ Q′ are all disjoint. If q ∈ Q(3r0) \ Q′ is at distance greater than2r0 from Q′, then B(q, r0) ∩ B(q′, r0) = ∅ for all q′ ∈ Q′, contradicting themaximality of Q′. Therefore Q(3r0) is contained in the union of the ballsB(q′, 2r0), with q′ ∈ Q′. This implies

|Q(3r0)| ≤ |Q′| · |B(2r0)|,

which gives (7.59).Let then Q′ ⊂ Q(3r0) be as above and set Q′ =

∐q′∈Q′ B(q′, r0). Note

that Q′ ⊂ Q(2r0) and that

|Q′| = |Q′| · |BS(r0)|. (7.60)

Given a subset E ⊂ Q we denote by πE : AQ → AE the projection map. Nowobserve that, for all q ∈ Q(2r0), we have a natural bijection πB(q,r0)(Z) →πBS(r0)(Y ) given by u �→ u ◦ψq,r0 , where ψq,r0 denotes, as above, the uniqueisomorphism of S-labeled graphs from BS(r0) to B(q, r0) sending 1G to q.Since πBS(r0)(Y ) � ABS(r0), this implies that

∣∣πB(q′,r0)(Z)∣∣ =

∣∣πBS(r0)(Y )∣∣ ≤ |A||BS(r0)| − 1, (7.61)

for all q′ ∈ Q′.We have

Z ⊂ πQ′(Z) × πQ(2r0)\Q′(Z)

⊂

⎛

⎝∏

q′∈Q′

πB(q′,r0)(Z)

⎞

⎠ × πQ(2r0)\Q′(Z)

⊂

⎛

⎝∏

q′∈Q′

πB(q′,r0)(Z)

⎞

⎠ × AQ(2r0)\Q′

Notes 275

and we deduce

|Z| ≤(|A||BS(r0)| − 1

)|Q′|· |A||Q(2r0)|−|Q′| (by (7.61))

≤(|A||BS(r0)| − 1

)|Q′|· |A||Q|−|Q′| (as Q(2r0) ⊂ Q)

=(1 − |A|−|BS(r0)|

)|Q′|· |A||Q

′|·|BS(r0)| · |A||Q|−|Q′|

=(1 − |A|−|BS(r0)|

)|Q′|· |A||Q| (by (7.60)).

By taking logarithms we get

log |Z| ≤ |Q′| · log(1 − |A|−|BS(r0)|

)+ |Q| · log |A|.

As we remarked earlier, we have log(1 − |A|−|BS(r0)|) < 0 so that, by (7.59),we obtain

log |Z| ≤ |Q(3r0)||B(2r0)|

· log(1 − |A|−|BS(r0)|

)+ |Q| · log |A|

and, by (7.58),

|Q(3r0)| · log |A| ≤ |Q(3r0)||B(2r0)|

· log(1 − |A|−|BS(r0)|) + |Q| · log |A|.

This gives us

|Q(3r0)| ≤|B(2r0)| · log |A|

|B(2r0)| · log |A| − log(1 − |A|−|BS(r0)|)|Q|

and, by (7.56),|Q(3r0)| < (1 − ε)|Q|

which contradicts (7.57). �

Notes

The problem of approximation of infinite groups by finite ones goes back, inthe framework of purely algebraic constructions, to A.I. Mal’cev [Mal1, Mal3].These ideas were developed further by A.M. Vershik [Ver] and A.M. Stepin[Ste1, Ste2] also in the wider context of operator algebras and ergodic theory.The notion of local embeddability was introduced by Vershik and E.I. Gordonin [VeG]. They studied the groups which are locally embeddable into theclass of finite groups (LEF groups). Among other things, they establisheda relationship between this notion of approximation by finite groups and


the convergence in the topological space of marked groups. Moreover, somestability properties for the class of LEF groups were presented. They alsoshowed that finitely presented LEF groups are residually finite. Nilpotentgroups and metabelian groups are LEF groups since they are locally residuallyfinite [Hall2]. Vershik and Gordon [VeG] gave an example of a solvable (andtherefore amenable) group which is not an LEF group.

The basic results on general local embeddability presented in Sect. 7.1 arenatural generalizations of those for LEF groups established in [VeG]. Vershikand Gordon also considered some notions of local approximation for groupactions, namely equivariant approximation of actions on groups, free approxi-mation of actions on arbitrary sets, and uniform free approximation of actionson spaces with quasi-invariant measures. Equivariant approximation was usedto show that semi-direct products of LEF groups with equivariantly approx-imable actions are LEF groups. It was then deduced that the non-residuallyfinite group G1 of Sect. 2.6 considered by Mal’cev in [Mal1] is an LEF group.Also, Vershik and Gordon proved the following characterizations of local em-beddability into the class of finite groups. A group admits a (faithful) freelyapproximable action if and only if it is an LEF group. A countable group ad-mits a uniformly freely approximable action on some measurable space if andonly if it is an LEF group (note that the “only if” part was already establishedwith slightly different terminology by A. Stepin in [Ste1], see also [Ste2]).

Groups which are locally embeddable into the class of amenable groupswhere considered by M. Gromov in [Gro5] under the name of initially suba-menable groups. More precisely, in Sect. 4.G of [Gro5], a finitely generatedgroup G is said to be initially subamenable if for any finite generating sub-set S ⊂ G there exists a sequence of F -marked amenable groups convergingto G, where F denotes the free group based on S. The fact that a finitelygenerated group is initially subamenable if and only if it is LEA follows fromProposition 7.3.7(4). In Sect. 6.E” of [Gro5], Gromov introduced the notionof an initially subamenable graph (which, for Cayley graphs, reduces to con-dition (b) in Theorem 7.7.1) and, in the Example in Sect. 6.E”’, he claimsthat the Cayley graph of a finitely generated group is initially subamenableif and only if the group is initially subamenable. Later, groups whose Cay-ley graphs are initially subamenable (according with Gromov’s definition inSect. 6.E” of [Gro5]) were called sofic groups by B. Weiss [Weiss], this ter-minology coming from the Hebrew word סופי� which means finite and derivesfrom סו�� which means end.

Gromov [Gro5] and Weiss [Weiss] proved that sofic groups are surjunctive(Theorem 7.8.1). G. Elek and E. Szabo [ES2] proved that a group is sofic if andonly if it is a subgroup of an ultraproduct of finite symmetric groups equippedwith their Hamming distances (Theorem 7.6.6). They used this result to provethat any countable sofic group can be embedded into a countable simple soficgroup. In [ES3] Elek and Szabo proved that the class of sofic groups is closedunder taking direct products, subgroups, projective limits and direct limits,free products, and extensions by amenable groups. Moreover, by modifying

Notes 277

the example of the group G1 in Sect. 2.6 (which is LEF but not residuallyfinite), Elek and Szabo constructed an example of a finitely generated LEF-group which is not residually amenable.

A detailed exposition of the theory of hyperlinear and sofic groups is pre-sented in V.G. Pestov’s survey [Pes] and in the notes by Pestov and A.Kwiatkowska [PeK]. To define hyperlinear groups, it suffices to replace thegroups Sym(F ), where F is a nonempty finite set, occurring in the definitionof sofic groups by the unitary groups U(H), where H is a finite-dimensionalHilbert space, equipped with their normalized Hilbert-Schmidt metric. Thisclass of groups has its origin in the theory of operator algebras and is relatedto the Connes Embedding Conjecture. Indeed, by a profound result of E.Kirchberg, F. Radulescu and N. Ozawa, a group is hyperlinear if and only ifit satisfies the Connes embedding conjecture for groups, that is, if and onlyif its von Neumann algebra embeds into an ultrapower of the hyperfinite II1factor R (see [Pes] and the references therein). Elek and Szabo [ES2] provedthat every sofic group is hyperlinear. Thus, we have the implications

finite resid. finite LEF

amenable resid. amenable LEA sofic hyperlinear

There exist finitely presented groups which are LEA but not LEF. Forexample, the Baumslag-Solitar group BS(2, 3) is known to be residuallysolvable. Thus, it is residually amenable and therefore LEA. As the groupBS(2, 3) is finitely presented and not residually finite, it is not LEF by Propo-sition 7.3.8. In [Abe], H. Abels gave an example of a finitely presented solvablegroup which is not residually finite. Abels’ group is amenable (and thereforeLEA) but not LEF.

There exist finitely presented groups which are not LEA. For example, if Gis a finitely presented non-amenable simple group, such as one of Thompson’sgroups V and T (see [CFP]) or one of the Burger-Mozes groups [BuM], thenG is not residually amenable and therefore not LEA by Proposition 7.3.8.

Y. de Cornulier [Cor] provided an example of a finitely presented soficgroup which is not LEA. Cornulier’s example is sofic because it is an extensionof a locally residually finite group by an abelian group and it is not LEAbecause it is non-amenable and isolated as a free-marked group. A. Thom[Tho] provided an example of a group which is hyperlinear but not LEA,but it is unknown whether Thom’s example is sofic or not. The existence ofnon-sofic groups, of non-hyperlinear groups, and of hyperlinear groups whichare not sofic, remain open problems.

L. Bowen [Bow] obtained the following extension of the Kolmogorov-Ornstein-Weiss theorem for Bernoulli shifts. Recall that given a finite setA, a strict probability distribution p = (pa)a∈A on A (that is, pa > 0 forall a ∈ A and

∑a∈A pa = 1) and a countable group G, then, denoting by


μp =∏

g∈G p the corresponding probability measure on the product spaceAG, the triple (G, AG, μp) is called a Bernoulli shift. The associated entropyis the nonnegative number H(p) = −

∑a∈A pa log pa. Two Bernoulli shifts

(G, AG, μp) and (G, BG, μq) are said to be isomorphic (or measurably con-jugate) if there exist two G-invariant subsets X ⊂ AG and Y ⊂ BG suchthat μp(AG \X) = μq(BG \Y ) = 0 and a G-equivariant bijective measurablemap θ : X → Y with measurable inverse θ−1 : Y → X such that θ∗μp = μq.Bowen [Bow, Theorem 1.1] proved that if G is a countable sofic group, thentwo isomorphic Bernoulli shifts (G, AG, μp) and (G, BG, μq) have the sameentropy, i.e. H(p) = H(q). This result had been established when G = Z

by N. Kolmogorov [Ko1, Ko2] in 1958–59 and then extended to countableamenable groups by D. Ornstein and B. Weiss [OrW] in 1987.

L. Glebsky and L.M. Rivera [GlR] introduced the concept of a weakly-sofic group. This is a natural extension of the definition of soficity wherethe Hamming metric on symmetric groups is replaced by general bi-invariantmetrics on finite groups. Glebsky and Rivera showed that the existence ofa non-weakly-sofic group is equivalent to a conjecture on the closure in theprofinite topology of products of conjugacy classes in free groups of finiterank.

Exercises

7.1. Let G and C be two groups. Suppose that K is a finite symmetric subsetof G such that 1G ∈ K. Show that a map ϕ : G → C is a K-almost homo-morphism if and only if it satisfies ϕ(k1k2) = ϕ(k1)ϕ(k2) for all k1, k2 ∈ Kand ϕ(k) �= 1C for all k ∈ K \ {1G}.

7.2. Show that every group which is locally embeddable into the class ofabelian groups is itself abelian.

7.3. Show that every group which is locally embeddable into the class ofmetabelian groups is itself metabelian.

7.4. Show that every abelian group is locally embeddable into the class offinite cyclic groups.

7.5. Let C be a class of groups. Suppose that a group G contains a family(Hi)i∈I of subgroups satisfying the following properties: (1) G =

⋃i∈I Hi; (2)

For all i, j ∈ I there exists k ∈ I such that Hi ∪ Hj ⊂ Hk; (3) Hi is locallyembeddable into C for all i ∈ I. Show that G is locally embeddable into C.

7.6. Let C be a class of groups which is closed under finite direct products.Show that every group which is residually locally embeddable into C is locallyembeddable into C.

Exercises 279

7.7. Let C be a class of groups which is closed under taking subgroups andtaking finite direct products. Let G be a group which is residually C. Showthat there exists a net (Ni)i∈I which converges to {1G} in N (G) such thatG/Ni ∈ C fro all i ∈ I. Hint: Take as I the set of finite subsets of G partiallyordered by inclusion and use Proposition 7.1.13.

7.8. Show that if G is a finitely presented infinite simple group then G is notLEF.

7.9. Show that if G is a finitely presented non-amenable simple group thenG is not LEA.

7.10. (cf. [VeG]) Let G1 and G2 be two groups. Recall that, by Proposi-tion 7.3.1, Gi (i = 1, 2) is an LEF-group if and only if the following holds:

(∗) for every finite subset Ki ⊂ Gi there exist a finite set Li such thatKi ⊂ Li ⊂ Gi and a binary operation �i : Li ×Li → Li such that (Li,�i)is a group and kik

′i = ki � k′

i for all ki, k′i ∈ Ki.

Suppose that G1 and G2 are LEF-groups. An action of G2 on G1 by groupautomorphisms, i.e., a group homomorphism ϕ : G2 → Aut(G1), is said tobe equivariantly approximable, if for all finite subsets K1 ⊂ G1 and K2 ⊂ G2

there exist finite groups (L1,�1) and (L2,�2) as in (∗) and an action of L2 onL1 by group automorphisms ψ : L2 → Aut(L1) such that if k1 ∈ K1, k2 ∈ K2

and ϕ(k2)(k1) ∈ K1 then ϕ(k2)(k1) = ψ(k2)(k1).Show that if G1 and G2 are LEF-groups and ϕ : G2 → Aut(G1) is an

equivariantly approximable action, then the semidirect product G1 �ϕ G2 isan LEF-group.

7.11. Use the previous exercise to show that the group G1 of Sect. 2.6 is anLEF-group (cf. Proposition 7.3.9).

7.12. Show that every residually free group is torsion-free.

7.13. A group G is called fully residually free if for any finite subset K ⊂ G,there exist a free group F and a group homomorphism φ : G → F whoserestriction to K is injective.

(a) Show that every fully residually free group is residually free.(b) Show that every fully residually free group is locally embeddable into

the class of free groups.(c) A group G is called commutative-transitive if whenever a, b, c ∈ G\{1G}

satisfy ab = ba and bc = cb, then ac = ca. Prove that every group which islocally embeddable into the class of free groups is commutative-transitive.Hint: First prove that if F is a free group and a, b ∈ F satisfy ab = ba, thenthere exist an element x ∈ F and integers m, n ∈ Z such that a = xm andb = xn.

(d) Show that if F is a nonabelian free group, then the group F × Z isresidually free but not locally embeddable into the class of free groups.


7.14. Give an example of a finitely generated LEF-group which is neitherresidually finite nor amenable. Hint: Take for instance the group G = G1×F2,where G1 is the group introduced in Sect. 2.6 and F2 denotes a free group ofrank two.

7.15. Let F be a nonempty finite set. Denote by GL(RF ) the automorphismgroup of the real vector space R

F = {x : F → R}. For α ∈ Sym(F ), defineλ(α) : R

F → RF by λ(α)(x) = x ◦ α−1 for all x ∈ R

F .(a) Show that λ(α) ∈ GL(RF ) for all α ∈ Sym(F ).(b) Show that the map λ : Sym(F ) → GL(RF ) is an injective group ho-

momorphism.(c) Show that the Hamming metric dF on Sym(F ) satisfies

dF (α, β) =1|F | Tr(λ(α−1β))

for all α, β ∈ Sym(F ), where Tr(·) denotes the trace.

7.16. Let H be a real or complex Hilbert space of finite dimension n ≥ 1.Let L(H) denote the vector space consisting of all linear maps u : H → H.If u ∈ L(H), we denote by Tr(u) the trace of u and by u∗ its adjoint. Foru, v ∈ L(H), we set

〈u, v〉HS =1n

Tr(u ◦ v∗).

(a) Show that 〈·, ·〉HS is a scalar product on L(H). Let ‖ · ‖HS denote theassociated norm.

(b) Let U(H) = {u ∈ L(H) : u ◦ u∗ = IdH}. Show that U(H) is a groupfor the composition of maps.

(c) Show that the map dHS : U(H) × U(H) → R defined by dHS(u, v) =‖u − v‖HS for all u, v ∈ U(H) is a bi-invariant metric on U(H). (The groupU(H) is called the unitary group of H and dHS is called the normalizedHilbert-Schmidt metric on U(H).)

7.17. Let G be a group, K ⊂ G a finite subset, C a finite group, andϕ : G → C a K-almost-homomorphism. Denote by L : C → Sym(C) theCayley homomorphism, that is, the map defined by L(g)(h) = gh for allg, h ∈ C and set Φ = L ◦ ϕ : G → Sym(C). Show that Φ is a (K, ε)-almost-homomorphism for all ε > 0.

7.18. Let G be a group and let K be a finite subset of G. Let F be anonempty finite set. Show that if 0 < ε < 2/|F | then every (K, ε)-almost-homomorphism ϕ : G → Sym(F ) is a K-almost-homomorphism of G into thegroup Sym(F ).

7.19. Suppose that a group G contains a family (Hi)i∈I of subgroups satisfy-ing the following properties: (1) G =

⋃i∈I Hi; (2) For all i, j ∈ I there exists

k ∈ I such that Hi ∪ Hj ⊂ Hk; (3) Hi is sofic for all i ∈ I. Show that G issofic.

Exercises 281

7.20. Show that every virtually sofic group is sofic. Hint: Use Proposi-tion 7.5.14.

7.21. By using Exercise 6.5 and Exercise 6.6, give a direct proof of the factthat the direct product of finitely many groups which satisfy condition (b)in Theorem 7.7.1 also satisfies it.

7.22. Let G be a finitely generated group and suppose that S and S′ are twofinite symmetric generating subsets of G. Show that the pair (G, S) satisfiescondition (b) in Theorem 7.7.1 if and only if (G, S′) does.

7.23. Give a direct proof of the fact that every finitely generated residuallyfinite group satisfies the condition (b) in Theorem 7.7.1.

7.24. Give a direct proof of the fact that every finitely generated amenablegroup satisfies the condition (b) in Theorem 7.7.1.

Chapter 8

Linear Cellular Automata

In this chapter we study linear cellular automata, namely cellular automatawhose alphabet is a vector space and which are linear with respect to theinduced vector space structure on the set of configurations. If the alphabetvector space and the underlying group are fixed, the set of linear cellularautomata is a subalgebra of the endomorphism algebra of the configurationspace (Proposition 8.1.4). An important property of linear cellular automatais that the image of a finitely supported configuration by a linear cellularautomaton also has finite support (Proposition 8.2.3). Moreover, a linearcellular automaton is entirely determined by its restriction to the space offinitely-supported configurations (Proposition 8.2.4) and it is pre-injective ifand only if this restriction is injective (Proposition 8.2.5). The algebra oflinear cellular automata is naturally isomorphic to the group algebra of theunderlying group with coefficients in the endomorphism algebra of the al-phabet vector space (Theorem 8.5.2). Linear cellular automata may be alsoregarded as endomorphisms of the space of finitely-supported configurations,viewed as a module over the group algebra of the underlying group with co-efficients in the ground field (Proposition 8.7.5). This representation of linearcellular automata is always one-to-one and, when the alphabet vector spaceis finite-dimensional, it is also onto (Theorem 8.7.6). The image of a linearcellular automaton is closed in the space of configurations for the prodiscretetopology, provided that the alphabet is finite dimensional (Theorem 8.8.1).We exhibit an example showing that if one drops the finite dimensionalityof the alphabet, then the image of a linear cellular automaton may fail tobe closed. In Sect. 8.9 we prove a linear version of the Garden of Eden theo-rem. For the proof, we introduce the mean dimension of a vector subspace ofthe configuration space. We show that for a linear cellular automaton withfinite-dimensional alphabet, both pre-injectivity and surjectivity are equiv-alent to the maximality of the mean dimension of the image of the cellularautomaton (Theorem 8.9.6). We exhibit two examples of linear cellular au-tomata with finite-dimensional alphabet over the free group of rank two,one which is pre-injective but not surjective, and one which is surjective but


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284 8 Linear Cellular Automata

not pre-injective. This shows that the linear version of the Garden of Edentheorem fails to hold for groups containing nonabelian free subgroups (seeSects. 8.10 and 8.11). Provided the alphabet is finite dimensional, the inverseof every bijective linear cellular automaton is also a linear cellular automaton(Corollary 8.12.2). In Sect. 8.13 we study the pre-injectivity and surjectiv-ity of the discrete Laplacian over the real numbers and prove a Garden ofEden type theorem (Theorem 8.13.2) for such linear cellular automata withno amenability assumptions on the underlying group. As an application, wededuce a characterization of locally finite groups in terms of real linear cellu-lar automata (Corollary 8.13.4). In Sect. 8.14 we define linear surjunctivityand prove that all sofic groups are linearly surjunctive (Theorem 8.14.4).The notion of stable finiteness for rings is introduced in Sect. 8.15. A sta-bly finite ring is a ring for which one-sided invertible square matrices arealso two-sided invertible. It is shown that linear surjunctivity is equivalentto stable finiteness of the associated group algebra (Corollary 8.15.6). As aconsequence, we deduce that group algebras of sofic groups are stably finitefor any ground field (Corollary 8.15.8). In the last section, we prove that theabsence of zero-divisors in the group algebra of an arbitrary group is equiva-lent to the fact that every non-identically-zero linear cellular automaton withone-dimensional alphabet is pre-injective (Corollary 8.16.12).

We recall that in this book all rings are assumed to be associative (but notnecessarily commutative) with a unity element, and that a field is a nonzerocommutative ring in which each nonzero element is invertible.

8.1 The Algebra of Linear Cellular Automata

Let G be a group and let V be a vector space over a field K.The set V G consisting of all configurations x : G → V over the group G

and the alphabet V has a natural structure of vector space over K in whichaddition and scalar multiplication are given by

(x + x′)(g) = x(g) + x′(g) and (kx)(g) = kx(g)

for all x, x′ ∈ V G, k ∈ K, and g ∈ G. With the prodiscrete topology, V G

becomes a topological vector space (cf. Sect. F.1). The G-shift (see Sect. 1.1)is then K-linear and continuous, that is, for each g ∈ G, the map x �→ gx isa continuous endomorphism of V G.

A linear cellular automaton over the group G and the alphabet V is acellular automaton τ : V G → V G which is K-linear, i.e., which satisfies

τ(x + x′) = τ(x) + τ(x′) and τ(kx) = kτ(x)

for all x, x′ ∈ V G and k ∈ K.

8.1 The Algebra of Linear Cellular Automata 285

Proposition 8.1.1. Let G be a group and let V be a vector space over afield K. Let τ : V G → V G be a cellular automaton with memory set S ⊂ Gand local defining map μ : V S → V . Then τ is linear if and only if μ isK-linear.

Proof. Suppose first that τ is linear. Let y, y′ ∈ V S and k ∈ K. Denote byx and x′ two configurations in V G extending y and y′ respectively, i.e., suchthat x|S = y and x′|S = y′. We then have

μ(y + y′) = τ(x + x′)(1G) = (τ(x) + τ(x′))(1G) = τ(x)(1G) + τ(x′)(1G)= μ(y) + μ(y′).

Similarly, as (kx)|S = ky, we have

μ(ky) = τ(kx)(1G) = kτ(x)(1G) = kμ(y).

This shows that μ is K-linear.Conversely, suppose that μ is K-linear. Then, for all x, x′ ∈ V G, k ∈ K

and g ∈ G we have

τ(x + x′)(g) = μ((g−1(x + x′))|S) = μ((g−1x)|S + (g−1x′)|S)

= μ((g−1x)|S) + μ((g−1x′)|S)= τ(x)(g) + τ(x′)(g)= (τ(x) + τ(x′))(g)

and

τ(kx)(g) = μ((g−1(kx))|S) = μ(k(g−1x)|S) = kμ((g−1x)|S) = kτ(x)(g).

This shows that τ(x + x′) = τ(x) + τ(x′) and τ(kx) = kτ(x). It follows thatτ is linear. ��

The following result is a linear analogue of the Curtis-Hedlund theorem(Theorem 1.8.1).

Theorem 8.1.2. Let G be a group and let V be a vector space over a field K.Let τ : V G → V G be a G-equivariant and K-linear map. Then the followingconditions are equivalent:

(a) the map τ is a linear cellular automaton;(b) the map τ is uniformly continuous (with respect to the prodiscrete uniform

structure on V G);(c) the map τ is continuous (with respect to the prodiscrete topology on V G);(d) the map τ is continuous (with respect to the prodiscrete topology on V G)

at the constant configuration x = 0.


Proof. The implication (a) ⇒ (b) immediately follows from Theorem 1.9.1.Since the topology associated with the prodiscrete uniform structure is theprodiscrete topology (cf. Example B.1.4(a)) and every uniformly continuousmap is continuous (cf. Proposition B.2.2), we also have (b) ⇒ (c). The im-plication (c) ⇒ (d) is trivial. Therefore, we are only left to show that (d) ⇒(a). Suppose that τ is continuous at 0. Then, the map V G → V defined byx �→ τ(x)(1G) is continuous at 0 since the projection maps V G → V definedby x �→ x(g) are continuous (for the prodiscrete topology) for all g ∈ G andthe composition of continuous maps is continuous. We deduce that there ex-ists a finite subset M ⊂ G such that if x ∈ V G satisfies x(m) = 0 for allm ∈ M , then τ(x)(1G) = 0. By linearity, we have that if two configurationsx and y coincide on M then τ(x)(1G) = τ(y)(1G). Thus there exists a linearmap μ : V M → V such that τ(x)(1G) = μ(x|M ) for all x ∈ V G. As τ isG-equivariant, we deduce that τ(x)(g) = τ(g−1x)(1G) = μ((g−1x)|M ) for allx ∈ V G and g ∈ G. This shows that τ is the (linear) cellular automaton withmemory set M and local defining map μ. Thus (d) implies (a). ��

Examples 8.1.3. (a) Let G be a group, let S be a nonempty finite subset ofG, and let K be a field. The discrete Laplacian ΔS : K

G → KG (cf. Exam-

ple 1.4.3(b)) is a linear cellular automaton.(b) Let G be a group, V a vector space over a field K, and f ∈ EndK(V ).

Then the map τ : V G → V G defined by τ(x) = f ◦x for all x ∈ V G is a linearcellular automaton (cf. Example 1.4.3(d)).

(c) Let G be a group, V a vector space over a field K, and s0 an el-ement of G. Let Rs0 : G → G be the right multiplication by s0 in G,that is, the map defined by Rs0(g) = gs0 for all g ∈ G. Then the mapτ : V G → V G defined by τ(x) = x ◦ Rs0 is a linear cellular automaton (cf.Example 1.4.3(e)).

(d) Let G = Z and K be a field. Consider the vector space V = K[t] ofall polynomials in the indeterminate t with coefficients in K. A configurationx ∈ V G may be viewed as a sequence x = (xn)n∈Z, where xn = xn(t) isa polynomial for all n ∈ Z. Let S = {0, 1} and consider the K-linear mapμ : V S → V defined by μ(p, q) = p − tq′ for all (p, q) ∈ V S = V × V ,where q′ ∈ V denotes the derivative of the polynomial q. The linear cellularautomaton τ : V Z → V Z with memory set S and local defining map μ is thengiven by τ(x) = y, where yn = xn − tx′

n+1 ∈ V , n ∈ Z, for all x = (xn)n∈Z ∈V Z.

We recall that an algebra over a field K (or a K-algebra) is a vector spaceA over K endowed with a product A × A → A such that A is a ring withrespect to the sum and the product and such that the following associativelaw holds for the product and the multiplication by scalars:

(ha)(kb) = (hk)(ab)

for all h, k ∈ K and a, b ∈ A.

8.1 The Algebra of Linear Cellular Automata 287

A subset B of a K-algebra A is called a subalgebra of A if B is both avector subspace and a subring of A.

If V is a vector space over a field K, the set EndK(V ) consisting of allendomorphisms of the vector space V has a natural structure of a K-algebrafor which

(f + f ′)(v) = f(v) + f ′(v),

(ff ′)(v) = (f ◦ f ′)(v) = f(f ′(v))

and(kf)(v) = kf(v)

for all f, f ′ ∈ EndK(V ), k ∈ K, and v ∈ V . The identity map IdV is the unityelement of EndK(V ).

Given a group G and a vector space V over a field K, we denote byLCA(G; V ) the set of all linear cellular automata over the group G and thealphabet V . It immediately follows from the definition of a linear cellularautomaton that LCA(G; V ) ⊂ EndK(V G).

Proposition 8.1.4. Let G be a group and let V be a vector space over a fieldK. Then, LCA(G;V ) is a subalgebra of EndK(V G).

Proof. Let τ1, τ2 ∈ LCA(G;V ). Let S1 and S2 be memory sets for τ1 and τ2.Then, the set S = S1 ∪ S2 is also a memory set for τ1 and τ2 (cf. Sect. 1.5).Let μ1 : V S → V and μ2 : V S → V be the corresponding local defining mapsand set μ = μ1 + μ2. For all x ∈ V G and g ∈ G we have

(τ1 + τ2)(x)(g) = τ1(x)(g) + τ2(x)(g)

= μ1(g−1x|S) + μ2(g−1x|S)

= μ(g−1x|S).

This shows that τ1 + τ2 is a cellular automaton with memory set S andlocal defining map μ. Since the map τ1 + τ2 is K-linear, we deduce thatτ1 + τ2 ∈ LCA(G;V ).

On the other hand, let k ∈ K and let τ ∈ LCA(G; V ) with memory setS and local defining map μ : V S → V . Then, for all x ∈ V G and g ∈ G, wehave

(kτ)(x)(g) = kτ(x)(g) = kμ(g−1x|S) = (kμ)(g−1x|S).

Therefore the K-linear map kτ is a cellular automaton with memory set Sand local defining map kμ. It follows that kτ ∈ LCA(G; V ).

We clearly have IdV G ∈ LCA(G;V ) (cf. Example 1.4.3(d)). Finally, itfollows from Proposition 1.4.9 that if τ1, τ2 ∈ LCA(G;V ) then the K-linearmap τ1τ2 = τ1 ◦ τ2 is also a cellular automaton and hence τ1τ2 ∈ LCA(G;V ).This shows that LCA(G; V ) is a subalgebra of EndK(V G). ��


8.2 Configurations with Finite Support

Let G be a group and let V be a vector space over a field K.The support of a configuration x ∈ V G is the set {g ∈ G : x(g) �= 0V }. We

denote by V [G] the subset of V G consisting of all configurations with finitesupport.

Proposition 8.2.1. Let G be a group and let V be a vector space over a fieldK. Then the set V [G] is a vector subspace of V G. Moreover, V [G] is densein V G for the prodiscrete topology.

Proof. If k1, k2 ∈ K and x1, x2 ∈ V G, then the support of k1x1 + k2x2 iscontained in the union of the support of x1 and the support of x2. Therefore,if x1 and x2 have finite support, so does k1x1 + k2x2. Consequently, V [G] isa vector subspace of V G.

Let x ∈ V G and let W ⊂ V G be a neighborhood of x for the prodiscretetopology. By definition of the prodiscrete topology, there exists a finite subsetΩ ⊂ G such that W contains all configurations which coincide with x onΩ. It follows that the configuration y ∈ V [G] which coincides with x onΩ and is identically zero outside of Ω is in W . This shows that V [G] isdense in V G. ��

Proposition 8.2.2. Let G be a group and let V be a vector space over a fieldK. Let x, x′ ∈ V G. Then the configurations x and x′ are almost equal if andonly if x − x′ ∈ V [G].

Proof. By definition, x and x′ are almost equal if and only if the set {g ∈G : x(g) �= x′(g)} is finite. This is equivalent to x − x′ ∈ V [G] since {g ∈ G :x(g) �= x′(g)} = {g ∈ G : (x − x′)(g) �= 0V }. ��

Proposition 8.2.3. Let G be a group and let V be a vector space over a fieldK. Let τ ∈ LCA(G; V ). Then one has τ(V [G]) ⊂ V [G].

Proof. Denote by S ⊂ G a memory set for τ and let μ : V S → V be the corre-sponding local defining map. Let x ∈ V [G] and let T ⊂ G denote the supportof x. For all g ∈ G, we have τ(x)(g) = μ((g−1x)|S). As μ is linear (Proposi-tion 8.1.1) and the support of g−1x is g−1T , we deduce that τ(x)(g) = 0 ifg−1T ∩ S = ∅. It follows that the support of τ(x) is contained in the finiteset TS−1 ⊂ G. This shows that τ(x) ∈ V [G]. ��

Observe that if τ ∈ LCA(G; V ), then the restriction map

τ |V [G] : V [G] → V [G]

is K-linear, that is, τ |V [G] ∈ EndK(V [G]).

8.3 Restriction and Induction of Linear Cellular Automata 289

Let A and B be two algebras over a field K. A map F : A → B is called aK-algebra homomorphism if F is both a vector space homomorphism (i.e., aK-linear map) and a ring homomorphism. This is equivalent to the fact thatF satisfies F (a + a′) = F (a) + F (a′), F (aa′) = F (a)F (a′), F (ka) = kF (a)for all a, a′ ∈ A and k ∈ K, and F (1A) = 1B.

Proposition 8.2.4. Let G be a group and let V be a vector space over a fieldK. Then the map Λ : LCA(G;V ) → EndK(V [G]) defined by Λ(τ) = τ |V [G],where τ |V [G] : V [G] → V [G] is the restriction of τ to V [G], is an injectiveK-algebra homomorphism.

Proof. The fact that Λ is an algebra homomorphism immediately followsfrom the definition of the algebra operations on LCA(G; V ) and EndK(V [G]).Suppose that τ ∈ LCA(G; V ) satisfies Λ(τ) = 0. This means that τ(y) = 0for all y ∈ V [G]. As τ : V G → V G is continuous by Proposition 1.4.8 andV [G] is dense in V G by Proposition 8.2.1 for the prodiscrete topology, wededuce that τ(x) = 0 for all x ∈ V G, that is, τ = 0. This shows that Λ isinjective. ��

Proposition 8.2.5. Let G be a group and let V be a vector space over a fieldK. Let τ ∈ LCA(G;V ). Then the following conditions are equivalent:

(a) τ is pre-injective;(b) τ |V [G] : V [G] → V [G] is injective.

Proof. Suppose that τ is pre-injective. Let x ∈ ker(τ |V [G]). Then x ∈ V [G]and τ(x) = τ |V [G](x) = 0. As the trivial configuration 0 and the configurationx are almost equal and τ(0) = 0 by linearity of τ , the pre-injectivity of τimplies that x = 0. This shows that τ |V [G] is injective.

Conversely, suppose that τ |V [G] is injective. Let x, x′ ∈ V G be two configu-rations which are almost equal and such that τ(x) = τ(x′). Then x−x′ ∈ V [G]by Proposition 8.2.2 and we have τ |V [G](x−x′) = τ(x−x′) = τ(x)−τ(x′) = 0.As τ |V [G] is injective, this implies x − x′ = 0, that is, x = x′. Therefore, τ ispre-injective. ��

8.3 Restriction and Induction of Linear CellularAutomata

In this section, we show that the operations of restriction and induction forcellular automata that were introduced in Sect. 1.7 preserve linearity.

Let G be a group and let V be a vector space over a field K. Let H bea subgroup of G. We denote by LCA(G, H; V ) = LCA(G; V ) ∩ CA(G, H; V )the set of all linear cellular automata τ : V G → V G admitting a memory setS such that S ⊂ H. Recall from Sect. 1.7, that, given a cellular automa-ton τ : V G → V G with memory set S ⊂ H, we denote by τH : V H → V H


the restriction cellular automaton. Similarly, given a cellular automatonσ : V H → V H , we denote by σG : V G → V G the induced cellular automa-ton.

Proposition 8.3.1. Let τ ∈ CA(G, H; V ). Then, τ ∈ LCA(G, H;V ) if andonly if τH ∈ LCA(H;V ).

Proof. It follows immediately from the definitions of restriction and inductionthat a cellular automaton τ ∈ CA(G, H;V ) is linear if and only if τH ∈CA(H; V ) is linear. ��

Let A and B be two algebras over a field K. A map F : A → B is calleda K-algebra isomorphism if F is a bijective K-algebra homomorphism. It isclear that if F : A → B is a K-algebra isomorphism then its inverse mapF−1 : B → A is also a K-algebra isomorphism.

Proposition 8.3.2. The set LCA(G, H;V ) is a subalgebra of LCA(G; V ).Moreover, the map τ �→ τH is a K-algebra isomorphism from LCA(G, H; V )onto LCA(H;V ) whose inverse is the map σ �→ σG.

Proof. Let τ1, τ2 ∈ LCA(G, H;V ) with memory sets S1, S2 ⊂ H respec-tively. Then the linear cellular automaton τ1 + τ2 admits S1 ∪ S2 as a mem-ory set. As S1 ∪ S2 ⊂ H, we have τ1 + τ2 ∈ LCA(G, H;V ). If k ∈ K andτ ∈ LCA(G, H;V ), with memory set S ⊂ H, then S is also a memory setfor kτ and therefore kτ ∈ LCA(G, H; V ). This shows that LCA(G, H;V )is a vector subspace of LCA(G;V ). Since LCA(G, H;V ) is a submonoid ofLCA(G;V ) by Proposition 1.7.1, we deduce that LCA(G, H;V ) is a subalge-bra of LCA(G;V ).

To simplify notation, denote by Φ : LCA(G, H; V ) → LCA(H;V ) andΨ : LCA(H;V ) → LCA(G, H;V ) the maps defined by Φ(τ) = τH and Ψ(σ) =σG respectively. It is clear from the definitions that Ψ ◦Φ : LCA(G, H; V ) →LCA(G, H; V ) and Φ ◦ Ψ : LCA(H;V ) → LCA(H; V ) are the identity maps.Therefore, Φ is bijective with inverse Ψ . It remains to show that Φ is a K-algebra homomorphism.

Let τ1, τ2 ∈ LCA(G, H;V ) and k1, k2 ∈ K. Let x ∈ V H and let x ∈ V G

extending x. By applying (1.13), we have

Φ(k1τ1 + k2τ2)(x)(h) = (k1τ1 + k2τ2)(x)(h) = k1τ1(x)(h) + k2τ2(x)(h)

for all h ∈ H. We deduce that Φ(k1τ1 + k2τ2)(x) = (k1Φ(τ1) + k2Φ(τ2))(x)for all x ∈ V H , that is, Φ(k1τ1 + k2τ2) = k1Φ(τ1) + k2Φ(τ2). This shows thatΦ is K-linear.

Finally, it follows from Proposition 1.7.2 that Φ(IdV G) = IdV H andΦ(τ1τ2) = Φ(τ1)Φ(τ2) for all τ1, τ2 ∈ LCA(G, H;V ). We have shown thatΦ is a K-algebra isomorphism. ��

8.4 Group Rings and Group Algebras 291

8.4 Group Rings and Group Algebras

Let G be a group and let R be a ring. We denote by 0R (resp. 1R) the zero(resp. unity) element of R. We regard R as a left R-module over itself. ThenRG = {α : G → R} has a natural structure of left R-module with additionand scalar multiplication given by

(α + β)(g) = α(g) + β(g) and (rα)(g) = rα(g)

for all α, β ∈ RG, r ∈ R, and g ∈ G. Define the support of an element α ∈ RG

as being the set {g ∈ G : α(g) �= 0R}. Let R[G] denote the set consisting of allelements α ∈ RG which have finite support. Then R[G] is a free submoduleof RG. We have

R[G] =⊕

g∈G

R ⊂∏

g∈G

R = RG,

so that the elements δg : G → R, g ∈ G, defined by

δg(h) =

{1R if h = g

0R if h �= g(8.1)

freely generate R[G] as a left R-module. Note that the decomposition of anelement α ∈ R[G] in this basis is simply given by the formula

α =∑

g∈G

α(g)δg.

Let now α and β be two elements of R[G] and denote their supports by Sand T . The convolution product of α and β is the element αβ ∈ RG definedby

(αβ)(g) =∑

h∈G

α(h)β(h−1g) =∑

h∈S

α(h)β(h−1g) (8.2)

for all g ∈ G. Note that αβ ∈ R[G] as the support of αβ is contained inST = {st : s ∈ S, t ∈ T}. By using the change of variables h1 = h andh2 = h−1g, the convolution product (8.2) may be also expressed as follows:

(αβ)(g) =∑

h1,h2∈Gh1h2=g

α(h1)β(h2). (8.3)

Proposition 8.4.1. Let G be a group and let R be a ring. Then the additionand the convolution product gives a ring structure to R[G].

Proof. We know that (R[G], +) is an abelian group. Let α, β, γ ∈ R[G] andg ∈ G. We have


[(αβ)γ](g) =∑

h∈G

(αβ)(h)γ(h−1g)

=∑

h∈G

∑

k∈G

α(k)β(k−1h)γ(h−1g)

(by setting s = k−1h) =∑

s∈G

∑

k∈G

α(k)β(s)γ(s−1k−1g)

=∑

k∈G

α(k)(∑

s∈G

β(s)γ(s−1k−1g))

=∑

k∈G

α(k)(βγ)(k−1g)

= [α(βγ)](g).

This shows that (αβ)γ = α(βγ). Thus, the convolution product is associative.On the other hand, given α ∈ R[G] we have, for all g ∈ G,

[δ1Gα](g) =

∑

h∈G

δ1G(h)α(h−1g) = α(g) =

∑

k∈G

α(k)δ1G(k−1g) = [αδ1G

](g).

Thus, δ1Gα = αδ1G

= α for all α ∈ R[G]. This shows that δ1Gis an identity

element for the convolution product. Finally,

[(α + β)γ](g) =∑

h∈G

(α + β)(h)γ(h−1g)

=∑

h∈G

[α(h) + β(h)]γ(h−1g)

=∑

h∈G

α(h)γ(h−1g) +∑

k∈G

β(k)γ(k−1g)

= (αγ)(g) + (βγ)(g).

Thus, we have (α + β)γ = αγ + βγ. Similarly, one shows that α(β + γ) =αβ+αγ. It follows that the distributive laws also hold in R[G]. Consequently,R[G] is a ring with unity element 1R[G] = δ1G

. ��

The ring R[G] is called the group ring of G with coefficients in R.Given a ring R, denote by U(R) the multiplicative group consisting of all

invertible elements in R.

Proposition 8.4.2. Let G be a group and let R be a ring. Then one hasδg ∈ U(R[G]) for all g ∈ G. Moreover the map φ : G → U(R[G]) given byφ(g) = δg is a group homomorphism.

Proof. Let g1, g2, g ∈ G. Then (δg1δg2)(g) =∑

h∈G δg1(h)δg2(h−1g) equals 1

if g−11 g = g2, that is, if g = g1g2, and equals 0 otherwise. This shows that

δg1δg2 = δg1g2 . (8.4)

8.4 Group Rings and Group Algebras 293

We deduce that δgδg−1 = δgg−1 = δ1G= 1R[G] and, similarly, δg−1δg =

1R[G]. This shows that δg belongs to U(R[G]). The fact that φ is a grouphomomorphism follows from (8.4). ��

Let G be a group and let R be a ring. For α ∈ R[G] define α∗ : G → Rby setting α∗(g) = α(g−1) for all g ∈ G. If S is the support of α, then thesupport of α∗ is S−1. Thus one has α∗ ∈ R[G].

Proposition 8.4.3. Let G be a group and let R be a ring. Let α, β ∈ R[G].Then one has

(i) (1R[G])∗ = 1R[G],(ii) (α∗)∗ = α,(iii) (α + β)∗ = α∗ + β∗.

Moreover, if the ring R is commutative, then one has

(iv) (αβ)∗ = β∗α∗.

Proof. (i) We have (1R[G])∗ = 1R[G] since the support of 1R[G] = δ1Gis {1G}.

(ii) For all g ∈ G, we have

(α∗)∗(g) = α((g−1)−1) = α(g).

Therefore (α∗)∗ = α.(iii) For all g ∈ G, we have

(α+β)∗(g) = (α+β)(g−1) = α(g−1)+β(g−1) = α∗(g)+β∗(g) = (α∗+β∗)(g).

Therefore (α + β)∗ = α∗ + β∗.(iv) Suppose that the ring R is commutative. For all g ∈ G, we have

(αβ)∗(g) = (αβ)(g−1)

=∑

h∈G

α(h)β(h−1g−1)

=∑

h∈G

β(h−1g−1)α(h) (since R is commutative)

=∑

h∈G

β∗(gh)α∗(h−1)

=∑

k∈G

β∗(k)α∗(k−1g)

= (β∗α∗)(g).

Therefore (αβ)∗ = β∗α∗. ��

If R = (R,+, ·) is a ring, its opposite ring is the ring Rop = (R,+, ∗)having the same underlying set and addition as R and with multiplication ∗defined by r ∗ s = sr for all r, s ∈ R.


From Proposition 8.4.3, we deduce that, if G is a group and R is a com-mutative ring, then the map α �→ α∗ is a ring isomorphism between the ringR[G] and its opposite ring (R[G])op. Thus we have

Corollary 8.4.4. Let G be a group and let R be a commutative ring. Thenthe ring R[G] is isomorphic to its opposite ring (R[G])op. ��

Remarks 8.4.5. Let G be a group and let R be a ring.(a) It is immediate to verify that the map R → R[G] defined by r �→ rδ1G

isan injective ring homomorphism. This is often used to regard R as a subringof R[G].

(b) Observe that (rα)β = r(αβ) for all r ∈ R and α, β ∈ R[G]. Therefore,we can use the notation rαβ = (rα)β = r(αβ). If the ring R is commutative,then one has the additional property (r1α)(r2β) = r1r2αβ for all r1, r2 ∈ Rand α, β ∈ R[G].

(c) Suppose that the group G is abelian and that the ring R is commuta-tive. Then the ring R[G] is commutative. Indeed, it immediately follows from(8.3) that we have αβ = βα for all α, β ∈ R[G] under these hypotheses.

Suppose now that we are given an algebra A over a field K. Then A[G] isboth a K-vector space and a ring; Moreover, we have (k1α)(k2β) = k1k2αβfor all k1, k2 ∈ K and α, β ∈ A[G]. Therefore, A[G] is an algebra over K.The K-algebra A[G] is called the group algebra of G with coefficients in theK-algebra A. In the particular case A = K, this gives the K-algebra K[G].

8.5 Group Ring Representation of Linear CellularAutomata

Let G be a group and let V be a vector space over a field K.Consider the K-algebra EndK(V )[G], that is, the group algebra of G with

coefficients in the K-algebra EndK(V ) (see Sect. 8.4).For each α ∈ EndK(V )[G], we define a map τα : V G → V G by setting

τα(x)(g) =∑

h∈G

α(h)(x(gh)) (8.5)

for all x ∈ V G and g ∈ G. Observe that α(h) ∈ EndK(V ) and x(gh) ∈ Vfor all x ∈ V G and g, h ∈ G. Note also that there is only a finite number ofnonzero terms in the sum appearing in the right hand side of (8.5) since αhas finite support. It follows that τα is well defined.

For each v ∈ V , define the configuration cv ∈ V [G] by

cv(g) ={

v if g = 1G,0 otherwise. (8.6)

8.5 Group Ring Representation of Linear Cellular Automata 295

Note that the map from V to V [G] given by v �→ cv is injective and K-linear.

Proposition 8.5.1. Let α ∈ EndK(V )[G]. Then one has:

(i) τα ∈ LCA(G; V );(ii) α(g)(v) = τα(cv)(g) for all v ∈ V and g ∈ G;(iii) the support of α is the minimal memory set of τα.

Proof. Let S ⊂ G denote the support of α. Consider the map μ : V S → Vgiven by

μ(y) =∑

s∈S

α(s)(y(s))

for all y ∈ V S . Then, for all x ∈ V G and g ∈ G, we have

τα(x)(g) =∑

h∈G

α(h)(x(gh))

=∑

h∈G

α(h)(g−1x(h))

=∑

s∈S

α(s)(g−1x(s))

= μ((g−1x)|S).

(8.7)

Thus τα is the cellular automaton with memory set S and local definingmap μ. It is clear from (8.5) that τα is a K-linear map. Thus, we have τα ∈LCA(G; V ). This shows (i).

Let v ∈ V and let cv ∈ V [G] as in (8.6). By applying (8.5), we get, for allg ∈ G,

τα(cv)(g−1) =∑

h∈G

α(h)(cv(g−1h)) = α(g)(v).

This gives us (ii).Finally, let S0 ⊂ G denote the minimal memory set of τα. We have seen

in the proof of (i) that S is a memory set for τα. Therefore, we have S0 ⊂ S.On the other hand, we deduce from (ii) that, for all v ∈ V and g ∈ G, wehave

α(g)(v) = τα(cv)(g) = μ0((g−1cv)|S0),

where μ0 : V S0 → V denote the local defining map for τα associated withS0. Since the configuration g−1cv is identically zero on G \ {g}, it followsthat α(g) = 0 if g /∈ S0. This implies S ⊂ S0. Thus, we have S = S0.This shows (iii). ��

Theorem 8.5.2. Let G be a group and let V be a vector space over a field K.Then the map Ψ : EndK(V )[G] → LCA(G;V ) defined by α �→ τα is a K-algebra isomorphism.

Proof. Let α, β ∈ EndK(V )[G] and k ∈ K. By applying (8.5), we get


τα+β(x)(g) =∑

h∈G

[(α + β)(h)](x(gh))

=∑

h∈G

(α(h) + β(h))(x(gh))

=∑

h∈G

[α(h)(x(gh)) + β(h)(x(gh))]

=∑

h∈G

α(h)(x(gh)) +∑

h∈G

β(h)(x(gh))

= τα(x)(g) + τβ(x)(g)

and, similarly,

τkα(x)(g) =∑

h∈G

[(kα)(h)](x(gh))

=∑

h∈G

k[α(h)](x(gh))

= k∑

h∈G

[α(h)](x(gh))

= kτα(x)(g)

for all x ∈ V G and g ∈ G. Thus τα+β = τα + τβ and τkα = kτα. This showsthat Ψ is a K-linear map.

Let us show that Ψ is a ring homomorphism. Let α, β ∈ EndK(V )[G]. Letx ∈ V G and set y = τβ(x). For all g, h ∈ G, we have

y(gh) =∑

t∈G

β(t)(x(ght)). (8.8)

It follows that

τα(τβ(x))(g) = τα(y)(g)

=∑

h∈G

α(h)(y(gh))

(by (8.8)) =∑

h∈G

α(h)(∑

t∈G

β(t)(x(ght)))

(by setting z = ht) =∑

h∈G

α(h)(∑

z∈G

β(h−1z)(x(gz)))

=∑

z∈G

(∑

h∈G

α(h)β(h−1z))

(x(gz))

=∑

z∈G

[αβ](z)(x(gz))

= ταβ(x)(g).

8.5 Group Ring Representation of Linear Cellular Automata 297

Thus we have ταβ = τα ◦ τβ , that is, Ψ(αβ) = Ψ(α)Ψ(β). Observe that1EndK(V )[G] = δ1G

, where δ1G: G → EndK(V ) is given by δ1G

(h) = IdV ifh = 1G and δ1G

(h) = 0 otherwise. Thus, if x ∈ V G we have

Ψ(1EndK(V )[G])(x)(g) = Ψ(δ1G)(x)(g)

=∑

h∈G

δ1G(h)(x(gh))

= x(g)

for all g ∈ G. It follows that Ψ(1EndK(V )[G]) = IdV G . This proves that Ψ is aring homomorphism.

Let us show now that Ψ is injective. Let α ∈ ker(Ψ). Then,

τα(x) = 0 (8.9)

for all x ∈ V G. Taking x = cv in (8.9), where, for v ∈ V , the elementcv ∈ V [G] is as in (8.6), we deduce from Proposition 8.5.1(ii) that α(g)(v) = 0for all v ∈ V and g ∈ G. Thus, we have α(g) = 0 for all g ∈ G and thereforeα = 0. It follows that ker(Ψ) = {0}, that is, Ψ is injective.

Let us show that Ψ is surjective. Suppose that τ ∈ LCA(G;V ) hasmemory set S and local defining map μ : V S → V . As μ is K-linear (cf.Proposition 8.1.1), there exist K-linear maps αs : V → V , s ∈ S, such thatμ(y) =

∑s∈S αs(y(s)) for all y ∈ V S . For all x ∈ V G and g ∈ G, we have

τ(x)(g) = μ((g−1x)|S) =∑

s∈S

αs

(g−1x(s)

)=

∑

s∈S

αs (x(gs)) .

This shows that τ = τα, where α ∈ EndK(V )[G] is defined by

α(h) ={

αs if h = s ∈ S,0 otherwise.

Thus Ψ is surjective. ��

If the vector space V is one-dimensional over the field K, then each endo-morphism of V is of the form v �→ kv, for some k ∈ K. It follows that theK-algebra EndK(V ) is canonically isomorphic to K in this case. Thus, we get:

Corollary 8.5.3. Let G be a group and let V be a one-dimensional vectorspace over a field K. Then the map Ψ : K[G] → LCA(G;V ) defined by

Ψ(α)(x)(g) =∑

h∈G

α(h)x(gh)

for all α ∈ K[G], x ∈ V G, and g ∈ G, is a K-algebra isomorphism. ��


When G is an abelian group, then the K-algebra K[G] is commutative forany field K by Remark 8.4.5(c). Thus, as an immediate consequence of thepreceding corollary, we get:

Corollary 8.5.4. Let G be an abelian group and let V be a one-dimensional vector space over a field K. Then the K-algebra LCA(G; V ) iscommutative. ��

Examples 8.5.5. (a) Let G be a group, S a nonempty finite subset of G, andlet K be a field. Consider the element

α = |S|δ1G−

∑

s∈S

δs ∈ K[G],

Then, taking V = K, we have, for all x ∈ KG and g ∈ G,

Ψ(α)(x)(g) =∑

h∈G

α(h)x(gh) = |S|x(g) −∑

s∈S

x(gs).

It follows that Ψ(α) is the discrete Laplacian ΔS : KG → K

G associated withS (cf. Example 8.1.3(a)).

(b) Let G be a group, V a vector space over a field K and f ∈ EndK(V ).Consider the element α ∈ EndK(V )[G] defined by α = fδ1G

. Let x ∈ V G andg ∈ G. We have

Ψ(α)(x)(g) =∑

h∈G

α(h)(x(gh)) = f(x(g)).

It follows that Ψ(α) is the linear cellular automaton τ ∈ LCA(G; V ) definedby τ(x) = f ◦ x for all x ∈ V G (cf. Example 8.1.3(b)).

(c) Let G be a group, V a vector space over a field K, and s0 an element ofG. Consider the group algebra element α ∈ EndK(V )[G] defined by α = δs0 .For all x ∈ V G and g ∈ G, we have

Ψ(α)(x)(g) =∑

h∈G

α(h)(x(gh)) = x(gs0).

It follows that Ψ(α) is the linear cellular automaton τ ∈ LCA(G; V ) definedby τ(x) = x ◦ Rs0 , where Rs0 : G → G is the right multiplication by s0 (cf.Example 8.1.3(c)).

(d) Let G = Z, let K be a field, and let V = K[t] be the vector spaceconsisting of all polynomials in the indeterminate t with coefficients in K.Denote by D and U the elements in EndK(V ) defined respectively by D(p) =p′ and U(p) = tp, for all p ∈ V . Consider the element

α = δ0 − (U ◦ D)δ1 ∈ EndK(V )[Z].

8.6 Modules over a Group Ring 299

For all x = (xn)n∈Z ∈ V Z, we have Ψ(α)(x) = (yn)n∈Z ∈ V Z, where

yn =∑

m∈Z

α(m)(xn+m) = xn − tx′n+1

for all n ∈ Z. It follows that Ψ(α) is the linear cellular automaton τ ∈LCA(Z; K[t]) described in Example 8.1.3(d).

8.6 Modules over a Group Ring

Let R be a ring and let M be a left R-module. We denote by EndR(M) theendomorphism ring of M . Recall that EndR(M) is the set consisting of allmaps f : M → M satisfying

f(x + x′) = f(x) + f(x′) and f(rx) = rf(x)

for all r ∈ R and x, x′ ∈ M . The ring operations in EndR(M) are given by

(f + f ′)(x) = f(x) + f ′(x) and (ff ′)(x) = (f ◦ f ′)(x) = f(f ′(x))

for all f, f ′ ∈ EndR(M) and x ∈ M , and the unity element of EndR(M) isthe identity map IdM .

Suppose that there is a group G which acts on M by endomorphisms. Wethen define a structure of left R[G]-module on M as follows.

For α ∈ R[G] and x ∈ M , define the element αx ∈ M by

αx =∑

g∈G

α(g)gx. (8.10)

Note that this definition makes sense. Indeed, one has α(g) ∈ R and gx ∈ Mfor each g ∈ G. On the other hand, there is only finitely many nonzero termsin the right hand side of (8.10) since the support of α is finite.


(i) 1R[G]x = x,(ii) α(x + x′) = αx + αx′,(iii) (α + β)x = αx + βx,(iv) α(βx) = (αβ)x

for all α, β ∈ R[G] and x, x′ ∈ M .

Proof. (i) We have1R[G]x = δ1G

x = x.


(ii) We have

α(x + x′) =∑

g∈G

α(g)g(x + x′) =∑

g∈G

(α(g)gx + α(g)gx′)

=∑

g∈G

α(g)gx +∑

g∈G

α(g)gx′ = αx + αx′.

(iii) We have

(α + β)x =∑

g∈G

(α + β)(g)gx =∑

g∈G

(α(g) + β(g))gx

=∑

g∈G

α(g)gx +∑

g∈G

β(g)gx = αx + βx.

(iv) We have, by using (8.3),

α(βx) =∑

h1∈G

α(h1)h1(βx)

=∑

h1∈G

α(h1)h1

( ∑

h2∈G

β(h2)h2x

)

=∑

h1∈G

∑

h2∈G

α(h1)β(h2)h1h2x

=∑

g∈G

( ∑

h1,h2∈Gh1h2=g

α(h1)β(h2))

gx

=∑

g∈G

(αβ)(g)gx

= (αβ)x.

��

It follows from Proposition 8.6.1 that the addition on M and the multi-plication (α, x) �→ αx defined by (8.10) gives us a left R[G]-module structureon M . Observe that this R[G]-module structure extends the R-module struc-ture on M if we regard R as a subring of R[G] via the map r �→ rδ1G

(cf.Remark 8.4.5(a)).

Proposition 8.6.2. Let f : M → M be a map. Then the following conditionsare equivalent:

(a) f ∈ EndR[G](M);(b) f ∈ EndR(M) and f is G-equivariant.

Proof. Suppose that f satisfies (b). Then f(x + x′) = f(x) + f(x′) for allx, x′ ∈ M since f ∈ EndR(M). On the other hand, if α ∈ R[G] and x ∈ M ,

8.7 Matrix Representation of Linear Cellular Automata 301

we have, by using the G-equivariance and the R-linearity of f ,

f(αx) = f

(∑

g∈G

α(g)gx

)=

∑

g∈G

α(g)f(gx) =∑

g∈G

α(g)gf(x) = αf(x).

It follows that f ∈ EndR[G](M). This shows that (b) implies (a).Conversely, suppose that f satisfies (a). Let x, x′ ∈ M and r ∈ R. Then

we have f(x + x′) = f(x) + f(x′), f(rx) = f(rδ1Gx) = rδ1G

f(x) = rf(x),and f(gx) = f(δgx) = δgf(x) = gf(x) since f ∈ EndR[G](M). This showsthat f ∈ EndR(M) and that f is G-equivariant. Thus, (a) implies (b). ��

Remark 8.6.3. Conversely, suppose that we are given a left R[G]-module M .Then, by applying the above construction, we recover the initial R[G]-modulestructure on M if we start from the R-module structure on M obtained byrestricting the scalars and the R-linear action of G on M defined by gx = δgxfor all g ∈ G and x ∈ M .

8.7 Matrix Representation of Linear Cellular Automata

Let G be a group and let V be a vector space over a field K.Since the G-shift action on V G is K-linear, it induces a structure of left

K[G]-module on V G which extends the K-vector space structure on V G (seeSect. 8.6). Note that by applying (8.10) we get, for all α ∈ K[G] and x ∈ V G,

αx =∑

h∈G

α(h)hx, (8.11)

that is,(αx)(g) =

∑

h∈G

α(h)x(h−1g)

for all g ∈ G. Thus, αx may be regarded as the “convolution product” of αand x (compare with (8.2)).

Let τ : V G → V G be a linear cellular automaton. Then τ is K-linearby definition. On the other hand, τ is G-equivariant by Proposition 1.4.4.Thus, it follows from Proposition 8.6.2 that τ is an endomorphism of theK[G]-module V G. Consequently, we have LCA(G;V ) ⊂ EndK[G](V G). It isclear that EndK[G](V G) is a subalgebra of the K-algebra EndK(V G). SinceLCA(G; V ) is a subalgebra of EndK(V G) by Proposition 8.1.4, we get:

Proposition 8.7.1. The set LCA(G; V ) is a subalgebra of the K-algebraEndK[G](V G). ��

Remark 8.7.2. When G is a finite group, we have LCA(G; V ) = EndK[G](V G).Indeed, in this case, every u ∈ EndK[G](V G) is a linear cellular automaton


with memory set G and local defining map μ : V G → V given by μ(y) =u(y)(1G), since u(x)(g) = g−1u(x)(1G) = u(g−1x)(1G) for all x ∈ V G andg ∈ G.

Consider now the vector subspace V [G] ⊂ V G consisting of all configura-tions with finite support. Observe that if x ∈ V G has support Ω ⊂ G andg ∈ G, then the support of the configuration gx is gΩ. Therefore, V [G] is asubmodule of the K[G]-module V G.

Recall that, for v ∈ V , the configuration cv ∈ V [G] is defined by

cv(g) =

{v if g = 1G,

0 otherwise.

Note that if g ∈ G and v ∈ V , then gcv = δgcv is the configuration whichtakes the value v at g and is identically 0 on G \ {g}. It follows that everyx ∈ V [G] can be written as

x =∑

g∈G

gcx(g). (8.12)

Proposition 8.7.3. Suppose that (ei)i∈I is a basis of the K-vector space V .Then the family of configurations (cei)i∈I is a free basis for the left K[G]-module V [G].

Proof. Let x ∈ V [G]. As (ei)i∈I is a K-basis for V , we can find elementsαi ∈ K[G], i ∈ I, such that

x(g) =∑

i∈I

αi(g)ei

for all g ∈ G. This gives us

x =∑

g∈G

gcx(g)

=∑

g∈G

g

(∑

i∈I

αi(g)cei

)

=∑

g∈G

(∑

i∈I

αi(g)gcei

)

=∑

i∈I

(∑

g∈G

αi(g)gcei

)

=∑

i∈I

αicei ,

8.7 Matrix Representation of Linear Cellular Automata 303

where the last equality follows from (8.11). If x = 0 ∈ V [G], then x(g) = 0for all g ∈ G and therefore αi = 0 for all i ∈ I. This shows that (cei)i∈I is abasis for the left K[G]-module V [G]. ��

As every vector space admits a basis, we deduce the following

Corollary 8.7.4. The left K[G]-module V [G] is free. ��

If τ : V G → V G is a linear cellular automaton, we have τ(V [G]) ⊂ V [G] byProposition 8.2.3. Moreover, it follows from Proposition 8.2.4 that the mapΛ : LCA(G;V ) → EndK(V [G]) which associates with each τ ∈ LCA(G; V ) itsrestriction τ |V [G] : V [G] → V [G] is an injective homomorphism of K-algebras.As we have τ |V [G] ∈ EndK[G](V [G]) ⊂ EndK(V [G]) for all τ ∈ LCA(G; V ),we get:

Proposition 8.7.5. Let G be a group and let V be a vector space over a fieldK. Then the map Φ : LCA(G;V ) → EndK[G](V [G]) defined by Φ(τ) = τ |V [G],where τ |V [G] : V [G] → V [G] is the restriction of τ to V [G], is an injectiveK-algebra homomorphism. ��

When the alphabet is finite-dimensional, we have the following:

Theorem 8.7.6. Let G be a group and let V be a finite-dimensional vectorspace over a field K. Then the map Φ : LCA(G;V ) → EndK[G](V [G]) definedby Φ(τ) = τ |V [G], where τ |V [G] : V [G] → V [G] is the restriction of τ to V [G],is a K-algebra isomorphism.

Proof. By the preceding proposition, it suffices to show that Φ is surjective.Let u ∈ EndK[G](V [G]). Consider an element x ∈ V [G]. We have

x =∑

g∈G

gcx(g)

by (8.12). This implies

u(x) = u

(∑

g∈G

gcx(g)

)=

∑

g∈G

gu(cx(g))

Since u is K[G]-linear. We deduce that

u(x)(1G) =∑

g∈G

u(cx(g))(g−1). (8.13)

Suppose now that dimK(V ) = d and let e1, e2, . . . , ed be a K-basis for V .Denote by Ti ⊂ G the support of u(cei), 1 ≤ i ≤ d, and consider the finitesubset S ⊂ G defined by S =

⋃1≤i≤d T−1

i .If v ∈ V , we can write v =

∑di=1 kiei with ki ∈ K, 1 ≤ i ≤ d. We then

have cv =∑d

i=1 kicei and hence


u(cv) = u

( d∑

i=1

kicei

)=

d∑

i=1

kiu(cei).

We deduce that u(cv)(g−1) = 0 if g /∈ S. Thus, using (8.13), we get

u(x)(1G) =∑

g∈S

u(cx(g))(g−1).

This shows that u(x)(1G) only depends on the restriction of x to S. Moreprecisely, there is a K-linear map μ : V S → V such that

u(x)(1G) = μ(x|S) for all x ∈ V [G].

Using the K[G]-linearity of u, we get

u(x)(g) = (g−1u(x))(1G)

= u(g−1x)(1G)

= μ((g−1x)|S)

for all x ∈ V [G] and g ∈ G. Therefore u is the restriction to V [G] of the linearcellular automaton τ : V G → V G with memory set S and local defining mapμ. This shows that Φ is surjective. ��

Remarks 8.7.7. (a) When G is a finite group and V is a (not necessarilyfinite-dimensional) vector space over a field K, one has V [G] = V G andΦ : LCA(G;V ) → EndK[G](V [G]) = EndK[G](V G) is the identity map byRemark 8.7.2.

(b) Suppose now that G is an infinite group and that V is an infinite-dimensional vector space over a field K. Then the map Φ : LCA(G;V ) →EndK[G](V [G]) is not surjective. To see this, take a K-basis (ei)i∈I for V . AsG and I are infinite, we can find a family (Ti)i∈I of finite subsets of G suchthat

⋃i∈I Ti is infinite. Choose, for each i ∈ I, a configuration zi ∈ V [G]

whose support is Ti. It follows from Proposition 8.7.3 that V [G] is a free leftK[G]-module admitting the family (cei)i∈I as a basis. Therefore, there is anelement u ∈ EndK[G](V [G]) such that u(cei) = zi for all i ∈ I. Let us showthat u is not in the image of Φ. We proceed by contradiction. Suppose thatu is in the image of Φ. This means that there is a linear cellular automatonτ : V G → V G such that u = τ |V [G]. If S is a memory set for τ , this impliesthat, for x ∈ V [G], the value of u(x)(1G) depends only on the restriction of xto S. As S is finite and

⋃i∈I T−1

i is infinite, we can find elements g0 ∈ G andi0 ∈ I such that g0 /∈ S and g0 ∈ T−1

i0. Consider the configuration x0 ∈ V [G]

which takes the value ei0 at g0 and is identically 0 on G \ {g0}. Observe thatx0 = g0cei0

, so that

u(x0) = u(g0cei0) = g0u(cei0

) = g0zi0 .

8.8 The Closed Image Property 305

Thus, we haveu(x0)(1G) = (g0zi0)(1G) = zi0(g

−10 ).

We deduce that u(x0)(1G) �= 0 as g−10 is in the support Ti0 of zi0 . This gives

us a contradiction since x0 coincides with the 0 configuration on S. Thisshows that Φ is not surjective.

(c) By combining the two preceding remarks with Theorem 8.7.6, we de-duce that the map Φ : LCA(G; V ) → EndK[G](V [G]) is surjective if and onlyif G is finite or V is finite-dimensional.

Let us recall the following basic facts from linear algebra. Let R be a ringand let d ≥ 1 be an integer. We denote by Matd(R) the ring consisting of alld×d matrices A = (aij)1≤i,j≤d with entries aij ∈ R. Recall that the additionand the multiplication in Matd(R) are given by

A + B = (aij + bij)1≤i,j≤d and AB =( d∑

k=1

aikbkj

)

1≤i,j≤d

for all A = (aij)1≤i,j≤d, B = (bij)1≤i,j≤d ∈ Matd(R). Suppose now that Mis a free left R-module with basis e1, e2, . . . , ed. If u ∈ EndR(M), then wehave u(ei) =

∑dj=1 aijej , where aij ∈ R for 1 ≤ i, j ≤ d. Then the map

u �→ (aij)1≤i,j≤d is a ring isomorphism from EndR(M) onto Matd(Rop),where Rop denotes the opposite ring of R. Moreover, when R is an algebraover a field K, it is an isomorphism of K-algebras.

Let G be a group and let V be a vector space of finite dimension d ≥ 1over a field K. By Proposition 8.7.3, the left K[G]-module V [G] admits a freebasis of cardinality d. As the K-algebras K[G] and (K[G])op are isomorphicby Corollary 8.4.4, we deduce from Theorem 8.7.6 the following:

Corollary 8.7.8. Let G be a group and let V be a vector space over a field K

of finite dimension dimK(V ) = d ≥ 1. Then the K-algebras LCA(G; V ) andMatd(K[G]) are isomorphic. ��

Remark 8.7.9. For d = 1, Corollary 8.7.8 tells us that if G is a group and V isa one-dimensional vector space over a field K, then the K-algebras LCA(G; V )and K[G] are isomorphic. Note that this last result also follows from Corol-lary 8.5.3.

8.8 The Closed Image Property

As we have seen in Lemma 3.3.2, the image of a cellular automaton τ : AG →AG is always closed in AG for the prodiscrete topology if the alphabet A isfinite. When A is infinite, the image of τ may fail to be closed in AG (cf.Example 8.8.3). However, it turns out that if A = V is a finite-dimensional


vector space over a field K and τ : V G → V G is a linear cellular automaton,then the image of τ is closed in V G even when the field K is infinite.

Theorem 8.8.1. Let G be a group and let V be a finite-dimensional vectorspace over a field K. Let τ : V G → V G be a linear cellular automaton. Thenτ(V G) is closed in V G for the prodiscrete topology.

Proof. We split the proof into two steps. Suppose first that G is countable.Then we can find a sequence (An)n∈N of finite subsets of G such that G =⋃

n∈NAn and An ⊂ An+1 for all n ∈ N. Let S be a memory set for τ and let

Bn = A−Sn denote the S-interior of An (cf. Sect. 5.4). Note that G =

⋃n∈N

Bn

and Bn ⊂ Bn+1 for all n ∈ N.It follows from Proposition 5.4.3 that if x and x′ are elements in V G such

that x and x′ coincide on An then the configurations τ(x) and τ(x′) coincideon Bn. Therefore, given xn ∈ V An and denoting by xn ∈ V G a configurationextending xn, the pattern

yn = τ(xn)|Bn ∈ V Bn

does not depend on the particular choice of the extension xn. Thus we candefine a map τn : V An → V Bn by setting τn(xn) = yn for all xn ∈ V An . It isclear that τn is K-linear.

Let y ∈ V G and suppose that y is in the closure of τ(V G). Then, for alln ∈ N there exists zn ∈ V G such that

y|Bn = τ(zn)|Bn . (8.14)

Consider, for each n ∈ N, the affine subspace Ln ⊂ V An defined by Ln =τ−1n (y|Bn). We have Ln �= ∅ for all n by (8.14). For n ≤ m, the restriction

map V Am → V An induces an affine map πn,m : Lm → Ln. Consider, for alln ≤ m, the affine subspace Kn,m ⊂ Ln defined by Kn,m = πn,m(Lm). Wehave Kn,m′ ⊂ Kn,m for all n ≤ m ≤ m′ since πn,m′ = πn,m ◦ πm,m′ . Asthe sequence Kn,m (m = n, n + 1, . . . ) is a decreasing sequence of finite-dimensional affine subspaces, it stabilizes, i.e., for each n ∈ N there exist anon-empty affine subspace Jn ⊂ Ln and an integer m0 = m0(n) ≥ n suchthat Kn,m = Jn if m0 ≤ m. For all n ≤ n′ ≤ m, we have πn,n′(Kn′,m) ⊂ Kn,m

since πn,n′ ◦ πn′,m = πn,m. Therefore, πn,n′ induces by restriction an affinemap ρn,n′ : Jn′ → Jn for all n ≤ n′. We claim that ρn,n′ is surjective. Tosee this, let u ∈ Jn. Let us choose m large enough so that Jn = Kn,m andJn′ = Kn′,m. Then we can find v ∈ Lm such that u = πn,m(v). We haveu = ρn,n′(w), where w = πn′,m(v) ∈ Kn′,m = Jn′ . This proves the claim.Now, using the surjectivity of ρn,n+1 for all n, we construct by inductiona sequence of elements xn ∈ Jn, n ∈ N, as follows. We start by choosingan arbitrary element x0 ∈ J0. Then, assuming xn has been constructed, wetake as xn+1 an arbitrary element in ρ−1

n,n+1(xn). Since xn+1 coincides withxn on An, there exists x ∈ V G such that x|An = xn for all n. We have

8.8 The Closed Image Property 307

τ(x)|Bn = τn(xn) = yn = y|Bn for all n. Since G = ∪n∈NBn, we deduce thatτ(x) = y. This ends the proof in the case when G is a countable group.

We now drop the countability assumption on G and prove the theorem inthe general case. Let S ⊂ G be a memory set for τ and denote by H ⊂ Gthe subgroup of G generated by S. Then H is countable since it is finitelygenerated. Consider the restriction cellular automaton τH : V H → V H andobserve that, by the first step of the proof,

τH(V H) is closed in V H (8.15)

for the prodiscrete topology.With the notation from Sect. 1.7 we have, by virtue of (1.16), that

τ(V G) =∏

c∈G/H

τc(V c). (8.16)

Also, it follows from (1.18) that, for all c ∈ G/H and g ∈ c,

τc(V c) = (φ∗g)

−1(τH(V H)

). (8.17)

As the map φ∗g : V c → V H is a uniform isomorphism and therefore a home-

omorphism, it follows from (8.15) and (8.17) that τc(V c) is closed in V c forall c ∈ G/H. As the product of closed subspaces is closed in the producttopology (cf. Proposition A.4.3), we deduce from (8.16) that τ(V G) is closedin V G. ��Corollary 8.8.2. Let G be a group and let V be a finite-dimensional vectorspace over a field K. Let τ : V G → V G be a linear cellular automaton. Supposethat every configuration with finite support y ∈ V [G] lies in the image of τ .Then τ is surjective.

Proof. By our hypothesis, we have V [G] ⊂ τ(V G) ⊂ V G. As V [G] is densein V G by Proposition 8.2.1 and τ(V G) is closed in V G by Theorem 8.8.1, wededuce that τ(V G) = V G. Thus, τ is surjective. ��

In the example below we show that if we drop the finite dimensionality ofthe alphabet vector space V , then the image of a linear cellular automatonmay fail to be closed in V G.

Example 8.8.3. Let G = Z and let K be a field. Consider the K-algebra V =K[t] consisting of all polynomials in the indeterminate t with coefficients in K.A configuration in V Z is therefore a sequence x = (xn)n∈Z, where xn = xn(t)is a polynomial for all n ∈ Z. Consider the K-linear map τ : V Z → V Z definedby setting τ(x) = y where y = (yn)n∈Z is given by

yn = xn+1 − txn

for all n ∈ Z. Then τ is a linear cellular automaton with memory set {0, 1}and local defining map μ : V {0,1} → V given by μ(x0, x1) = x1 − tx0.


Consider the configuration z = (zn)n∈Z, where zn = 1 for all n ∈ Z.Let Ω be a finite subset of Z and choose an integer M ∈ Z such that

Ω ⊂ [M,∞). Consider the configuration x = (xn)n∈Z defined by

xn =

{1 + t + · · · + tn−M if n ≥ M,

0 if n < M.

Observe that xn+1 = txn+1 for all n ≥ M , so that the configuration y = τ(x)coincides with z on [M,∞) and hence on Ω. Thus z is in the closure of τ(V Z)in V Z. On the other hand, the configuration z is not in the image of τ .Indeed, z = τ(x) for some x ∈ V Z would imply xn+1 = txn + 1 and hencedeg(xn+1) > deg(xn) for all n ∈ Z, which is clearly impossible. This showsthat τ(V Z) is not closed in V Z for the prodiscrete topology.

Observe that every y ∈ V [Z] is in the image of τ . Indeed, if y = (yn)n∈Z ∈V [Z] has support contained in [M,∞) for some M ∈ Z, we can constructx = (xn)n∈Z ∈ V Z with τ(x) = y inductively by setting xn = 0 for all n ≤ Mand xn+1 = txn + yn for all n ≥ M . This shows that we cannot omit thehypothesis saying that V is finite-dimensional in Corollary 8.8.2.

8.9 The Garden of Eden Theorem for Linear CellularAutomata

In this section, we present a linear version of Theorem 5.8.1. We first in-troduce the notion of mean dimension, which plays here the role which wasplayed by the entropy in the finite alphabet case, and we present some of itsbasic properties. In the definition of mean dimension, the dimension of finite-dimensional vector spaces replaces the cardinality of finite sets. The proof ofthe properties of the mean dimension and the proof of the linear version ofthe Garden of Eden Theorem follow the same lines as their finite alphabetcounterparts (see Sects. 5.7 and 5.8).

From now on, in this section, G is an amenable group, F = (Fj)j∈J a rightFølner net for G, and V a finite-dimensional vector space over a field K.

For E ⊂ G, we denote by πE : V G → V E the canonical projection (re-striction map). We thus have πE(x) = x|E for all x ∈ V G. Note that πE isK-linear so that, if X is a vector subspace of V G, then πE(X) is a vectorsubspace of V E .

Definition 8.9.1. Let X be a vector subspace of V G. The mean dimensionmdimF (X) of X with respect to the right Følner net F = (Fj)j∈J is definedby

mdimF (X) = lim supj

dim(πFj (X))|Fj |

, (8.18)

8.9 The Garden of Eden Theorem for Linear Cellular Automata 309

where dim(πFj (X)) denotes the dimension of the K-vector subspace πFj (X) ⊂V Fj .

Remark 8.9.2. In the particular case when K is a finite field, every finite-dimensional vector space W over K is finite of cardinality |W | = |K|dim(W ).Therefore, in this case, we have

mdimF (X) =1

log |K| entF (X)

for every vector subspace X ⊂ V G.

Here are some immediate properties of mean dimension.


(i) mdimF (V G) = dim(V );(ii) mdimF (X) ≤ mdimF (Y ) if X ⊂ Y are vector subspaces of V G;(iii) mdimF (X) ≤ dim V for all vector subspaces X ⊂ V G.

Proof. (i) If X = V G, then, for every j, we have πFj (X) = V Fj and therefore

dim(πFj (X))|Fj |

=dim(V Fj )

|Fj |=

|Fj | dim(V )|Fj |

= dim(V ).

This implies that mdimF (X) = dim(V ).(ii) If X ⊂ Y are vector subspaces of V G, then πFj (X) ⊂ πFj (Y ) are

vector subspaces of V Fj and hence dim(πFj (X)) ≤ dim(πFj (Y )) for all j.This implies mdimF (X) ≤ mdimF (Y ).

(iii) This follows immediately from (i) and (ii). ��

An important property of linear cellular automata is the fact that applyinga linear cellular automaton to a vector subspace of configurations cannotincrease the mean dimension of the subspace. More precisely, we have thefollowing:

Proposition 8.9.4. Let τ : V G → V G be a linear cellular automaton and letX be a vector subspace of V G. Then one has

mdimF (τ(X)) ≤ mdimF (X).

Proof. Let us set Y = τ(X) and observe that Y is a vector subspace of V G,by linearity of τ . Let S ⊂ G be a memory set for τ . After replacing S byS ∪{1G}, we can assume that 1G ∈ S. Let Ω be a finite subset of G. Observefirst that τ induces a map

τΩ : πΩ(X) → πΩ−S (Y )

defined as follows. If u ∈ πΩ(X), then


τΩ(u) = (τ(x))|Ω−S ,

where x is an element of X such that x|Ω = u. Note that the fact that τΩ(u)does not depend on the choice of such an x follows from Proposition 5.4.3.Clearly τΩ is surjective. Indeed, if v ∈ πΩ−S (Y ), then there exists x ∈ Xsuch that (τ(x))|Ω−S = v. Then, setting u = πΩ(x) we have, by construction,τΩ(u) = v. Since τΩ is K-linear and surjective, we have

dim(πΩ−S (Y )) ≤ dim(πΩ(X)). (8.19)

Observe now that Ω−S ⊂ Ω, since 1G ∈ S (cf. Proposition 5.4.2(iv)). ThusπΩ(Y ) is a vector subspace of πΩ−S (Y ) × V Ω\Ω−S

. This implies

dim(πΩ(Y )) ≤ dim(πΩ−S (Y ) × V Ω\Ω−S

)

= dim(πΩ−S (Y )) + dim(V Ω\Ω−S

)

= dim(πΩ−S (Y )) + |Ω \ Ω−S | dim(V )

≤ dim(πΩ(X)) + |Ω \ Ω−S | dim(V ),

by (8.19). As Ω \ Ω−S ⊂ ∂S(Ω), we deduce that

dim(πΩ(Y )) ≤ dim(πΩ(X)) + |∂S(Ω)| dim(V ).

By taking Ω = Fj , this gives us

dim(πFj (Y ))|Fj |

≤dim(πFj (X))

|Fj |+

|∂S(Fj)||Fj |

dim(V ).

Since

limj

|∂S(Fj)||Fj |

= 0

by Proposition 5.4.4, we finally get

mdimF (Y ) = lim supj

dim(πFj (Y ))|Fj |

≤ lim supj

dim(πFj (X))|Fj |

= mdimF (X).

��

By Proposition 8.9.3, the maximal value for the mean dimension of a vectorsubspace X ⊂ V G is dim(V ). The following result gives a sufficient conditionon X which guarantees that its mean dimension is strictly less than dim(V ).

Proposition 8.9.5. Let X be a G-invariant vector subspace of V G. Supposethat there exists a finite subset E ⊂ G such that πE(X) � V E. Then one hasmdimF (X) < dim(V ).

Proof. Let E′ = {g1g−12 : g1, g2 ∈ E}. By Proposition 5.6.3, we may find an

(E, E′)-tiling T ⊂ G.


For each j ∈ J , let us define, as in Proposition 5.6.4, the subset Tj ⊂ Tby Tj = T ∩ F−E

j = {g ∈ T : gE ⊂ Fj} and set

F ∗j = Fj \

∐

g∈Tj

gE.

Since πE(X) � V E and X is G-invariant, we have πgE(X) � V gE for allg ∈ G. We thus have

dim(πgE(X)) ≤ dim(V gE) − 1 = |gE|dim(V ) − 1 for all g ∈ T. (8.20)

AsπFj (X) ⊂ V F∗

j ×∏

g∈Tj

πgE(X),

we get

dim(πFj (X)) ≤ dim(V F∗j ×

∏

g∈Tj

πgE(X))

= |F ∗j | dim(V ) +

∑

g∈Tj

dim(πgE(X))

≤ |F ∗j | dim(V ) +

∑

g∈Tj

(|gE| dim(V ) − 1) (by (8.20))

=(|F ∗

j | +∑

g∈Tj

|gE|)

dim(V ) − |Tj |

= |Fj | dim(V ) − |Tj |,

since|Fj | = |F ∗

j | +∑

g∈Tj

|gE| and |gE| = |E|.

Now, by Proposition 5.6.4, there exist α > 0 and j0 ∈ J such that |Tj | ≥ α|Fj |for all j ≥ j0. Thus

dim(πFj (X))|Fj |

≤ dim(V ) − α for all j ≥ j0.

This implies that

mdimF (X) = lim supj

dim(πFj (X))|Fj |

≤ dim(V ) − α < dim(V ).

��


We are now in position to state the linear version of the Garden of Edentheorem.

Theorem 8.9.6. Let G be an amenable group and let V be a finite-dimen-sional vector space over a field K. Let F = (Fj)j∈J be a right Følner netfor G. Let τ : V G → V G be a linear cellular automaton. Then the followingconditions are equivalent:

(a) τ is surjective;(b) mdimF (τ(V G)) = dim(V );(c) τ is pre-injective.

Since every injective cellular automaton is pre-injective, we immediatelydeduce the following:

Corollary 8.9.7. Let G be an amenable group and let V be a finite-dimen-sional vector space over a field K. Then every injective linear cellular au-tomaton τ : V G → V G is surjective. ��

We divide the proof of Theorem 8.9.6 into several lemmas.

Lemma 8.9.8. Let τ : V G → V G be a linear cellular automaton. Supposethat τ is not surjective. Then mdimF (τ(V G)) < dim(V ).

Proof. Let X = τ(V G) and choose a configuration y ∈ V G \X. Since V G \Xis an open subset of V G for the prodiscrete topology by Theorem 8.8.1, wecan find a finite subset Ω ⊂ G such that πΩ(y) /∈ πΩ(X). Therefore we haveπΩ(X) � V Ω. Observe that X is a G-invariant vector subspace of V G, as τis G-equivariant and linear. By applying Proposition 8.9.5, we deduce thatmdimF (X) < dim(V ). ��

Lemma 8.9.9. Let τ : V G → V G be a linear cellular automaton. Supposethat

mdimF (τ(V G)) < dim(V ). (8.21)

Then τ is not pre-injective.

Proof. Let us set X = τ(V G). Let S be a memory set for τ such that 1G ∈ S.As πF+S

j(X) is a vector subspace of πFj (X) × V F+S

j \Fj , we have

dim(πF+Sj

(X)) ≤ dim(πFj (X)) + |F+Sj \ Fj | dim(V )

≤ dim(πFj (X)) + |∂S(Fj)| dim(V ).

We thus have

dim(πF+Sj

(X))

|Fj |≤

dim(πFj (X))|Fj |

+|∂S(Fj)||Fj |

dim(V ).


It follows from (8.21) and Proposition 5.4.4 that we can find j0 ∈ J suchthat dim(πF+S

j0(X)) < |Fj0 | dim(V ). Let Z denote the (finite-dimensional)

vector subspace of V G consisting of all configurations whose support is con-tained in Fj0 . Observe that τ(x) vanishes outside of F+S

j0for every x ∈ Z by

Proposition 5.4.3. Thus we have

dim(τ(Z)) = dim(πF+Sj0

(τ(Z)))

≤ dim(πF+Sj0

(X)) < |Fj0 | dim(V ) = dim(Z).

This implies that the restriction of τ to Z is not injective. As all configu-rations in Z have finite support, we deduce that τ is not pre-injective (cf.Proposition 8.2.5). ��

Lemma 8.9.10. Let τ : V G → V G be a linear cellular automaton. Supposethat τ is not pre-injective. Then mdimF (τ(V G)) < dim(V ).

Proof. Since the linear cellular automaton τ is not pre-injective, we can find,by Proposition 8.2.5, an element x0 ∈ V G with nonempty finite supportΩ ⊂ G such that τ(x0) = 0. Let S be a memory set for τ such that 1G ∈ S

and S = S−1. Let E = Ω+S2. By Proposition 5.6.3, we can find a finite subset

F ⊂ G such that G contains a (E, F )-tiling T . Note that for each g ∈ G,the support of gx0 is gΩ ⊂ gE. Let us choose, for each g ∈ T , a hyperplaneHg ⊂ V gΩ which does not contain the restriction to gΩ of gx0. Consider thevector subspace X ⊂ V G consisting of all x ∈ V G such that the restrictionof x to gΩ belongs to Hg for each g ∈ T . We claim that τ(V G) = τ(X).Indeed, let z ∈ V G. Then, for each g ∈ T , there exists a scalar kg ∈ K suchthat the restriction to gΩ of z + kg(gx0) belongs to Hg. Let z′ ∈ V G besuch that πgΩ(z′) = πgΩ(z + kg(gx0)) for each g ∈ T and z′ = z outsideof

∐g∈T gΩ. We have z′ ∈ X by construction. On the other hand, since z′

and z coincide outside∐

g∈T gΩ, we have τ(z′) = τ(z) outside∐

g∈T gΩ+S .Now, if h ∈ gΩ+S for some g ∈ T , then hS ⊂ gΩ+S2

= gE and thereforeτ(z′)(h) = τ(z +kg(gx0))(h) = τ(z)(h) since gx0 lies in the kernel of τ . Thusτ(z) = τ(z′) and the claim follows.

We deduce that

mdimF (τ(V G)) = mdimF (τ(X)) ≤ mdimF (X) < dim(V )

where the first inequality follows from Proposition 8.9.4 and the second onefrom Proposition 8.9.5. ��

Proof of Theorem 8.9.6. Condition (a) implies (b) since we havemdimF (V G) = dim(V ) by Proposition 8.9.3(i). The converse implicationfollows from Lemma 8.9.8. On the other hand, condition (c) implies (b) byLemma 8.9.9. Finally, (b) implies (c) by Lemma 8.9.10. ��


We end this section by showing that Corollary 8.9.7 as well as both im-plications (a) ⇒ (c) and (c) ⇒ (a) in Theorem 8.9.6 fail to hold when thevector space V is infinite-dimensional.

Example 8.9.11. Let G be any group and let V be an infinite-dimensionalvector space over a field K. Let us choose a basis subset B for V . Every mapϕ : B → B uniquely extends to a K-linear map ϕ : V → V . The productmap τϕ = ϕG : V G → V G is a linear cellular automaton with memory setS = {1G} and local defining map ϕ. Since B is infinite, we can find a mapϕ1 : B → B which is surjective but not injective and a map ϕ2 : B → B whichis injective but not surjective.

Consider first the cellular automaton τ1 = τϕ1 . Let us show that τ1 issurjective but not pre-injective. Observe that ϕ1 is surjective. As a productof surjective maps is a surjective map, it follows that τ1 is surjective. On theother hand, as ϕ1 is not injective, there exists v ∈ ker(ϕ1)\{0}. Consider theconfiguration x ∈ V G defined by x(1G) = v and x(g) = 0 for all g ∈ G\{1G}.Then x ∈ V [G] ∩ ker(τ1) and x �= 0. This shows that τ1 is not pre-injective.

Consider now the cellular automaton τ2 = τϕ2 . Let us show that τ2 isinjective (and therefore pre-injective) but not surjective. Suppose that x ∈ker(τ2), that is, 0 = τ2(x)(g) = ϕ2(x(g)) for all g ∈ G. As ϕ2 is injective,we deduce that x(g) = 0 for all g ∈ G, in other words, x = 0. This showsthat τ2 is injective. On the other hand, as ϕ2 is not surjective, there existsv ∈ V \ ϕ2(V ). Consider the constant configuration x ∈ V G where x(g) = vfor all g ∈ G. It is clear that x ∈ V G \ τ2(V G). This shows that τ2 is notsurjective.

8.10 Pre-injective but not Surjective Linear CellularAutomata

In this section, we give examples of linear cellular automata with finite-dimensional alphabet which are pre-injective but not surjective. We recallthat the underlying groups for such automata cannot be amenable by Theo-rem 8.9.6.

Proposition 8.10.1. Let G = F2 be the free group of rank two and let Vbe a two-dimensional vector space over a field K. Then there exists a linearcellular automaton τ : V G → V G which is pre-injective but not surjective.

Proof. We may assume V = K2. Let a and b denote the canonical generators

of G = F2. Let p1 and p2 be the elements of EndK(V ) defined respectivelyby p1(v) = (k1, 0) and p2(v) = (k2, 0) for all v = (k1, k2) ∈ V . Consider thelinear cellular automaton τ : V G → V G given by

τ(x)(g) = p1(x(ga)) + p2(x(gb)) + p1(x(ga−1)) + p2(x(gb−1))

8.11 Surjective but not Pre-injective Linear Cellular Automata 315

for all x ∈ V G and g ∈ G. This cellular automaton is the one describedin Sect. 5.11 for H = K. By Proposition 5.11.1, τ is pre-injective but notsurjective. ��

If H is a subgroup of a group G and τ : V H → V H is a linear cellularautomaton over H which is pre-injective and not surjective, then the inducedcellular automaton τG : V G → V G is also linear, pre-injective and not surjec-tive (see Proposition 1.7.4 and Proposition 5.2.2). Therefore, an immediateconsequence of Proposition 8.10.1 is the following:

Corollary 8.10.2. Let G be a group containing a free subgroup of rank twoand let V be a two-dimensional vector space over a field K. Then there existsa linear cellular automaton τ : V G → V G which is pre-injective but not sur-jective. ��

This shows that implication (c) ⇒ (a) in Theorem 8.9.6 becomes false ifG is a group containing a free subgroup of rank two.

8.11 Surjective but not Pre-injective Linear CellularAutomata

We now present examples of linear cellular automata with finite-dimensionalalphabet which are surjective but not pre-injective. We recall that, by The-orem 8.9.6, the underlying groups for such examples are necessarily non-amenable.

Proposition 8.11.1. Let G = F2 be the free group of rank two and let Vbe a two-dimensional vector space over a field K. Then there exists a linearcellular automaton τ : V G → V G which is surjective but not pre-injective.

Proof. Let a, b denote the canonical generators of G. We can assume thatV = K

2. Consider the elements q1, q2 ∈ EndK(V ) respectively defined byq1(v) = (k1, 0) and q2(v) = (0, k1) for all v = (k1, k2) ∈ V .

Let τ : V G → V G be the linear cellular automaton given by

τ(x)(g) = q1(x(ga)) + q1(x(ga−1)) + q2(x(gb)) + q2(x(gb−1))

for all x ∈ V G, g ∈ G. A memory set for τ is the set S = {a, b, a−1, b−1}.Let k0 be any nonzero element in K. Consider the configuration x0 ∈ V G

defined by

x0(g) =

{(0, k0) if g = 1G

(0, 0) otherwise.

Then, x0 is almost equal to 0 but x0 �= 0. As τ(x0) = 0, we deduce that τ isnot pre-injective.


However, τ is surjective. To see this, let z = (z1, z2) ∈ KG × K

G = V G.Let us show that there exists x ∈ V G such that τ(x) = z. We define x(g)by induction on the graph distance (cf. Sect. 6.2), which we denote by |g|, ofg ∈ G from 1G in the Cayley graph of G. We first set x(1G) = (0, 0).

Then, for s ∈ S we set

x(s) =

⎧⎨

⎩

(z1(1G), 0) if s = a(z2(1G), 0) if s = b(0, 0) otherwise.

(8.22)

Suppose that x(g) has been defined for all g ∈ G with |g| ≤ n, for somen ≥ 1. For g ∈ G with |g| = n, let g′ ∈ G and s′ ∈ S be the unique elementssuch that |g′| = n − 1 and g = g′s′. Then, for s ∈ S with s′s �= 1G, we set

x(gs) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

(z1(g) − x1(g′), 0) if s′ ∈ {a, a−1} and s = s′

(z2(g), 0) if s′ ∈ {a, a−1} and s = b(z1(g), 0) if s′ ∈ {b, b−1} and s = a(z2(g) − x2(g′), 0) if s′ ∈ {b, b−1} and s = s′

(0, 0) otherwise.

(8.23)

Let us check that τ(x) = z. We have

τ(x)(1G) = q1(x(a)) + q1(x(a−1)) + q2(x(b)) + q2(x(b−1))(by (8.22)) = q1(z1(1G), 0) + q1(0, 0) + q2(z2(1G), 0) + q2(0, 0)

= (z1(1G), 0) + (0, 0) + (0, z2(1G)) + (0, 0)= (z1(1G), z2(1G))= z(1G).

Let now g = g′s′ ∈ G with |g| = |g′| + 1 > 2. Suppose, for instance, thats′ = a. We then have:

τ(x)(g) = τ(x)(g′a)

(by (8.23)) = q1(x(g′a2)) + q1(x(g′)) + q2(x(g′ab)) + q2(x(g′ab−1))= q1(z1(g) − x1(g′), 0) + q1(x1(g′), x2(g′)) + q2(z2(g), 0) + q2(0, 0)= (z1(g) − x1(g′), 0) + (x1(g′), 0) + (0, z2(g)) + (0, 0)= (z1(g), z2(g))= z(g).

The cases when s′ = a−1, b, b−1 are similar. It follows that τ(x) = z. Thisshows that τ is surjective. ��

If H is a subgroup of a group G, and τ : V H → V H is a linear cellularautomaton over H which is surjective and not pre-injective, then the in-duced cellular automaton τG : V G → V G is also linear, surjective and not

8.12 Invertible Linear Cellular Automata 317

pre-injective (see Proposition 1.7.4 and Proposition 5.2.2). Therefore, an im-mediate consequence of Proposition 8.11.1 is the following:

Corollary 8.11.2. Let G be a group containing a free subgroup of rank twoand let V be a two-dimensional vector space over a field K. Then there existsa linear cellular automaton τ : V G → V G which is surjective but not pre-injective. ��

As a consequence, we deduce that implication (a) ⇒ (c) in Theorem 8.9.6becomes false if the group G contains a free subgroup of rank two.

8.12 Invertible Linear Cellular Automata

Theorem 8.12.1. Let G be a group and let V be a finite-dimensional vectorspace over a field K. Let τ : V G → V G be an injective linear cellular automa-ton. Then there exists a linear cellular automaton σ : V G → V G such thatσ ◦ τ = IdV G .

Proof. We split the proof into two steps. Suppose first that G is countable.Since τ is injective, it induces a bijective map τ : V G → Y where Y = τ(V G)is the image of τ . The fact that τ is K-linear and G-equivariant impliesthat Y is a G-invariant vector subspace of V G and that the inverse mapτ−1 : Y → V G is K-linear and G-equivariant. Let us show that the followinglocal property is satisfied by τ−1: there exists a finite subset T ⊂ G such that

(∗) for y ∈ Y , the element τ−1(y)(1G) only depends on the restrictionof y to T .

Let us assume by contradiction that there exists no such T .Let S be a memory set for τ such that 1G ∈ S. Since G is countable, we

can find a sequence (An)n∈N of finite subsets of G such that G =⋃

n∈NAn,

S ⊂ A0 and An ⊂ An+1 for all n ∈ N. Let Bn = A−Sn denote the S-interior

of An (cf. Sect. 5.4). Note that G =⋃

n∈NBn and Bn ⊂ Bn+1 for all n ∈ N.

Since there exists no finite subset T ⊂ G satisfying condition (∗), we canfind, for each n ∈ N, two configurations y′

n, y′′n ∈ Y such that y′

n|Bn = y′′n|Bn

and τ−1(y′n)(1G) �= τ−1(y′′

n)(1G). By linearity of τ−1, the configuration yn =y′

n − y′′n ∈ Y satisfies yn|Bn = 0 and τ−1(yn)(1G) �= 0.

It follows from Proposition 5.4.3 that if x and x′ are elements in V G suchthat x and x′ coincide on An then the configurations τ(x) and τ(x′) coincideon Bn. Therefore, given xn ∈ V An and denoting by xn ∈ V G a configurationextending xn, the pattern

un = τ(xn)|Bn ∈ V Bn

does not depend on the particular choice of the extension xn of xn. Thus wecan define a map τn : V An → V Bn by setting τn(xn) = un. It is clear that τn

is K-linear.


Consider, for each n ∈ N, the vector subspace Ln ⊂ V An defined byLn = Ker(τn). where for n ≤ m, the restriction map V Am → V An induces aK-linear map πn,m : Lm → Ln. Indeed, if u ∈ Lm, then we have u|An ∈ Ln

since τm(u) = 0 and therefore τn(u|An) = (τm(u))|Bn = 0. Consider now, forall n ≤ m, the vector subspace Kn,m ⊂ Ln defined by Kn,m = πn,m(Lm). Wehave Kn,m′ ⊂ Kn,m for all n ≤ m ≤ m′ since πn,m′ = πn,m◦πm,m′ . Therefore,if we fix n, the sequence Kn,m, where m = n, n+1, . . ., is a decreasing sequenceof vector subspaces of Ln. As Ln ⊂ V An is finite-dimensional, this sequencestabilizes, i.e., there exist a vector subspace Jn ⊂ Ln and an integer kn ≥ nsuch that Kn,m = Jn for all m ≥ kn.

For all n ≤ n′ ≤ m, we have πn,n′(Kn′,m) ⊂ Kn,m since πn,n′ ◦ πn′,m =πn,m. Therefore, πn,n′ induces by restriction a linear map ρn,n′ : Jn′ → Jn

for all n ≤ n′. We claim that ρn,n′ is surjective. To see this, let u ∈ Jn.Let us choose m large enough so that Jn = Kn,m and Jn′ = Kn′,m. Thenwe can find v ∈ Lm such that u = πn,m(v). We have u = ρn,n′(w), wherew = πn′,m(v) ∈ Kn′,m = Jn′ . This proves the claim.

Now, using the surjectivity of ρn,n+1 for all n, we construct by inductiona sequence of elements zn ∈ Jn, n ∈ N, as follows. We start by taking asz0 the restriction of xk0 to A0. Observe that xk0 ∈ Lk0 and hence z0 =π0,k0(xk0) ∈ J0. Then, assuming that zn has been constructed, we take aszn+1 an arbitrary element in ρ−1

n,n+1(zn). Since zn+1 coincides with zn onAn, there exists a unique element z ∈ V G such that z|An = zn for all n.We have z ∈ Y , since zn ∈ πAn(Y ) for all n and X is closed in V G. Asz(1G) = z0(1G) = xk0(1G) �= 0, we have z �= 0. On the other hand, τ(z) =0 since τ(z)|Bn = τn(zn) = 0 for all n by construction. This contradictsthe injectivity of τ . Thus, there exists a finite subset T ⊂ G satisfying (∗).Consider the linear map ν : V T → V defined by ν(z) = τ−1(z)(1G), wherez ∈ V G is any configuration extending the pattern z ∈ V T . Then the linearcellular automaton σ : V G → V G with memory set T and local definingmap ν clearly satisfies σ ◦ τ = IdV G . Indeed, if x ∈ V G, then denoting byy = τ(x) ∈ Y its image by τ , we have x = τ−1(y) and

(σ ◦ τ)(x)(1G) = σ(y)(1G)= ν(y|T )

= τ−1(y)(1G)= x(1G)

showing that (σ ◦ τ)(x) = x, by G-equivariance of σ ◦ τ . It follows thatσ ◦ τ = IdV G and this proves the statement when G is a countable group.

We now drop the countability assumption on G and prove the theorem inthe general case. Let S ⊂ G be a memory set for τ and denote by μ : V S → Vthe corresponding local defining map. Let H ⊂ G be the subgroup generatedby S. Note that H is countable. Consider the set G/H of all left cosets of Hin G. For c ∈ G/H denote by

8.12 Invertible Linear Cellular Automata 319

πc : V G → V c =∏

g∈c

V

the projection map. We have V G =∏

c∈G/H V c and, for every x ∈ V G wewrite x = (xc)c∈G/H , where xc = πc(x) ∈ V c.

For c ∈ G/H and g ∈ c denote by φg : H → c the linear map definedby φg(h) = gh for all h ∈ H. Consider the map φ∗

g : V c → V H defined byφ∗

g(z) = z ◦ φg. Then, if x = (xc)c∈G/H ∈ V G and c ∈ G/H, we have

(φg ∗ (xc))(h) = xc(gh) = x(gh) = (g−1x)(h) = (g−1x)H(h)

for all h ∈ H, that is,φ∗

g(xc) = (g−1x)H . (8.24)

For c ∈ G/H, define the map τc : V c → V c by setting

τc(z)(g) = μ((φ∗g(z))|S)

for all z ∈ V c and g ∈ c. Note that τH : V H → V H is the restriction of τ tothe subgroup H. We then have

τ =∏

c∈G/H

τc. (8.25)

From (8.25) we immediately deduce that Y =∏

c∈G/H Yc, where Yc =πc(Y ) = τc(V c), and that τc is injective for all c ∈ G/H.

By the first part of the present proof, there exists a linear cellular automa-ton σH : V H → V H such that

σH ◦ τH = IdV H . (8.26)

Let T ⊂ H be a memory set for σH and let ν : V T → V be the correspondinglocal defining map. Consider the linear cellular automaton σ : V G → V G

defined by settingσ(y)(g) = ν((g−1y)|T )

for all y ∈ V G and g ∈ G. Note that σ = (σH)G is the induced cellularautomaton of σH from H to G, so that, if σc : V c → V c is the map definedby setting σc(z)(g) = ν((φ∗

g(z))|T ) for all z ∈ V c and g ∈ c then

σ =∏

c∈G/H

σc. (8.27)

Given c ∈ G/H, z ∈ V c and g ∈ c, we have


(σc ◦ τc)(z)(g) = σc(τc(z))(g)= [φ∗

gσc(τc(z))](1G)

(by (8.24)) = σH(φ∗g(τc(z))(1G)

(again by (8.24)) = σH(τH(φ∗g(z)))(1G)

(by (8.26)) = (φ∗g(z))(1G)

= z(g).

This shows that (σc ◦τc)(z) = z for all z ∈ V c. We deduce that σc ◦τc = IdV c

for all c ∈ C. It follows from (8.25) and (8.27) that σ ◦ τ = IdV G . ��

We recall that given a group G and a set A, a cellular automaton τ : AG →AG is said to be invertible if τ is bijective and the inverse map τ−1 : AG → AG

is also a cellular automaton. When A is a finite set, every bijective cellularautomaton τ : AG → AG is invertible by Theorem 1.10.2. The following is alinear analogue.

Corollary 8.12.2. Let G be a group and let V be a finite-dimensional vectorspace over a field K. Then every bijective linear cellular automaton τ : V G →V G is invertible.

Proof. Let τ : V G → V G be a bijective linear cellular automaton. It followsfrom Theorem 8.12.1 that there exists a linear cellular automaton σ : V G →V G such that σ◦τ = IdV G . This implies that τ−1 = σ is a cellular automaton.Thus τ is an invertible cellular automaton. ��

Remark 8.12.3. Let G = Z and let K be a field. Consider the infinite di-mensional K-vector space V = K[[t]] consisting of all formal power series inone indeterminate t with coefficients in K. Thus, an element of V is just asequence v = (ki)i∈N of elements of K written in the form

v = k0 + k1t + k2t2 + k3t

3 + · · · =∑

i∈N

kiti.

Consider the map τ : V Z → V Z defined by

τ(x)(n) = x(n) − tx(n + 1)

for all x ∈ V Z, n ∈ Z. In Example 1.10.3 we have showed that τ is a bijec-tive cellular automaton whose inverse map τ−1 : V Z → V Z is not a cellularautomaton. It is clear that τ is K-linear. This shows that Corollary 8.12.2becomes false if we omit the finite-dimensionality of the alphabet V .

Let G be a group and let V be a vector space over a field K. We haveseen in Sect. 1.10 that the set ICA(G;V ) consisting of all invertible cellularautomata τ : V G → V G is a group for the composition of maps. The subset ofICA(G;V ) consisting of all invertible linear cellular automata is a subgroup

8.13 Pre-injectivity and Surjectivity of the Discrete Laplacian 321

of ICA(G;V ) since it is the intersection of ICA(G; V ) with the automorphismgroup of the K-vector space V G.

Given a ring R and an integer d ≥ 1, we denote by GLd(R) the group ofinvertible elements of the matrix ring Matd(R).

An immediate consequence of Corollary 8.12.2 and Corollary 8.7.8 is thefollowing

Corollary 8.12.4. Let G be a group and let V be a vector space over a field K

of finite dimension dimK(V ) = d ≥ 1. Then the set consisting of all bijectivelinear cellular automata τ : V G → V G is a subgroup of ICA(G;V ) isomorphicto GLd(K[G]). ��

8.13 Pre-injectivity and Surjectivity of the DiscreteLaplacian

Let G be a group and let K be a field. Given a nonempty finite subset S ofG, we recall that the discrete Laplacian associated with G and S is the linearcellular automaton ΔS : K

G → KG (with memory set S ∪ {1G}) defined by

ΔS(f)(g) = |S|f(g) −∑

s∈S

f(gs)

for all f ∈ KG and g ∈ G, where |S| denotes the cardinality of S (cf. Exam-

ple 1.4.3(b) and Example 8.1.3(a)).Let us observe that ΔS is never injective since all constant maps f : G → K

are in the kernel of ΔS .

Proposition 8.13.1. Let G be a group and let K be a field. Let S ⊂ G be anon-empty finite subset and suppose that the subgroup H ⊂ G generated byS is finite. Then ΔS is neither pre-injective nor surjective.

Proof. Let (ΔS)H : KH → K

H denote the restriction cellular automaton ofΔS to H. Observe that (ΔS)H is the discrete Laplacian on K

H associatedwith H and S. As we have seen above, (ΔS)H is not injective. As H is finite, itfollows that (ΔS)H is not pre-injective. On the other hand, the non-injectivityof (ΔS)H implies its non-surjectivity since (ΔS)H is K-linear and K

H isfinite-dimensional. By applying Proposition 5.2.2 (resp. Proposition 1.7.4),we deduce that ΔS is not pre-injective (resp. not surjective). ��

In the remaining of this section, we assume that the field K is the field R

of real numbers. The following result is a Garden of Eden type theorem forthe discrete laplacian. Note that there is no amenability hypothesis for theunderlying group.


Theorem 8.13.2. Let G be a group and let S be a nonempty finite subset ofG. Let ΔS : R

G → RG denote the associated discrete real laplacian. Then the

following conditions are equivalent:

(a) ΔS is surjective;(b) the subgroup of G generated by S is infinite;(c) ΔS is pre-injective.

Let us first establish the following simple fact. Recall that given a groupG, the subsemigroup generated by a subset S ⊂ G is the smallest subset Pof G containing S which is closed under the group operation, i.e., such thatp1p2 ∈ P for all p1, p2 ∈ P . Clearly, P consists of all elements of the formp = s1s2 · · · sn where n ≥ 1 and si ∈ S for 1 ≤ i ≤ n.

Lemma 8.13.3. Let G be a group and let S be a subset of G. Then thefollowing conditions are equivalent:

(a) the subsemigroup of G generated by S is infinite;(b) the subgroup of G generated by S is infinite.

Proof. Let P (resp. H) denote the subsemigroup (resp. subgroup) of G gen-erated by S. The implication (a) ⇒ (b) is obvious since P ⊂ H.

Suppose that S is nonempty and that P is finite. Then, for each s ∈ S,the set {sn : n ≥ 1} ⊂ P is also finite. This implies that every element ofS has finite order. Therefore, for each s ∈ S, we can find an integer n ≥ 2such that sn = 1G. It follows that s−1 = sn−1 ∈ S. Consequently, we haveP−1 = P and 1G ∈ P . This implies H = P , so that H is finite. This shows(b) ⇒ (a). ��

Proof of Theorem 8.13.2. The implication (a) ⇒ (b) immediately followsfrom Proposition 8.13.1.

Suppose that the subgroup generated by S is infinite. Let f be an elementof R[G] such that ΔS(f) = 0. Let us show that f = 0. Let g0 ∈ G such that|f(g0)| = maxg∈G |f(g)|. Note that such a g0 exists since f takes only finitelymany values. As

0 = ΔS(f)(g0) = |S|f(g0) −∑

s∈S

f(g0s),

we get

|f(g0)| ≤1|S|

∑

s∈S

|f(g0s)| (8.28)

by applying the triangle inequality. We deduce from (8.28) and the definitionof g0 that |f(g0s)| = |f(g0)| for all s ∈ S. By iterating the previous argument,we get |f(g0p)| = |f(g0)| for all p ∈ P , where P denotes the subsemigroupof G generated by S. As P is infinite (cf. Lemma 8.13.3), this implies that|f(h)| = |f(g0)| = maxg∈G |f(g)| for infinitely many h ∈ G. Since f has finite

8.13 Pre-injectivity and Surjectivity of the Discrete Laplacian 323

support, it follows that f = 0. Therefore, ΔS is pre-injective. This proves theimplication (b) ⇒ (c).

To complete the proof, it suffices to prove that (c) implies (a).Suppose first that G is an amenable group. It follows from the implication

(c) ⇒ (a) in Theorem 8.9.6 that if ΔS is pre-injective, then it is surjective.This shows the implication (c) ⇒ (a) for G amenable.

Suppose now that G is non-amenable (and hence infinite) and that Sis a generating subset for G. This implies that G is countable. Consider theHilbert space �2(G) ⊂ R

G consisting of all square-summable real functions onG and the continuous linear map Δ

(2)S : �2(G) → �2(G) obtained by restriction

of ΔS to �2(G) (cf. Sect. 6.12).By the Kesten-Day amenability criterion (Theorem 6.12.9), the nona-

menability of G implies that 0 does not belong to the spectrum σ(Δ(2)S )

of Δ(2)S . Thus, Δ

(2)S is bijective and hence

R[G] ⊂ �2(G) = Δ(2)S (�2(G)) = ΔS(�2(G)) ⊂ ΔS(RG).

By applying Corollary 8.8.2, we deduce that ΔS(RG) = RG. This shows that

ΔS is surjective in the case when G is non-amenable and S generates G.Finally, suppose now that G is an arbitrary non-amenable group and that

ΔS is pre-injective. Denote by H the subgroup of G generated by S. Thenthe restriction cellular automaton (ΔS)H : R

H → RH , which is the discrete

laplacian associated with H and S, is pre-injective by Proposition 5.2.2. Thesubgroup H may be amenable or not. However, it follows from the two preced-ing cases that (ΔS)H is surjective. By applying Proposition 1.7.4, we deducethat ΔS is surjective as well. This completes the proof that (a) implies (c).

��

As a consequence of Theorem 8.13.2, we obtain the following characteri-zation of locally finite groups in terms of real linear cellular automata:

Corollary 8.13.4. Let G be a group and let V be a real vector space of finitedimension d ≥ 1. Then the following conditions are equivalent:

(a) G is locally finite;(b) every surjective linear cellular automaton τ : V G → V G is injective.

Proof. Suppose (a). Let τ : V G → V G be a surjective linear cellular automa-ton with memory set S ⊂ G. As G is locally finite, the subgroup H generatedby S is finite. Consider the linear cellular automaton τH : V H → V H obtainedfrom τ by restriction. Observe that τH is surjective by Proposition 1.7.4(ii).Since V H is finite-dimensional, it follows that τH is injective. By applyingProposition 1.7.4(i), we deduce that τ is also injective. This shows that (a)implies (b).

Now, suppose that G is not locally finite. Let us show that there exists alinear cellular automaton τ : V G → V G which is surjective but not injective.


We can assume that V = Rd. Since G is not locally finite, we can find a

finite subset S ⊂ G such that the subgroup H generated by S is infinite.By Theorem 8.13.2, the discrete Laplacian ΔS : R

G → RG is surjective. As

mentioned above ΔS is not injective since all constant configurations arein its kernel. Consider now the product map τ = (ΔS)d : (Rd)G → (Rd)G,where we use the natural identification (Rd)G = (RG)d. Clearly, τ is a linearcellular automaton admitting S ∪{1G} as a memory set. On the other hand,τ is surjective but not injective since any product of surjective (resp. non-injective) maps is a surjective (resp. non-injective) map. This shows that (b)implies (a). ��

8.14 Linear Surjunctivity

In analogy with the finite alphabet case, we introduce the following definition.

Definition 8.14.1. A group G is said to be L-surjunctive if, for any fieldK and any finite-dimensional vector space V over K, every injective linearcellular automaton τ : V G → V G is surjective.

Proposition 8.14.2. Every subgroup of an L-surjunctive group is L-surjunc-tive.

Proof. Suppose that H is a subgroup of a L-surjunctive group G. Let V bea finite-dimensional vector space over a field K and let τ : V H → V H be aninjective linear cellular automaton over H. Consider the cellular automatonτG : V G → V G over G obtained from τ by induction (see Sect. 1.7). The factthat τ is injective implies that τG is injective by Proposition 1.7.4(i). Also,τG is linear by Proposition 8.3.1. Since G is L-surjunctive, it follows that τG

is surjective. By applying Proposition 1.7.4(ii), we deduce that τ is surjective.This shows that H is L-surjunctive. ��


(a) G is L-surjunctive;(b) every finitely generated subgroup of G is L-surjunctive.

Proof. The fact that (a) implies (b) follows from Proposition 8.14.2. Con-versely, let G be a group all of whose finitely generated subgroups are L-surjunctive. Let V be a finite dimensional vector space over a field K andlet τ : V G → V G be an injective linear cellular automaton with memory setS. Let H denote the subgroup of G generated by S and consider the linearcellular automaton τH : V H → V H obtained by restriction of τ (see Sect. 1.7and Proposition 8.3.1). The fact that τ is injective implies that τH is injec-tive by Proposition 1.7.4(i). As H is finitely generated, it is L-surjunctive by

8.14 Linear Surjunctivity 325

our hypothesis on G. It follows that τH is surjective. By applying Proposi-tion 1.7.4(ii), we deduce that τ is also surjective. This shows that (b) im-plies (a). ��

Note that the preceding proposition may be stated by saying that a groupis L-surjunctive if and only if it is locally L-surjunctive.

Every finite group is obviously L-surjunctive. Indeed, if G is a finite groupand V is a finite-dimensional vector space, then the vector space V G is alsofinite-dimensional and therefore every injective endomorphism of V G is sur-jective. More generally, it follows from Corollary 8.9.7 that every amenablegroup is L-surjunctive. In fact, we have the following result, which is a linearanalogue of Theorem 7.8.1,

Theorem 8.14.4. Every sofic group is L-surjunctive.

Proof. Let G be a sofic group. Let V be a finite-dimensional vector spaceover a field K of dimension dimK(V ) = d ≥ 1 and let τ : V G → V G be aninjective linear cellular automaton. We want to show that τ is surjective.Every subgroup of a sofic group is sofic by Proposition 7.5.4. On the otherhand, it follows from Proposition 3.2.2 that a group is L-surjunctive if all itsfinitely generated subgroups are L-surjunctive. Thus we can assume that Gis finitely generated.

Let then S ⊂ G be a finite symmetric generating subset of G. As usual,for r ∈ N, we denote by BS(r) ⊂ G the ball of radius r centered at 1G inthe Cayley graph associated with (G, S). We set Y = τ(V G). Observe thatY is a G-invariant vector subspace of V G. On the other hand, it followsfrom Theorem 8.8.1 that Y is closed in V G with respect to the prodiscretetopology.

By Theorem 8.12.1, there exists a linear cellular automaton σ : V G → V G

such thatσ ◦ τ = IdV G .

Choose r0 large enough so that the ball BS(r0) is a memory set for bothτ and σ. Let μ : V BS(r0) → V and ν : V BS(r0) → V denote the correspondinglocal defining maps for τ and σ respectively.

We proceed by contradiction. Suppose that τ is not surjective, that is,Y � V G. Then, since Y is closed in V G, there exists a finite subset Ω ⊂ Gsuch that Y |Ω � V Ω. It is not restrictive, up to taking a larger r0, again ifnecessary, to suppose that Ω ⊂ BS(r0). Thus, Y |BS(r0) � V BS(r0).

Let ε > 0 be such that

ε <1

d|B(2r0)| + 1. (8.29)

Note that from (8.29) we have 1 − ε > 1 − 1d|B(2r0)|+1 which yields

(1 − ε)−1 < 1 +1

d|B(2r0)|. (8.30)


Since G is sofic, we can find a finite S-labeled graph (Q, E, λ) such that

|Q(3r0)| ≥ (1 − ε)|Q|, (8.31)

where we recall that Q(r), r ∈ N, denotes the set of all q ∈ Q such thatthere exists an S-labeled graph isomorphism ψq,r : BS(r) → B(q, r) satisfyingψq,r(1G) = q (cf. Theorem 7.7.1).

Note the inclusions

Q(r0) ⊃ Q(2r0) ⊃ · · · ⊃ Q(ir0) ⊃ Q((i + 1)r0) ⊃ · · · .

(cf. (7.41); see also Fig. 7.2). Also recall from Lemma 7.7.2 that B(q, r0) ⊂Q(ir0) for all Q((i + 1)r0) and i ≥ 0.

For each integer i ≥ 1, we define the map μi : V Q(ir0) → V Q((i+1)r0) bysetting, for all u ∈ V Q(ir0) and q ∈ Q((i + 1)r0),

μi(u)(q) = μ(u|B(q,r0) ◦ ψq,r0

)(1G),

where ψq,2r0 is the unique isomorphism of S-labeled graphs from BS(r0) ⊂ Gto B(q, r0) ⊂ Q sending 1G to q (cf. 7.39 and 7.40).

Similarly, we define the map νi : V Q(ir0) → V Q((i+1)r0) by setting, for allu ∈ V Q(ir0) and q ∈ Q((i + 1)r0),

νi(u)(q) = ν(u|B(q,r0) ◦ ψq,r0

)(1G).

From the fact that τ−1 ◦ τ is the identity map on V G, we deduce that thecomposite νi+1◦μi : V Q(ir0) → V Q((i+2)r0) is the identity on V Q((i+2)r0). Moreprecisely, denoting by ρi : V Q(ir0) → V Q((i+2)r0) the restriction map, we havethat νi+1 ◦ μi = ρi for all i ≥ 1. In particular, we have ν2 ◦ μ1 = ρ1. Thus,setting Z = μ1(V Q(r0)) ⊂ V Q(2r0), we deduce that ν2(Z) = ρ1(V Q(r0)) =V Q(3r0). It follows that

dim(Z) ≥ d|Q(3r0)|. (8.32)

Let Q′ ⊂ Q(3r0) be as in (7.59) and set Q′ =∐

q′∈Q′ B(q′, r0). Note thatQ′ ⊆ Q(2r0) so that

|Q(2r0)| = |Q′| · |BS(r0)| + |Q(2r0) \ Q′|. (8.33)

Now observe that, for all q ∈ Q(2r0), we have a natural isomorphism ofvector spaces Z|B(q,r0) → Y |BS(r0) given by u �→ u◦ψq,r0 , where ψq,r0 denotesas above the unique isomorphism of S-labeled graphs from BS(r0) to B(q, r0)such that ψq,r0(1G) = q. Since Y |BS(r0) � V BS(r0), this implies that

dim(Z|B(q,r0)

)= dim(Y |BS(r0)) ≤ d · |BS(r0)| − 1, (8.34)

for all q ∈ Q′.Thus we have

8.15 Stable Finiteness of Group Algebras 327

dim(Z) ≤ dim(Z|Q′

)+ dim

(Z|Q(2r0)\Q′

)

≤ |Q′| · (d · |BS(r0)| − 1) + d · |Q(2r0) \ Q′|

= d(|Q(2r0)| −

|Q′|d

)

where the last equality follows from (8.33). Comparing this with (8.32) weobtain

|Q(3r0)| ≤ |Q(2r0)| −|Q′|d

.

Thus,

|Q| ≥ |Q(2r0)| ≥ |Q(3r0)| +|Q′|d

≥ |Q(3r0)| +|Q(3r0)|d|B(2r0)|

by (7.59),

= |Q(3r0)|(1 +

1d|B(2r0)|

)

> |Q(3r0)|(1 − ε)−1

where the last inequality follows from (8.30). This yields

|Q(3r0)| < (1 − ε)|Q|

which contradicts (8.31). This shows that τ(V G) = Y = V G, that is, τ issurjective. It follows that the group G is L-surjunctive. ��

8.15 Stable Finiteness of Group Algebras

Let R be a ring and denote by 1R its unity element.If a, b ∈ R satisfy ab = 1R, then one says that b is a right-inverse of a and

that a is a left-inverse of b. An element a ∈ R is said to be right-invertible(resp. left-invertible) if it admits a right-inverse (resp. a left-inverse). Everyinvertible element in R is both right-invertible and left-invertible. Conversely,suppose that an element a ∈ R admits a right-inverse b′ and a left-inverse b′′.Then, we have ab′ = 1R and b′′a = 1R, so that b′ = (b′′a)b′ = b′′(ab′) = b′′.This shows that a is invertible with inverse a−1 = b′ = b′′. Thus, if an elementis both right-invertible and left-invertible, then it is invertible.

One says that the ring R is directly finite if every right-invertible element(or, equivalently, every left-invertible element) is invertible. This is equivalentto saying that if any two elements a, b ∈ R satisfy ab = 1R, then they alsosatisfy ba = 1R.


The ring R is said to be stably finite if the matrix ring Matd(R) is directlyfinite for any d ≥ 1. Observe that every stably finite ring R is directly finitesince Mat1(R) = R.

Proposition 8.15.1. Every finite ring is stably finite.

Proof. Let R be a finite ring. Suppose that a, b ∈ R satisfy ab = 1R. Considerthe map f : R → R defined by f(r) = ar. We have f(br) = a(br) = (ab)r =1Rr = r for all r ∈ R. Therefore, the map f is surjective. As R is finite, thisimplies that f is also injective. Since f(ba) = a(ba) = (ab)a = a = f(1R), wededuce that ba = 1R. This shows that every finite ring is directly finite.

If the ring R is finite, then the ring Matd(R) is also finite for every d ≥ 1.Consequently, every finite ring is stably finite. ��

Proposition 8.15.2. Every commutative ring is stably finite.

Proof. Let d ≥ 1 and suppose that a ∈ Matd(R) is right-invertible. Thenthere exists b ∈ Matd(R) such that ab = Id. This implies that

det(a) det(b) = det(ab) = det(Id) = 1R.

It follows that det(a) is an invertible element in R. Therefore a is invertiblein Matd(R). This shows that Matd(R) is directly finite for any d ≥ 1. Conse-quently, R is stably finite. ��

Let us give an example of a ring which is not directly finite.

Example 8.15.3. Let R be a nonzero ring and consider the free left R-moduleM = ⊕n∈NR. Every element in M can be represented in the form m =(mn)n∈N where mn ∈ R for all n ∈ N and mn = 0R for all but finitely manyn ∈ N. Consider the maps a : M → M and b : M → M defined by settinga(m) = m′ (resp. b(m) = m′′) where m′

n = mn−1 for n ≥ 1 and m′0 = 0R

(resp. m′′n = mn+1 for all n ∈ N). Then a and b are obviously R-linear,

in other words, a, b ∈ EndR(M) and ab = IdM = 1EndR(M). Consider theelement m ∈ M such that m0 = 1R and mn = 0R for all n ≥ 1. As a(m) = 0we have ba(m) = 0. This shows that ba �= 1EndR(M). It follows that the ringEndR(M) is not directly finite.

Let G be a group and let V be a vector space over a field K. Recall thatthe set LCA(G;V ) consisting of all linear cellular automata τ : V G → V G

has a natural structure of K-algebra in which the multiplication is given bythe composition of maps.

Proposition 8.15.4. Let G be a group and let V be a vector space over afield K. Let τ : V G → V G be a linear cellular automaton. Then the followinghold:

(i) if τ is left-invertible in LCA(G; V ), then τ is injective;

8.15 Stable Finiteness of Group Algebras 329

(ii) if τ is right-invertible in LCA(G;V ), then τ is surjective;(iii) if τ is invertible in LCA(G; V ), then τ is bijective.

If, in addition, the vector space V is finite-dimensional, then:

(iv) τ is left-invertible in LCA(G; V ) if and only if τ is injective;(v) τ is invertible in LCA(G;V ) if and only if τ is bijective.

Proof. (i) Suppose that τ ∈ LCA(G; V ) is left-invertible. This means thatthere exists σ ∈ LCA(G; V ) such that σ ◦τ = IdV G . As IdV G is injective, thisimplies that τ is injective.

(ii) Suppose that τ ∈ LCA(G; V ) is right-invertible. This means that thereexists σ ∈ LCA(G;V ) such that τ ◦ σ = IdV G . As IdV G is surjective, thisimplies that τ is surjective.

(iii) This immediately follows from (i) and (ii).(iv) This immediately follows from (i) and Theorem 8.12.1.(v) This immediately follows from (iii) and Corollary 8.12.2. ��

Remarks 8.15.5. (a) If V is infinite-dimensional, it may happen that a bi-jective linear cellular automaton τ : V G → V G is not left-invertible inLCA(G; V ). Therefore, the converses of assertions (i) and (iii) in Propo-sition 8.15.4 are false if we do not add the hypothesis that V is finite-dimensional. To see this, consider the bijective linear cellular automatonτ : V Z → V Z described in Remark 8.12.3. Then, there is no σ ∈ LCA(G; V )such that σ ◦ τ = IdV G since otherwise the inverse map of τ would coincidewith σ, which is impossible as τ is not an invertible cellular automaton.

(b) The converse of assertion (ii) in Proposition 8.15.4 does not hold evenunder the additional hypothesis that V is finite-dimensional. For instance,take an arbitrary field K and consider the linear cellular automaton τ : K

Z →K

Z defined by τ(x)(n) = x(n + 1) − x(n) for all x ∈ KZ and n ∈ Z. Observe

that τ is surjective. Indeed, given y ∈ KZ, the configuration x ∈ K

Z definedby

x(n) =

⎧⎪⎨

⎪⎩

0 if n = 0,

y(0) + y(1) + · · · + y(n − 1) if n > 0,

y(n) + y(n + 1) + · · · + y(−1) if n < 0,

clearly satisfies τ(x) = y. However, τ is not right-invertible in LCA(Z; K).To see this, observe that, as the K-algebra LCA(Z; K) is commutative byCorollary 8.5.4, the right-invertibility of τ would imply the bijectivity of τ byProposition 8.15.4(iii). But τ is not injective as all constant configurationsare mapped to 0.

Corollary 8.15.6. Let G be a group and let K be a field. Let V be a vectorspace over K of finite dimension dimK(V ) = d ≥ 1. Then the followingconditions are equivalent:

(a) every injective linear cellular automaton τ : V G → V G is surjective;(b) the K-algebra LCA(G; V ) is directly finite;


(c) the K-algebra Matd(K[G]) is directly finite.

Proof. The equivalence between conditions (b) and (c) follow from the factthat the K-algebras LCA(G; V ) and Matd(K[G]) are isomorphic by Corol-lary 8.7.8.

Suppose (a). Let τ ∈ LCA(G; V ) be a left-invertible element in LCA(G;V ).Then τ is injective by Proposition 8.15.4(i). Condition (a) implies then thatτ is surjective and hence bijective. By applying Proposition 8.15.4(v), wededuce that τ is invertible in LCA(G;V ). This shows that (a) implies (b).

Conversely, suppose (b). Let τ : V G → V G be an injective linear cellularautomaton. Then τ is left-invertible in LCA(G;V ) by Proposition 8.15.4(iv).It follows from condition (b) that τ is invertible in LCA(G;V ). By usingProposition 8.15.4(ii), we deduce that τ is surjective. This shows that (b)implies (a). ��


(a) the group G is L-surjunctive;(b) for any field K, the group algebra K[G] is stably finite. ��

From Theorem 8.14.4 and Corollary 8.15.7, we deduce the following:

Corollary 8.15.8. Let G be a sofic group and let K be a field. Then the groupalgebra K[G] is stably finite. ��

8.16 Zero-Divisors in Group Algebrasand Pre-injectivity of One-Dimensional LinearCellular Automata

Let R be a ring.A nonzero element a ∈ R is said to be a left zero-divisor (resp. a right

zero-divisor) if there exists a nonzero element b in R such that ab = 0 (resp.ba = 0). Note that the existence of a left zero-divisor in R is equivalent to theexistence of a right zero-divisor. One says that the ring R has no zero-divisorsif R admits no left (or, equivalently, no right) zero-divisors.

Examples 8.16.1. (a) Let R = Z/nZ, n ≥ 2, be the (commutative) ring ofintegers modulo n. Then R has no zero-divisors if and only if n is a primenumber.

(b) Let G be a group and let R be a nonzero ring. Suppose that G containsan element g0 of finite order n ≥ 2 and let H = {1G, g0, g

20 , . . . , gn−1

0 } denotethe subgroup of G generated by g0. Consider the elements α, β ∈ R[G] definedrespectively by

8.16 Zero-Divisors in Group Algebras 331

α(g) =

⎧⎪⎨

⎪⎩

1 if g = 1G,

−1 if g = g0,

0 if g /∈ {1G, g0}.and β(g) =

{1 if g ∈ H,

0 otherwise.

One easily checks that α �= 0, β �= 0, and αβ = βα = 0. Thus, α and β areboth left and right zero-divisors in R[G].

Let R be a ring. An element x ∈ R is called an idempotent if it satisfiesx2 = x. An idempotent x ∈ R is said to be proper if x �= 0 and x �= 1.

Proposition 8.16.2. Let R be a ring. Then the following hold:

(i) if R has no zero-divisors, then R has no proper idempotents;(ii) if R has no proper idempotents, then R is directly finite.

Proof. (i) Every idempotent x ∈ R satisfies x(x − 1) = x2 − x = 0. If R hasno zero-divisors, this implies x = 0 or x = 1.

(ii) Suppose that R has no proper idempotents. Let a, b ∈ R such thatab = 1. Then we have (ba)2 = b(ab)a = ba so that ba is an idempotent.Therefore, ba = 0 or ba = 1. If ba = 0, then a = (ab)a = a(ba) = 0 so that0 = 1 and R is reduced to 0. Therefore, we have ba = 1 in all cases. Thisshows that R is directly finite. ��

Remarks 8.16.3. (a) A ring without proper idempotents may admit zero-divisors. For example, the ring Z/8Z has no proper idempotents. However,the classes of 2, 4, and 6 are zero-divisors in Z/8Z.

(b) A directly finite ring may admit proper idempotents. For example,take a nonzero commutative ring R. Then the ring Mat2(R) is directly fi-nite by Proposition 8.15.2. However, the matrices

(1 00 0

)and

(0 00 1

)are proper

idempotents in Mat2(R).

A group G is called a unique-product group if, given any two non-emptyfinite subsets A, B ⊂ G, there exists an element g ∈ G which can be uniquelyexpressed as a product g = ab with a ∈ A and b ∈ B.

Remark 8.16.4. A unique-product group is necessarily torsion-free. Indeed,suppose that G is a group containing an element g0 of order n ≥ 2. TakeA = B = {1G, g0, g

20 , . . . , gn−1

0 }. Then AB = A and there is no g ∈ AB whichcan be uniquely written in the form g = ab with a ∈ A and b ∈ B.

Recall that a total ordering on a set X is a binary relation ≤ on X whichis reflexive, antisymmetric, transitive, and such that one has x ≤ y or y ≤ xfor all x, y ∈ X. A group G is called orderable if it admits a left-invarianttotal ordering, that is, a total ordering ≤ such that g1 ≤ g2 implies gg1 ≤ gg2

for all g, g1, g2 ∈ G.


Examples 8.16.5. (a) Every subgroup of an orderable group is orderable. In-deed, if H is a subgroup of a group G and ≤ is a left-invariant total orderingon G, then the restriction of ≤ to H is a left-invariant total ordering on H.

(b) The additive groups Z, Q, and R are orderable since the usual orderingon R is translation-invariant.

(c) The direct product of two orderable groups is an orderable group.Indeed, let G1 and G2 be two groups and suppose that ≤1 and ≤2 are left-invariant total orderings on G1 and G2 respectively. Then the lexicographicordering ≤ on G1 × G2 defined by

(g1, g2) ≤ (h1, h2) ⇐⇒

⎧⎪⎨

⎪⎩

g1 ≤1 g2

org1 = g2 and h1 ≤2 h2

is a left-invariant total ordering on G1 × G2.(d) More generally, the direct product of any family (finite or infinite) of

orderable groups is an orderable group. Indeed, let (Gi)i∈I be a family oforderable groups. Let ≤i be a left-invariant total ordering on Gi for eachi ∈ I. Let us fix a well-ordering on I, i.e., a total ordering such that everynonempty subset of I admits a minimal element (the fact that any set canbe well-ordered is a basic fact in set theory which may be deduced from theAxiom of Choice). Then the lexicographic ordering on G =

∏i∈I Gi, which

is defined by setting g ≤ h for g = (gi), h = (hi) ∈ G if and only if g = h orgi0 < hi0 where i0 = min{i ∈ I : gi �= hi}, is a left-invariant total ordering onG. As ⊕i∈IGi is a subgroup of

∏i∈I Gi, it follows that the direct sum of any

family of orderable groups is an orderable group. Since a free abelian group isisomorphic to a direct sum of copies of Z, we deduce in particular that everyfree abelian group is orderable.

(e) In fact, every torsion-free abelian group is orderable. Indeed, if G isa torsion-free abelian group then G embeds in the Q-vector space G ⊗Z Q

via the map g �→ g ⊗ 1. On the other hand, the additive group underlying aQ-vector space V is isomorphic to a direct sum of copies of Q (since V admitsa Q-basis) and hence orderable.

(f) Suppose that G is a group containing a normal subgroup N such thatboth N and G/N are orderable groups. Then the group G is orderable. In-deed, let ≤1 (resp. ≤2) be a left-invariant total ordering on N (resp. G/N).Then one easily checks that the binary relation ≤ on G defined by setting

g ≤ h ⇐⇒

⎧⎪⎨

⎪⎩

ρ(g) ≤2 ρ(h)or

ρ(g) = ρ(h) and 1G ≤1 g−1h

is a left-invariant total ordering on G.(g) Let R be any ring. Consider the Heisenberg group

8.16 Zero-Divisors in Group Algebras 333

HR =

⎧⎨

⎩M(x, y, z) =

⎛

⎝1 y z0 1 x0 0 1

⎞

⎠ : x, y, z ∈ R

⎫⎬

⎭ .

(cf. Example 4.6.5). The kernel of the group homomorphism of HR onto R2

given by M(x, y, z) �→ (x, y) is the normal subgroup

N =

⎧⎨

⎩

⎛

⎝1 0 z0 1 00 0 1

⎞

⎠ : z ∈ R

⎫⎬

⎭ .

As the groups N and HR/N are both abelian, it follows from Examples (e)and (f) above that HR is orderable.

Given a set X equipped with a total ordering ≤, we denote by Sym(X,≤)the group of order-preserving permutations of X, that is, the subgroup ofSym(X) consisting of all bijective maps f : X → X such that x ≤ y impliesf(x) ≤ f(y) for all x, y ∈ X.


(a) G is orderable;(b) there exists a set X equipped with a total ordering ≤ such that G is iso-

morphic to a subgroup of Sym(X,≤).

Proof. Suppose that ≤ is a left invariant total ordering on G. Then the ac-tion of G on itself given by left multiplication is order-preserving. Thus G isisomorphic to a subgroup of Sym(G,≤). This shows that (a) implies (b).

Conversely, suppose that X is a set equipped with a total ordering ≤.Choose a well ordering ≤W on X. Then we can define a lexicographic ordering≤L on Sym(X,≤) by setting f ≤L g for f, g ∈ Sym(X,≤) if and only if f = gor f(x0) ≤ g(x0) where x0 = min{x ∈ X : f(x) �= g(x)}. Clearly ≤L is aleft invariant total ordering on Sym(X,≤). It follows that Sym(X,≤) is anorderable group. As any subgroup of an orderable group is orderable, weconclude that (b) implies (a). ��

Corollary 8.16.7. The group Homeo+(R) of orientation-preserving homeo-morphisms of R is orderable. ��

Proposition 8.16.8. Every orderable group is a unique-product group.

Proof. Let G be an orderable group and let ≤ be a left-invariant total orderingon G. Let A and B be nonempty finite subsets of G. Set bm = minB andlet am ∈ A be the unique element such that ambm = minAbm. Consider theelement g = ambm. For all a ∈ A and b ∈ B, we have ambm ≤ abm ≤ ab.Thus, if g = ab for some a ∈ A and b ∈ B, then ambm = abm = ab, so thatwe get am = a and bm = b after right and left cancellation. This shows thatG is a unique-product group. ��


Proposition 8.16.9. Let G be a unique-product group and let R be a ringwith no zero-divisors. Then the group ring R[G] has no zero-divisors.

Proof. Let α and β be nonzero elements in R[G], and denote by A and Btheir supports. As G is a unique-product group, there is an element g ∈ ABsuch that there exists a unique element (a, b) ∈ A×B such that g = ab. Thenwe have

(αβ)(g) =∑

h1∈A,h2∈Bh1h2=g

α(h1)β(h2) = α(a)β(b).

As R has no zero-divisors, this implies (αβ)(g) �= 0. Thus, we have αβ �= 0.This shows that R[G] has no zero-divisors. ��

Corollary 8.16.10. Let G be a unique-product group and let K be a field.Then the group algebra K[G] has no zero-divisors. ��

Let G be a group and let K be a field. We have seen in Corollary 8.5.3that the map Ψ : K[G] → LCA(G; K) defined by

Ψ(α)(x)(g) =∑

h∈G

α(h)x(gh)

for all α ∈ K[G], x ∈ KG, and g ∈ G, is a K-algebra isomorphism.

Proposition 8.16.11. Let G be a group and let K be a field. Let α be anonzero element in K[G]. Then the following conditions are equivalent:

(a) α is not a left zero-divisor in K[G];(b) the linear cellular automaton Ψ(α) : K

G → KG is pre-injective.

Proof. By Proposition 8.2.5, the linear cellular automaton Ψ(α) is not pre-injective if and only if there exists a nonzero element β ∈ K[G] such thatΨ(α)(β) = 0. As

Ψ(α)(β)(g) =∑

h∈G

α(h)β(gh) =∑

h∈G

α(h)β∗(h−1g−1) = (αβ∗)(g−1),

for all β ∈ K[G] and g ∈ G, we deduce that Ψ(α) is not pre-injective if andonly if there exists a nonzero element β ∈ K[G] such that αβ∗ = 0, that is,if and only if α is a left zero-divisor in K[G]. This shows that conditions (a)and (b) are equivalent. ��

Corollary 8.16.12. Let G be a group and let K be a field. Then the followingconditions are equivalent:

(a) the group algebra K[G] has no zero-divisors;(b) every non-identically-zero linear cellular automaton τ : K

G → KG is pre-

injective. ��

Notes 335

From Corollary 8.16.12 and Theorem 8.9.6 we deduce the following.

Corollary 8.16.13. Let G be an amenable group and let K be a field suchthat K[G] has no zero-divisors. Then every non-identically-zero linear cellularautomaton τ : K

G → KG is surjective. ��

Observe that it follows from Theorem 4.6.1, Example 8.16.5(e), Proposi-tion 8.16.8, and Corollary 8.16.10 that torsion-free abelian groups satisfy thehypotheses of Corollary 8.16.13 for any field K. This implies in particularthat if G is a torsion-free abelian group and S is a nonempty subset of Gwhich is not reduced to the identity element, then the discrete LaplacianΔS : K

G → KG is surjective for any field K.

Notes

In the literature, the term linear is used by some authors with a differentmeaning, namely to designate a cellular automaton τ : AG → AG for whichthe alphabet A is a finite abelian group and τ is a group endomorphism ofAG. Such cellular automata are also called additive cellular automata.

Linear cellular automata with vector spaces as alphabets were consideredby the authors in a series of papers starting with [CeC1]. The fact that thealgebra of linear cellular automata over a group G whose alphabet is a d-dimensional vector space over a field K is isomorphic to the algebra of d × dmatrices with coefficients in K[G] (Corollary 8.7.8) was proved in Sect. 6 of[CeC1] for d = 1 and in Sect. 4 of [CeC2] for all d ≥ 1. The representationsof linear cellular automata both as elements in EndK(V )[G] and as elementsin EndK[G](V [G]) were given in Sect. 4 of [CeC5].

Recall that a Laurent polynomial over a field K is a polynomial in thevariable t and its inverse t−1 with coefficients in K. The K-algebra of Lau-rent polynomials is thus denoted by K[t, t−1]. Also, a Laurent polynomialmatrix is a matrix whose entries are Laurent polynomials. For d ≥ 1, de-note by Matd(K[t, t−1]) the K-algebra of d × d Laurent polynomial matri-ces over K. When G = Z, there are canonical isomorphisms of K-algebrasK[Z] ∼= K(t, t−1] and LCA(Z; Kd) ∼= Matd(K[t, t−1]). In [LiM, Sect. 1.6]cellular automata τ ∈ LCA(Z; Kd) are called (d × d) convolutional en-coders.

The notion of mean dimension for vector subspaces of V G, where G is anamenable group and V is a finite-dimensional vector space, was introduced byGromov in [Gro5] and [Gro6]. Mean dimension was used by Elek [Ele] to provethat, given any amenable group G and any field K, there exists a non-trivialhomomorphism from the Grothendieck group of finitely generated modulesover the group algebra K[G] into the additive group of real numbers. As forentropy, it can be shown, as an application of the Ornstein-Weiss convergence


theorem (see [OrW], [LiW], [Gro6], [Ele], [Kri]) that the lim sup in (8.18) isin fact a true limit and does not depend on the Følner net F .

The fact that the image of a linear cellular automaton with finite dimen-sional alphabet is closed (Theorem 8.8.1) was proved in Sect. 3 of [CeC1].The linear version of the Garden of Eden theorem (Theorem 8.9.6) was firstproved for countable amenable groups in [CeC1] and then extended to allamenable groups in [CeC8] using induction and restriction for linear cellularautomata. The linear version of the Garden of Eden theorem was generalizedto R-linear cellular automata with coefficients in semisimple left R-modulesof finite length over a ring R and over amenable groups in [CeC4]. Invertibil-ity of linear cellular automata with finite dimensional alphabet V was provedin Sect. 3 of [CeC2] for bijective linear cellular automata τ : X → Y betweenclosed linear subshifts X,Y ⊂ V G over countable groups and in [CeC8] forbijective linear cellular automata τ : V G → V G over any group G.

In [CeC11] it is shown that if G is a non-periodic group, then for everyinfinite-dimensional vector space V over a field K there exist a bijective cellu-lar automaton τ : V G → V G which is not invertible (cf. Theorem 8.12.1 andRemark 8.12.3) and a cellular automaton τ ′ : V G → V G whose image τ ′(V G)is not closed in V G with respect to the prodiscrete topology (cf. Theorem8.8.1 and Example 8.8.3).

Theorem 8.13.2 and Corollary 8.13.4 were proved in [CeC6] and [CeC9].Directly finite rings are sometimes called Dedekind finite rings or von

Neumann finite rings. In [Coh], P.M. Cohn constructed, for each integerd ≥ 1, a ring R such that Matd(R) is directly finite but Matd+1(R) is notdirectly finite. I. Kaplansky [Kap2, p. 122], [Kap3, Problem 23] observed thattechniques from the theory of operator algebras could be used to prove that,for any group G and any field K of characteristic 0, the group algebra K[G]is stably finite and asked whether this property remains true for fields ofcharacteristic p > 0. The stable finiteness of K[G] in arbitrary characteristicwas established for free-by-abelian groups G by P. Ara, K.C. O’Meara andF. Perera in [AOP]. This was extended to all sofic groups by Elek and Szabo[ES1], using the notion of von Neumann dimension for continuous regularrings. The proof of Elek and Szabo’s result via linear cellular automata whichis presented in this chapter (cf. Corollary 8.15.8) was given in [CeC2]. Thenotion of L-surjunctivity was also introduced in [CeC2] and the equivalencebetween L-surjunctivity and stable finiteness was established in Corollary 4.3therein. In [CeC3], the authors proved that if R is a ring and G is a residuallyfinite group, then every injective R-linear cellular automaton over G whosealphabet is an Artinian left R-module is surjunctive. In [CeC5], it was shownthat if R is a ring, then R-linear cellular automata with coefficients in leftR-modules of finite length (thus a stronger condition than being Artinian)over sofic groups (thus a weaker condition than being residually finite) aresurjunctive. This last result was used to show in [CeC5] that the group ringR[G] is stably finite whenever R is a left (or right) Artinian ring and G is a

Notes 337

sofic group. This yields an extension of the Elek and Szabo stable finitenessresult since any division ring is Artinian.

Unique-product groups were introduced by W. Rudin and H. Schneider in[RuS] under the name of Ω-groups. The question of the existence of a torsion-free group which is not unique-product was raised by Rudin and Schneider[RuS, p. 592]. This question was answered in the affirmative by E. Rips andY. Segev [RiS] (see also [Pro]). More information on orderable groups maybe found for example in [BoR], [Pas], and [Gla]. A group G is said to belocally indicable if any nontrivial finitely generated subgroup of G admitsan infinite cyclic quotient. Every locally indicable group is orderable (seefor example [BoR, Theorem 7.3.1] or [Gla, Lemma 6.9.1]). This implies inparticular that locally nilpotent torsion-free groups, free groups, and fun-damental groups of surfaces not homeomorphic to the real projective planeare all orderable groups. It was observed by G.M. Bergman [Ber] that the

universal covering group SL2(R) of SL2(R) is orderable but not locally indi-

cable. The group SL2(R) is orderable since it has a natural faithful action byorientation-preserving homeomorphisms of the real line but it is not locallyindicable since it contains nontrivial finitely generated perfect groups. By arecent result due to D.W. Morris [Morr, Theorem B], every amenable order-able group is locally indicable. The orderability of the braid groups Bn wasestablished independently by P. Dehornoy [Deh] and W. Thurston. It followsfrom a result of E.A. Gorin and V.Ja. Lin [GoL] that the group Bn is notlocally indicable for n ≥ 5.

A group is called bi-orderable if it admits a total ordering which is bothleft and right invariant. All free groups and all locally nilpotent torsion-freegroups are bi-orderable. For n ≥ 3, the braid group Bn is not bi-orderable.However, the pure braid groups Pn (i.e., the kernel of the natural epimorphismof Bn onto Symn) is bi-orderable for all n. A theorem due to A.I. Mal’cev[Mal2] and B.H. Neumann [Neu1] says that, given a group G equipped with atotal ordering which is both left and right invariant and a field K, the vectorsubspace of K

G consisting of all maps x : G → K whose support is a well-ordered subset of G is a division K-algebra for the convolution product (see[Pas, Theorem 2.11 in Chap. 13]). When G = Z, this division algebra is thefield of Laurent series with coefficients in K. The Mal’cev-Neumann theoremimplies in particular that, for any bi-orderable group G and any field K, thegroup algebra K[G] embeds in a division K-algebra and is therefore stablyfinite.

A famous conjecture attributed to Kaplansky is the zero-divisor conjec-ture which states that if G is a torsion-free group then the group algebraK[G] has no zero-divisors for any field K (see [Kap1], [Kap2, p. 122], [Pas,Chap. 13]). The observation that unique-product groups (and hence orderablegroups) satisfy the Kaplansky zero-divisor conjecture (see Corollary 8.16.10)was made by Rudin and Schneider [RuS, Theorem 3.2]. The class of ele-mentary amenable groups is the smallest class of groups containing all finite


and all abelian groups that is closed under taking subgroups, quotients, ex-tensions, and directed unions. It is known (see [KLM, Theorem 1.4]) thattorsion-free elementary amenable groups satisfy the Kaplansky zero-divisorsconjecture. This implies in particular that all torsion-free virtually solvablegroups satisfy the Kaplansky zero-divisors conjecture. According to Corol-lary 8.16.12, the zero-divisor conjecture is equivalent to saying that if G isa torsion-free group and K is a field then every non-identically-zero linearcellular automaton τ : K

G → KG is pre-injective. This reformulation of the

Kaplansky zero-divisor conjecture in terms of linear cellular automata wasgiven in Sect. 6 of [CeC1].

Exercises

8.1. Let G be a group and let R be a nonzero ring. Show that the ring R[G]is commutative if and only if both G and R are commutative.

8.2. Recall that the center of a ring A is the subring B of A consisting of theelements x ∈ A which satisfy ax = xa for all a ∈ A. Let G be a group andlet R be a commutative ring. Let α ∈ R[G]. Show that α is in the center ofR[G] if and only if α is constant on each conjugacy class of G.

8.3. One says that a group G has the ICC-property if every conjugacy classof G except {1G} is infinite.

(a) Show that if a group G has the ICC-property then the center of G isreduced to the identity element.

(b) Show that if a group G has the ICC-property then every normal sub-group of G which is not reduced to the identity element is infinite.

(c) Let X be an infinite set. Show that the group Sym0(X), which consistsof all permutations of X with finite support, has the ICC-property.

(d) Show that every nonabelian free group has the ICC-property.(e) Let G be a group with the ICC-property and let R be a commutative

ring. Show that the center of R[G] consists of the maps x : G → R whichsatisfy x(g) = 0R for all g ∈ G \ {1G}. Hint: Use the result of Exercise 8.2.

(f) Let G be a group and let R be a nonzero commutative ring. Show thatG has the ICC-property if and only if the center of R[G] is isomorphic to thering R.

8.4. Let G be a group, K a field, and d ≥ 1 an integer. Show that the K-algebras Matd(K)[G] and Matd(K[G]) are isomorphic.

8.5. Show that Theorem 8.13.2 remains valid if the field R is replaced by thefield C of complex numbers. Hint: Write every f ∈ C

G in the form f = f0+if1,where f0, f1 ∈ R

G.

Exercises 339

8.6. Let G be a group and let S ⊂ G be a non-empty finite subset. Denoteby 〈·, ·〉 the standard scalar product on R[G] ⊂ �2(G) and let ‖ · ‖ denote theassociated norm.

(a) Show that for all x ∈ R[G] and λ ∈ R, one has

〈x,ΔS(x) + λx〉 =12

∑

g∈G

∑

s∈S

|x(g) − x(gs)|2 + λ‖x‖2.

(b) Show that if the subgroup generated by S is infinite then for anynon–empty finite subset F ⊂ G there exists s ∈ S such that Fs �⊂ F .

(c) Deduce from (a) and (b) that if λ ≥ 0 and the subgroup generated byS is infinite, then the linear cellular automaton ΔS + λ IdRG : R

G → RG is

pre-injective.

8.7. Show that if (Ri)i∈I is a family of directly finite rings then the productring P =

∏i∈I Ri is directly finite.

8.8. Show that every subring of a directly finite (resp. stably finite) ring isdirectly finite (resp. stably finite).

8.9. Let K be a field and let V be an infinite-dimensional vector space overK. Show that the K-algebra EndK(V ) is not directly finite.

8.10. Let R be a ring and let M be a left R-module. One says that the moduleM is Hopfian if every surjective endomorphism of M is injective. Show thatif M is Hopfian, then the ring EndR(M) is directly finite.

8.11. Let R be a ring and let M be a left R-module. One says that the moduleM is Noetherian if its submodules satisfy the ascending chain condition, i.e.,every increasing sequence

N1 ⊂ N2 ⊂ . . .

of submodules of M stabilizes (there is an integer i0 ≥ 1 such that Ni = Ni0

for all i ≥ i0). Show that if M is Noetherian, then M is Hopfian. Hint:Suppose that f is a surjective endomorphism of M and consider the sequenceof submodules Ni = Ker(f i), i ≥ 1.

8.12. Let R be a ring and let P be a left R-module. One says that the moduleP is projective if for every homomorphism f : P → M and every surjectivehomomorphism g : M → M of left R-modules, there exists a homomorphismh : P → M such that f = g ◦h. Show that if the module P is projective, thenP is Hopfian if and only if the ring EndR(P ) is directly finite.

8.13. Let R be a ring and let d ≥ 1 be an integer. Equip Rd with its naturalstructure of left R-module. Show that Rd is Hopfian if and only if the ringMatd(R) is directly finite.


8.14. One says that a ring R is left Noetherian if R is Noetherian as a leftmodule over itself.

(a) Let R be a left Noetherian ring. Show by induction that Rd is Noethe-rian as a left R-module for each integer d ≥ 1.

(b) Show that every left Noetherian ring is stably finite. Hint: Use Exer-cises 8.11 and 8.13.

8.15. One says that a ring R has the unique rank property if it satisfies thefollowing condition: if m and n are positive integers such that Rm and Rn areisomorphic as left R-modules, then one has m = n. Show that every stablyfinite ring has the unique rank property.

8.16. Let K be a field and let V be an infinite-dimensional vector space overK. Show that the ring R = EndK(V ) does not have the unique rank property.Hint: Observe that the vector spaces V and V ⊕ V are isomorphic and thenprove that the set consisting of all K-linear maps f : V ⊕ V → V , with itsnatural structure of left R-module, is isomorphic to both R and R2.

8.17. Show that every division ring is stably finite.

8.18. A ring R is said to be unit-regular if for any a ∈ R there exists aninvertible element u ∈ R such that a = aua.

(a) Show that every division ring is unit-regular.(b) Show that if K is a field then the ring Matd(K) is unit-regular for every

d ≥ 1. Hint: Prove that if A ∈ Matd(K) has rank r then there exist invertiblematrices U, V ∈ GLd(K) such that A = UDrV , where Dr ∈ Matd(K) is thediagonal matrix defined by δij = 1 if 1 ≤ i = j ≤ r and δij = 0 otherwise,and then observe that A = AXA, where X = (UV )−1.

(c) Show that every unit-regular ring is directly finite.(d) Prove that the ring Z is not unit-regular.(e) A ring R is called a Boolean ring if a2 = a for all a ∈ R. Show that

every Boolean ring is commutative and unit-regular.

8.19. One says that a ring R is a right Ore ring if R has no zero-divisorsand if for any pair a, b of nonzero elements in R there exist nonzero elementsu, v ∈ R such that au = bv. Left Ore rings are defined similarly. Show thatany right (or left) Ore ring is directly finite.

Hint: Prove that ab = 1R implies ba = 1R by a direct argument, or showthat R can be embedded as a subring of a division ring by adapting theconstruction of the field of fractions of an integral domain.

8.20. Let K be a field. Show that every finite-dimensional K-algebra is stablyfinite.

8.21. Let G be an amenable group and let F be a Følner net for G. Let Vbe a finite-dimensional vector space. Suppose that X (resp. Y ) is a vectorsubspace of V G. Show that mdimF (X ∪ Y ) ≤ mdimF (X) + mdimF (Y ).

Exercises 341

8.22. Use Corollary 8.15.7 and the result of Exercise 8.20 to recover the factthat every finite group is L-surjunctive.

8.23. Use Proposition 8.15.2 and Corollary 8.15.7 to recover the fact thatevery abelian group is L-surjunctive.

8.24. It follows from Theorem 8.14.4 that every residually finite group is L-surjunctive. The goal of this exercise is to present an alternative proof of thisresult. Let G be a residually finite group and let V be a finite-dimensionalvector space over a field K. Let τ : V G → V G be an injective linear cellularautomaton. Fix a family (Γi)i∈I of subgroups of finite index of G such that⋂

i∈I Γi = {1G} (the existence of such a family follows from the residualfiniteness of G).

(a) Show that, for each i ∈ I, the set Fix(Γi) = {x ∈ V G : gx =x for all g ∈ Γi} is a finite-dimensional vector subspace of V G.

(b) Show that τ(Fix(Γi)) = Fix(Γi) for all i ∈ I.(c) Prove that

⋃i∈I Fix(Γi) is dense in V G and conclude.

8.25. Let R be a ring and let x ∈ R. Show that x is an idempotent if andonly if 1R − x is an idempotent.

8.26. Show that any subgroup of a unique-product group is a unique-productgroup.

8.27. Let G be a group. Suppose that G contains a normal subgroup Nsuch that N and G/N are both unique-product groups. Show that G is aunique-product group. Hint: See for example [RuS, Theorem 6.1].

8.28. Show that every residually orderable group is orderable.

8.29. Show that the limit of a projective system of orderable groups is or-derable.

8.30. A group G is called bi-orderable if it admits a total ordering ≤ whichis both left and right invariant, i.e., such that g1 ≤ g2 implies gg1 ≤ gg2 andg1g ≤ g2g for all g, g1, g2 ∈ G. Let G be a bi-orderable group. Suppose thatan element g ∈ G satisfies the following property: there exist an integer n ≥ 1and elements h1, h2, · · · , hn ∈ G such that

h1gh−11 h2gh−1

2 · · ·hngh−1n = 1G.

Show that g = 1G.

8.31. Let G be a bi-orderable group. Show that if g, h ∈ G are such thatgn = hn for some integer n ≥ 1, then g = h.

8.32. The Klein bottle group is the group K given by the presentation K =〈x, y; xyx−1y〉. Thus, K is the quotient group K = F/N , where F is the free


group based on x and y, and N is the normal closure of xyx−1y in F . Letρ : F → K denote the quotient homomorphism.

(a) Let H denote the subgroup of K generated by ρ(y). Show that H isnormal in K and that H and K/H are both infinite cyclic.

(b) Show that K is an orderable group.(c) Show that the group K is not bi-orderable.

8.33. Let G be an amenable group, F a right Følner net for G, and V a finite-dimensional vector space over some field K. Let X be a vector subspace of V G

and let X denote the closure of X in V G for the prodiscrete topology. Showthat X is a vector subspace of V G and that one has mdimF (X) = mdimF (X).

Appendix A

Nets and the Tychonoff ProductTheorem

A.1 Directed Sets

Recall that a partially ordered set is a set I equipped with a binary relation≤ which is both reflexive (i ≤ i for all i ∈ I) and transitive (i ≤ j and j ≤ kimplies i ≤ k for all i, j, k ∈ I).

A directed set is a partially ordered set I which satisfies the followingcondition: for all i, j ∈ I, there exists an element k ∈ I such that i ≤ k andj ≤ k.

Examples A.1.1. (a) The set Z equipped with the relation ≤ defined by

i ≤ j ⇐⇒ i divides j

is a directed set.(b) If E is an arbitrary set, then the set P(E) of all subsets of E is a

directed set for inclusion.(c)) If X is a topological space and x is a point of X, then the set of

neighborhoods of x, equipped with the relation ≤ defined by

V ≤ W ⇐⇒ W ⊂ V,

is a directed set. Indeed, if V and W are neighborhoods of x, then V ∩ W isa neighborhood of x satisfying V ≤ V ∩ W and W ≤ V ∩ W .

A.2 Nets in Topological Spaces

Let X be a set. A net of points of X (or net in X) is a family (xi)i∈I ofpoints of X indexed by some directed set I.


343

http://dx.doi.org/10.1007/978-3-642-14034-1_9

344 A Nets and the Tychonoff Product Theorem

Let (xi)i∈I and (yj)j∈J be nets in a set X indexed by directed sets I andJ respectively. One says that the net (yj) is a subnet of the net (xi) if thereis a map ϕ : J → I which satisfies the following conditions:

(SN-1) yj = xϕ(j) for all j ∈ J ;(SN-2) for each i ∈ I, there exists j ∈ J such that if k ∈ J and j ≤ k then

i ≤ ϕ(k).

Let X be a topological space. Let (xi)i∈I be a net in X and let a ∈. Onesays that the net (xi) converges to a, or that a is a limit point of the net(xi), if the following condition is satisfied: for each neighborhood V of a inX, there exists an element i0 ∈ I such that xi ∈ V for all i ≥ i0. One saysthat the net (xi) is convergent if there exists a point a in X such that (xi)converges to a.

It is clear that if the net (xi) converges to a, then every subnet of (xi) alsoconverges to a.

Proposition A.2.1. Let X be a topological space, Y ⊂ X and a ∈ X. Let Ydenote the closure of Y in X. Then the following conditions are equivalent:

(a) a ∈ Y ;(b) there exists a net (yi)i∈I of points of Y which converges to a in X.

Proof. Suppose that (yi)i∈I is a net of points of Y which converges to a.Then, for each neighborhood V of a, there is an element i0 ∈ I such thatyi ∈ V for all i ≥ i0. Thus, every neighborhood of a meets Y . This showsthat a ∈ Y .

Conversely, suppose that a ∈ Y . Let I denote the directed set consistingof all neighborhoods of a in X partially ordered by reverse inclusion, that is,

V ≤ W ⇐⇒ W ⊂ V.

Since a is in the closure of Y , we can find for each neighborhood V ∈ I apoint yV in Y such that yV ∈ V . It is clear that the net (yV )V ∈I convergesto a. �

Proposition A.2.2. A topological space X is Hausdorff if and only if everyconvergent net in X admits a unique limit.

Proof. Suppose that X is Hausdorff. Consider a net (xi)i∈I in X which con-verges to some point a ∈ X. Let b be a point in X with a �= b. Since X isHausdorff, there exist a neighborhood V of a and a neighborhood W of bsuch that V ∩W . For i large enough, the point xi is in V and therefore not inW . Therefore the net (xi)i∈I does not converge to b. This shows that everyconvergent net in X has a unique limit.

Suppose now that X is not Hausdorff. Then there exist distinct points aand b in X such that each neighborhood of a meets each neighborhood of b.Consider the set I consisting of all pairs (V,W ), where V is a neighborhood of

A.2 Nets in Topological Spaces 345

a and W is a neighborhood of b. partially ordered by declaring that (V ′, W ′) ≤(V,W ) if and only if V ⊂ V ′ and W ⊂ W ′. Clearly I is a directed set. If wechoose, for each i = (V,W ) ∈ I a point xi ∈ V ∩ W , then the net (xi)i∈I

admits both points a and b as limits. This proves the converse implication.�

Let (xi)i∈I be a net in a topological space X. One says that a point a ∈ Xis a cluster point of the net (xi) if it satisfies the following condition: for eachneighborhood V of a in X and each i ∈ I, there exists an element j ∈ I suchthat i ≤ j and xj ∈ V .

Proposition A.2.3. Let X be a topological space, (xi)i∈I a net in X, anda ∈ X. Then the following conditions are equivalent:

(a) the point a is a cluster point of the net (xi);(b) the net (xi) admits a subnet converging to a.

Proof. Suppose that (yj)j∈J is a subnet of the net (xi)i∈I converging to a.Let ϕ : J → I be a map satisfying conditions (SN-1) and (SN-2) above.Consider a neighborhood V of a and an element i0 ∈ I. By (SN-2), we mayfind j0 ∈ J such that i0 ≤ ϕ(k) for all k ∈ J satisfying j0 ≤ k. Since the net(yj) converges to a, there exists k0 ∈ J such that j0 ≤ k0 and yk0 ∈ V . Thenwe have i0 ≤ ϕ(k0) and xϕ(k0) = yk0 ∈ V . This shows that a is a clusterpoint for the net (xi). Thus (b) implies (a).

Conversely, suppose that a is a cluster point for the net (xi). Denote byNa the set of all neighborhoods of a, partially ordered by reverse inclusion.Let J be the subset of the Cartesian product I × Na consisting of all pairs(i, V ) ∈ I ×Na such that xi ∈ V . The fact that a is a cluster point of the net(xi) implies that J is a directed set for the partial ordering ≤ defined by

(i, V ) ≤ (i′, V ′)def⇐⇒ (i ≤ i′ and V ≤ V ′).

Consider the non decreasing map ϕ : J → I given by ϕ((i, V )) = i. If we setyj = xϕ(j) for all j ∈ J , it is clear that ϕ satisfies conditions (SN-1) and(SN-2) above and that the net (yj)j∈J converges to a. This shows that (a)implies (b). �

Note that if f : X → Y is a continuous map between topological spaces anda ∈ X is a limit (resp. cluster) point of the net (xi), then f(a) is a limit (resp.cluster) point of the net (f(xi). This immediately follows from the definitionand the fact that f−1(W ) is a neighborhood of a for each neighborhood Wof f(a).


A.3 Initial Topology

Let X be a set and let (Yλ)λ∈Λ be a family of topological spaces indexedby an arbitrary set Λ. Suppose that we are given, for each λ ∈ Λ, a mapfλ : X → Yλ. Then one constructs a topology on X in the following way.

Let F denote the set of all subsets of X of the form f−1λ (Uλ), where λ ∈ Λ

and Uλ is an open subset of Yλ. Let B be the set of all subsets of X which maybe written as a finite intersection of elements of F . Finally, let T denote theset of all subsets of X which may be written as a (finite or infinite) union ofelements of B. It is straightforward to verify that the set T is the set of opensets of a topology on X which admits B as a base, and that this topology isthe smallest topology on X for which all maps fλ : X → Yλ are continuous.The topology on X whose open sets are the elements of T is called the initialtopology on X associated with the topological spaces (Yλ)λ∈Λ and the maps(fλ)λ∈Λ.

If Z is a topological space and g : Z → X is a map, then g is continuouswith respect to the initial topology on X if and only if all composite mapsfλ◦g : Z → Yλ, λ ∈ Λ, are continuous. If (xi)i∈I is a net in X and a is a pointin X, then the net (xi)i∈I converges to a if and only if the net (fλ(xi))i∈I

converges to fλ(a) for every λ ∈ Λ. Similarly, a is a cluster point of the net(xi)i∈I if and only if fλ(a) is a cluster point of the net (fλ(xi))i∈I for everyλ ∈ Λ.

A.4 Product Topology

Let (Xλ)λ∈Λ be a family of topological spaces indexed by a set Λ. The initialtopology on the cartesian product X =

∏λ∈Λ Xλ associated with the pro-

jection maps πλ : X → Xλ is called the product topology on X. A base forthe product topology on X consists of all subsets of the form V =

∏λ∈Λ Uλ,

where Uλ is an open subset of Xλ for each λ ∈ Λ and Uλ = Xλ for all butfinitely many λ ∈ Λ.

In the case when each Xλ is endowed with the discrete topology, the prod-uct topology on X =

∏λ∈Λ Xλ is called the prodiscrete topology .

Proposition A.4.1. Let (Xλ)λ∈Λ be a family of Hausdorff topological spaces.Then X =

∏λ∈Λ Xλ is Hausdorff for the product topology.

Proof. Let x = (xλ) and y = (yλ) be distinct points of X. Then there existsλ0 ∈ Λ such that xλ0 �= yλ0 . Since Xλ0 is Hausdorff, we may find disjointopen subsets U and V of Xλ0 containing xλ0 and yλ0 respectively. The pull-backs of U and V by the projection map πλ0 : X → Xλ0 are disjoint opensubsets of X containing x and y respectively. Therefore X is Hausdorff. �

Recall that a topological space X is called totally disconnected if anynonempty connected subset of X is reduced to a single point.

A.5 The Tychonoff Product Theorem 347

Proposition A.4.2. Let (Xλ)λ∈Λ be a family of totally disconnected topo-logical spaces. Then X =

∏λ∈Λ Xλ is totally disconnected for the product

topology.

Proof. Let C be a nonempty connected subset of X. Then, for each λ ∈ Λ,the image of C by the projection map πλ : X → Xλ is a nonempty connectedsubset of Xλ. Since Xλ is totally disconnected, the set πλ(C) is reduced to asingle point for every λ ∈ Λ. This implies that C is reduced to a single point.Therefore, X is totally disconnected. �

Proposition A.4.3. Let (Xλ)λ∈Λ be a family of topological spaces. Supposethat Fλ is a closed subset of Xλ for each λ ∈ Λ. Then F =

∏λ∈Λ Fλ is a

closed subset of X =∏

λ∈Λ Xλ for the product topology.

Proof. We haveF =

⋂

λ∈Λ

π−1λ (Fλ).

Thus F is closed in X as it is the intersection of a family of closed subsetsof X. �

A.5 The Tychonoff Product Theorem

Recall that a topological space X is called compact if every open cover ofX admits a finite subcover. This means that if (Uα)α∈A is a family of opensubsets of X with X =

⋃α∈A Uα, then there exists a finite subset B ⊂ A such

that X =⋃

α∈B Uα. By taking complements, one sees that the compactnessof X is equivalent to the fact that every family (Fα)α∈A of closed subsetsof X with the finite intersection property,, that is,

⋂α∈B Fα �= ∅ for every

finite subset B ⊂ A, has a nonempty intersection.It is well known that a metric space X is compact if and only if every

sequence in X admits a convergent subsequence. There is an analogous char-acterization of compactness for general topological spaces using nets:

Theorem A.5.1. Let X be a topological space. Then the following conditionsare equivalent:

(a) X is compact;(b) every net in X admits a cluster point;(c) every net in X admits a convergent subnet.

Proof. The equivalence between conditions (b) and (c) follows from Proposi-tion A.2.3. Thus, it suffices to prove that conditions (a) and (b) are equivalent.

Suppose first that X is compact. Let (xi)i∈I be a net in X. For each i ∈ I,define the subset Yi ⊂ X by


Yi = {xj : j ∈ I and i ≤ j}.

Let J be a finite subset of I. Since I is a directed set, we may find an elementk ∈ I such that i ≤ k for all i ∈ J . This implies xk ∈ Yi for all i ∈ J andhence

⋂i∈J Yi ⊃

⋂i∈J Yi �= ∅. It follows that the family of closed sets (Yi)i∈I

has the finite intersection property. By compactness of X, we deduce that⋂i∈I Yi �= ∅. Let a be a point in

⋂i∈I Yi. Then, any neighborhood of a meets

Yi for all i ∈ I. This means that a is a cluster point of the net (xi). Therefore,(a) implies (b).

Conversely, suppose that every net in X admits a cluster point. Let(Fα)α∈A be a family of closed subsets of X with the finite intersection prop-erty. Consider the directed set E consisting of all finite subsets of A partiallyordered by inclusion. Choose, for each E ∈ E , an element xE ∈

⋂α∈E Fα. By

our hypothesis, the net (xE)E∈E admits a cluster point a ∈ X. Let α0 ∈ A.If V is a neighborhood of a, then there exists a finite set E0 ⊂ A such that{α0} ⊂ E0 and xE0 ∈ V . Since xE0 ∈

⋂α∈E0

Fα ⊂ Fα0 , it follows thatany neighborhood of a meets Fα0 . Since the set Fα0 is closed, we deduce thata ∈ Fα0 . As α0 was an arbitrary element in A, we conclude that a ∈

⋂α∈A Fα.

This shows that⋂

i∈I Fi �= ∅. Consequently, the space X is compact. Thisproves that (b) implies (a). �

Theorem A.5.2 (Tychonoff theorem). Let (Xλ)λ∈Λ be a family of com-pact topological spaces. Then X =

∏λ∈Λ Xλ is compact for the product topol-

ogy.

Proof. By Theorem A.5.1, it suffices to prove that every net in X admits acluster point. Let (xi) be net in X. We shall prove that (xi) admits a clusterpoint by applying Zorn’s Lemma. Let us first introduce some notation. Givena subset A ⊂ Λ, we set X(A) =

∏λ∈A Xλ and equip X(A) with the product

topology. If A and B are subsets of Λ such that A ⊂ B, we denote by πBA

the projection map X(B) → X(A). Consider the set E consisting of all pairs(A, a), where A is a subset of Λ and a ∈ X(A) is a cluster point of the net(πΛ

A(xi))i∈I . We partially order E by declaring that two elements (A, a) and(B, b) satisfy (A, a) ≤ (B, b) if and only if A ⊂ B and πB

A (b) = a. The set E isnot empty since it contains the pair (A, a), where A = ∅ and a is the uniqueelement of X(∅). On the other hand, E is inductive. Indeed, suppose that Fis a totally ordered subset of E . Let B =

⋃(A,a)∈F A and consider the unique

element b ∈ X(B) such that πBA (b) = a for all (A, a) ∈ F . Clearly (B, b) ∈ E

and (B, b) is an upper bound for F . By applying Zorn’s Lemma, we deducethat E admits a maximal element (M,m). To prove that the net (xi) admitsa cluster point, it suffices to show that M = Λ. Suppose not and choose anelement λ0 ∈ Λ\M . Since the space Xλ0 is compact, the net (πλ0(xi)) admitsa cluster point a0 ∈ Xλ0 . Let us set M ′ = M ∪{λ0} and consider the elementm′ ∈ X(M ′) defined by πM ′

M (m′) = m and πM ′

{λ0}(m′) = a0. Clearly m′ is a

cluster point of the net (πΛM ′(xi)) and (M,m) ≤ (M ′, m′). This contradicts

the maximality of (M,m) and completes the proof. �

Notes 349

It follows from the definition of compactness that if a topological space Xhas only finitely many open subsets, then X is compact. In particular, everyfinite topological space is compact. Therefore, an immediate consequence ofTychonoff theorem is the following:

Corollary A.5.3 (Tychonoff theorem). Let (Xλ)λ∈Λ be a family of finitetopological spaces. Then X =

∏λ∈Λ Xλ is compact for the product topology.

�

Notes

The original Tychonoff theorem was only stated for product of compact in-tervals. The proof of the general Tychonoff theorem we have presented hereis based on the one given by P. Chernoff in [Che]. Three other proofs maybe found in Kelley’s book [Kel]: a proof using Alexander’s subbase theo-rem, Bourbaki’s proof [Bou] using ultrafilters, and a proof based on universalnets. There is also another proof using non-standard analysis in the book ofA. Robinson [RobA].

Appendix B

Uniform Structures

B.1 Uniform Spaces

Let X be a set. We shall use the following notation. We denote by ΔX thediagonal in X × X, that is,

ΔX = {(x, x) : x ∈ X} ⊂ X × X.

Suppose that R is a subset of X × X (in other words, R is a binary relationon X). For y ∈ X, we define the set R[y] ⊂ X by

R[y] = {x ∈ X : (x, y) ∈ R}.

The inverse−1

R ⊂ X × X of R is defined by

−1

R = {(x, y) : (y, x) ∈ R}.

One says that R is symmetric if it satisfies−1

R = R.If R and S are subsets of X ×X, we define their composite R◦S ⊂ X ×X

by

R ◦ S = {(x, y) : there exists z ∈ X such that (x, z) ∈ R and (z, y) ∈ S}.

Definition B.1.1. Let X be a set. A uniform structure on X is a non–emptyset U of subsets of X × X satisfying the following conditions:

(UN-1) if V ∈ U , then ΔX ⊂ V ;(UN-2) if V ∈ U and V ⊂ V ′ ⊂ X × X, then V ′ ∈ U ;(UN-3) if V ∈ U and W ∈ U , then V ∩ W ∈ U ;

(UN-4) if V ∈ U , then−1

V ∈ U ;(UN-5) if V ∈ U , then there exists W ∈ U such that W ◦ W ⊂ V .


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352 B Uniform Structures

A set X equipped with a uniform structure U is called a uniform spaceand the elements of U are called the entourages of X (see Fig. B.1).

Fig. B.1 An entourage in a uniform space X

Examples B.1.2. (a) Let X be a set. Then U = {X×X} is a uniform structureon X. This uniform structure is called the trivial uniform structure on X. Itis the smallest uniform structure on X.

(b) The discrete uniform structure on a set X is the uniform structurewhose entourages consist of all subsets of X × X containing ΔX . This isthe largest uniform structure on X. It follows from (UN-2) that the discreteuniform structure on X is the only uniform structure on X admitting thediagonal ΔX ⊂ X × X as an entourage.

(c) Suppose that d is a metric on X. For each ε > 0 let Vε ⊂ X×X denotethe set of pairs (x, y) such that d(x, y) < ε. Let U be the set of all subsetsW ⊂ X × X such that one can find ε > 0 for which Vε ⊂ W . Then U is auniform structure on X which is called the uniform structure associated withthe metric d.

A uniform structure U on a set X is said to be metrizable if U is theuniform structure on X associated with some metric on X.

Example B.1.3. If d is the discrete metric on a set X, that is, the metric givenby d(x, y) = 0 if x = y and d(x, y) = 1 otherwise, then the uniform structuredefined by d is the discrete uniform structure on X. Thus the discrete uniformstructure on X is metrizable.

Let X be a uniform space. One easily verifies that is possible to define atopology on X by taking as open sets the subsets Ω ⊂ X which satisfy thefollowing property: for each x ∈ Ω, there exists an entourage V ⊂ X × Xsuch that V [x] = Ω. One says that this topology is the topology associatedwith the uniform structure on X. A subset N ⊂ X is a neighborhood of apoint x ∈ X for this topology if and only if there exists an entourage V such

B.2 Uniformly Continuous Maps 353

that N = V [x]. This topology is Hausdorff if and only if the intersection ofthe entourages of X coincides with the diagonal ΔX ⊂ X × X.

Examples B.1.4. (a) The topology associated with the discrete uniform struc-ture on a set X is the discrete topology on X (every subset is open).

(b) If U is the uniform structure associated with a metric d on a set X,then the topology defined by U coincides with the topology defined by d.

Let U be a uniform structure on a set X.If Y is a subset of X, then UY = {V ∩(Y ×Y ) : V ∈ U} is a uniform struc-

ture on Y , which is said to be induced by U . The topology on Y associatedwith UY is the topology induced by the topology on X associated with U .

A subset B ⊂ U is called a base of U if for each W ∈ U there exists V ∈ Bsuch that V ⊂ W .

Example B.1.5. If d is a metric on X, then B = {Vε : ε > 0}, where Vε ={(x, y) ∈ X × X : d(x, y) < ε}, is a base for the uniform structure on Xassociated with d.

The proof of the following statement is straightforward.

Proposition B.1.6. Let X be a set and let B be a nonempty set of subsetsof X ×X. Then B is a base for some (necessarily unique) uniform structureon X if and only if it satisfies the following properties:

(BU-1) if V ∈ B, then ΔX ⊂ V ;(BU-2) if V ∈ B and W ∈ B, then there exists U ∈ B such that U ⊂ V ∩ W ;

(BU-3) if V ∈ B, then there exists W ∈ B such that W ⊂−1

V ;(BU-4) if V ∈ B, then there exists W ∈ B such that W ◦ W ⊂ V .

��

B.2 Uniformly Continuous Maps

Let X and Y be uniform spaces. A map f : X → Y is called uniformlycontinuous if it satisfies the following condition: for each entourage W of Y ,there exists an entourage V of X such that (f×f)(V ) ⊂ W . Here f×f denotesthe map from X ×X into Y × Y defined by (f × f)(x1, x2) = (f(x1), f(x2))for all (x1, x2) ∈ X × X.

If B (resp. B′) is a base of the uniform structure on X (resp. Y ), then amap f : X → Y is uniformly continuous if and only if it satisfies the followingcondition: for each W ∈ B′, there exists V ∈ B such that (f × f)(V ) ⊂ W .Note that this condition is equivalent to the fact that (f × f)−1(W ) is anentourage of Y for each entourage W of X.


Example B.2.1. Let (X, dX) and (Y, dY ) be metric spaces. Then a mapf : X → Y is uniformly continuous if and only if it satisfies the followingcondition: for each ε > 0, there exists δ > 0 such that dX(x1, x2) < δ impliesdY (f(x1), f(x2)) < ε.

Proposition B.2.2. Let X and Y be uniform spaces. Then every uniformlycontinuous map f : X → Y is continuous (with respect to the topologies onX and Y associated with the uniform structures).

Proof. Suppose that f : X → Y is uniformly continuous. Let x ∈ X and letN ⊂ Y be a neighborhood of f(x). Then there exists an entourage W ofY such that W [f(x)] = N . Since f is uniformly continuous, the set V =(f × f)−1(W ) is an entourage of X. The set V [x] is a neighborhood of x andsatisfies f(V [x]) ⊂ W [f(x)] = N . This shows that f is continuous. ��

A continuous map between uniform spaces may fail to be uniformly con-tinuous. For example, the map x → x2 is not uniformly continuous on R

(equipped with the uniform structure associated with its usual metric). How-ever, this is true when the source space is compact:

Theorem B.2.3. Let X and Y be uniform spaces and suppose that X iscompact. Then every continuous map f : X → Y is uniformly continuous.


Lemma B.2.4 (Lebesgue lemma). Let (Ωi)i∈I be an open cover of a com-pact uniform space X. Then there exists an entourage Λ of X satisfying thefollowing property: for each x ∈ X, there exists an index i ∈ I such thatΛ[x] ⊂ Ωi.

Proof. Let us choose, for each x ∈ X, an index i(x) ∈ I such that x ∈ Ωi(x).Since Ωi(x) is a neighborhood of x, there is an entourage Vx such that Vx[x] =Ωi(x). By (UN-5), we may find an entourage Wx such that Wx ◦ Wx ⊂ Vx.The set Wx[x] is a neighborhood of x for each x ∈ X. By compactness of X,there exists a finite subset A ⊂ X such that X =

⋃a∈A Wa[a]. Let us show

that the entourageΛ =

⋂

a∈A

Wa

has the required property. Let x ∈ X. Choose a point a ∈ A such thatx ∈ Wa[a]. Suppose that y ∈ Λ[x]. Since (x, a) ∈ Wa and (y, x) ∈ Λ ⊂ Wa,we have (y, a) ∈ Wa ◦ Wa ⊂ Va. Thus y ∈ Va[a]. Since Va[a] ⊂ Ωi(a), thisshows that Λ[x] ⊂ Ωi(a). ��

Proof of Theorem B.2.3. Let f : X → Y be a continuous map and let W bean entourage of Y . By (UN-3), (UN-4), and (UN-5), we may find a symmetricentourage S of Y such that S ◦ S ⊂ W . Since f is continuous, there exists,for each x ∈ X, an open neighborhood Ωx of X such that f(Ωx) ⊂ S[f(x)].

B.3 Product of Uniform Spaces 355

By Lemma B.2.4, we may find an entourage Λ of X such that, for eachy ∈ X, there exists x ∈ X such that Λ[y] ⊂ Ωx. Suppose that (x1, x2) ∈ Λ.Choose a ∈ X such that Λ[x2] ⊂ Ωa. Since x1 and x2 are in Λ[x2], we deducethat the points f(x1) and f(x2) are in S[f(a)]. It follows that (f(x1), f(a))and (f(a), f(x2)) are in S, and hence (f(x1), f(x2)) ∈ S ◦ S ⊂ W . Thus(f × f)(Λ) ⊂ W . This shows that f is uniformly continuous. ��

Let X and Y be uniform spaces.One says that a map f : X → Y is a uniform isomorphism if f is bijective

and both f and f−1 are uniformly continuous.One says that a map f : X → Y is a uniform embedding if f is injective

and induces a uniform isomorphism between X and f(X) ⊂ Y .

Proposition B.2.5. Let X and Y be uniform spaces with X compact and YHausdorff. Suppose that f : X → Y is a continuous injective map. Then f isa uniform embedding.

Proof. As X is compact and Y is Hausdorff, f induces a homeomorphismfrom X onto f(X). This homeomorphism is a uniform isomorphism by The-orem B.2.3. ��

B.3 Product of Uniform Spaces

Let X be a set. Suppose that we are given a family (Xλ)λ∈Λ of uniformspaces and a family (fλ)λ∈Λ of maps fλ : X → Xλ. Then the initial uniformstructure associated with these data is the smallest uniform structure on Xsuch that all maps fλ : X → Xλ, λ ∈ Λ, are uniformly continuous.

In the particular case when X =∏

λ∈Λ Xλ and fλ : X → Xλ is the projec-tion map, the associated initial uniform structure on X is called the productuniform structure. A base of entourages for the product uniform structureon X is obtained by taking all subsets of X × X which are of the form

∏

λ∈Λ

Vλ ⊂∏

λ∈Λ

Xλ × Xλ

=

(∏

λ∈Λ

Xλ

)×

(∏

λ∈Λ

Xλ

)

= X × X,

where Vλ ⊂ Xλ × Xλ is an entourage of Xλ and Vλ = Xλ × Xλ for all butfinitely many λ ∈ Λ.

When each Xλ is endowed with the discrete uniform structure, the prod-uct uniform structure on X =

∏λ∈Λ Xλ is called the prodiscrete uniform

structure.


B.4 The Hausdorff-Bourbaki Uniform Structureon Subsets

Let X be a uniform space with uniform structure U . In this section, weconstruct a uniform structure on the set P(X) of all subsets of X.

We shall use the following notation. Suppose that R is a subset of X ×X.Given a subset Y ⊂ X, we set

R[Y ] =⋃

y∈Y

R[y] = {x ∈ X : (x, y) ∈ R for some y ∈ Y }, (B.1)

and we define the subset R ⊂ P(X) × P(X) by

R = {(Y,Z) ∈ P(X) × P(X) : Y ⊂ R[Z] and Z ⊂ R[Y ]} . (B.2)

Proposition B.4.1. The set {V : V ∈ U} is a base for a uniform structureon P(X).

Proof. Let us check that the conditions of Proposition B.1.6 are satisfied.Property (BU-1) follows from the fact that we have Y ⊂ V [Y ] for all Y ∈P(X) and V ∈ U since ΔX ⊂ V . Then observe that

(V ∩ W )[Y ] = {x ∈ X : (x, y) ∈ V ∩ W for some y ∈ Y }⊂ {x ∈ X : (x, yV ) ∈ V and (x, yW ) ∈ W for some yV , yW ∈ Y }= V [Y ] ∩ W [Y ],

for all V,W ∈ U and Y ⊂ X. Therefore we have V ∩ W ⊂ V ∩ W for allV,W ∈ U , so that (BU-2) is satisfied. Property (BU-3) follows from the factthat the set V is a symmetric subset of P(X)×P(X) for each V ∈ U . Finally,let us verify (BU-4). Let V ∈ U and take W ∈ U such that W ◦ W ⊂ V . Weclaim that W ◦ W ⊂ V . To see this, let (Y,Z) ∈ W ◦ W . This means thatthere exists T ∈ P(X) such that (Y, T ) ∈ W and (T,Z) ∈ W . In particularwe have Y ⊂ W [T ] and T ⊂ W [Z]. Thus, given y ∈ Y , there exist t ∈ T andz ∈ Z such that (y, t) ∈ W and (t, z) ∈ W . We have (y, z) ∈ W ◦W ⊂ V . Thisshows that Y ⊂ V [Z]. Similarly, we get Z ⊂ V [Y ] by using Z ⊂ W [T ] andT ⊂ W [Y ]. We deduce that (Y,Z) ∈ V . This proves the claim. Consequently,(BU-4) is satisfied. ��

The uniform structure on P(X) admitting {V : V ∈ U} as a base is calledthe Hausdorff-Bourbaki uniform structure on P(X) and the topology asso-ciated with this uniform structure is called the Hausdorff-Bourbaki topologyon P(X).

Remarks B.4.2. (a) The empty set is an isolated point in P(X).(b) One easily checks that the map i : X → P(X) defined by i(x) = {x}

is a uniform embedding.

B.4 The Hausdorff-Bourbaki Uniform Structure on Subsets 357

Proposition B.4.3. Let X be a uniform space. Let Y and Z be closed subsetsof X. Suppose that there is a net (Ti)i∈I of subsets of X which converges toboth Y and Z with respect to the Hausdorff-Bourbaki topology on P(X). Thenone has Y = Z.

Proof. Let y ∈ Y and let Ω be a neighborhood of y in X. Then there is asymmetric entourage V of X such that V [y] ⊂ Ω. Choose an entourage Wof X such that W ◦ W ⊂ V . Since the net (Ti)i∈I converges to both Y andZ, we can find an element i0 ∈ I such that ⊂ W [Ti0 ] and Ti0 ⊂ W [Z]. Thus,there exist t ∈ Ti0 and z ∈ Z such that (y, t) ∈ W and (t, z) ∈ W . Thisimplies (y, z) ∈ W ◦ W ⊂ V . As V is symmetric, it follows that (z, y) ∈ Vand hence z ∈ V [y] ⊂ Ω. This shows that y is in the closure of Z. Since Zis closed in X, we deduce that Y ⊂ Z. By symmetry, we also have Z ⊂ Y .Consequently, Y = Z. ��

By using Proposition A.2.2, we immediately deduce from Proposition B.4.3the following:

Corollary B.4.4. Let X be a uniform space. Then the topology induced bythe Hausdorff-Bourbaki topology on the set of closed subsets of X is Haus-dorff. ��

Remark B.4.5. Suppose that (X, d) is a metric space and let Cb(X) denotethe set consisting of all closed bounded subsets of X. For x ∈ X and r > 0,denote by B(x, r) the open ball of radius r centered at x. Then it is notdifficult to verify that the map δ : Cb(X) × Cb(X) → R defined by

δ(Y,Z) = inf{r > 0 : Z ⊂⋃

y∈Y

B(y, r) and Y ⊂⋃

z∈Z

B(z, r)}

is a metric on Cb(X) and that the uniform structure associated with δ is theuniform structure induced by the Hausdorff-Bourbaki structure on the set ofsubsets of X. The metric δ is called the Hausdorff metric on Cb(X).

Proposition B.4.6. Let X and Y be uniform spaces and let f : X → Y bea uniformly continuous map. Then the map f∗ : P(X) → P(Y ) which sendseach subset A ⊂ X to its image f(A) ⊂ Y is uniformly continuous withrespect to the Hausdorff-Bourbaki uniform structures on P(X) and P(Y ).

Proof. Let W be an entourage of Y and let

W = {(B1, B2) ∈ P(Y ) × P(Y ) : B2 ⊂ W [B1] and B1 ⊂ W [B2]}

be the associated entourage of P(Y ). Since f is uniformly continuous, thereis an entourage V of X such that

(f × f)(V ) ⊂ W.


Suppose that (A1, A2) ∈ V , that is, A2 ⊂ V [A1] and A1 ⊂ V [A2]. If a1 ∈ A1,then there exists a2 ∈ A2 such that (a1, a2) ∈ V and hence (f(a1), f(a2)) ∈W . Therefore, we have f∗(A1) ⊂ W [f∗(A2)]. Similarly, we get f∗(A2) ⊂W [f∗(A1)]. It follows that (f∗(A1), f∗(A2)) ∈ W . This shows that (f∗ ×f∗)(V ) ⊂ W . Consequently, f∗ is uniformly continuous. ��

Notes

Uniform structures were introduced by Andre Weil [Weil]. The reader is re-ferred to [Bou, Ch. 2], [Kel, Ch. 6], and [Jam] for a detailed exposition ofthe general theory of uniform spaces. The Hausdorff-Bourbaki uniform struc-ture on the set of subsets of a uniform space was introduced in exercises byBourbaki (see [Bou, ch. II exerc. 5 p. 34 and exerc. 6 p. 36]).

Appendix C

Symmetric Groups

C.1 The Symmetric Group

Let X be a set. A permutation of X is a bijective map σ : X → X. Let Sym(X)denote the set of all permutations of X. We equip Sym(X) with a groupstructure by defining the product σ1σ2 of two elements σ1, σ2 ∈ Sym(X)as the composite map σ1 ◦ σ2. The associative property follows from theassociativity of the composition of maps. The identity map IdX : X → X isthe identity element and the inverse of σ ∈ Sym(X) is the inverse map σ−1.The group Sym(X) is called the symmetric group on X.

Remark C.1.1. Suppose that f : X → Y is a bijection from a set X onto aset Y . Then the map f∗ : Sym(X) → Sym(Y ) defined by f∗(σ) = f ◦ σ ◦ f−1

is a group isomorphism. As a consequence, symmetric groups on equipotentsets are isomorphic.

Theorem C.1.2 (Cayley’s theorem). Every group G is isomorphic to asubgroup of Sym(G).

Proof. Let G be a group. Given g ∈ G denote by Lg : G → G the left multi-plication by g, that is, the map defined by Lg(h) = gh for all h ∈ G. Observethat Lg ∈ Sym(G). Indeed, Lg is bijective since, given h, h′ ∈ G, we haveLg(h) = h′ if and only if h = g−1h′. Let us show that the map

L : G → Sym(G)g �→ Lg

is a group homomorphism. Given g, g′, h ∈ G we have Lgg′(h) = gg′h =Lg(g′h) = Lg(Lg′(h)) which shows that Lgg′ = LgLg′ . Moreover, L is injec-tive. Indeed, if g ∈ ker(L), that is, Lg = IdG, we have g = g · 1G = Lg(1G) =1G. This shows that ker(L) = {1G}, and therefore L is injective. It followsthat G is isomorphic to L(G) ⊂ Sym(G). ��


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360 C Symmetric Groups

C.2 Permutations with Finite Support

Let X be a set. The support of a permutation σ ∈ Sym(X) is the set S(σ) ⊂ Xconsisting of all x ∈ X such that σ(x) = x.

Proposition C.2.1. Let σ, τ ∈ Sym(X). Then

(i) σ(S(σ)) = S(σ);(ii) S(σ) = S(σ−1);(iii) S(στ) ⊂ S(σ) ∪ S(τ);(iv) if S(σ) ∩ S(τ) = ∅, then στ = τσ;(v) S(τστ−1) = τ(S(σ)).

Proof. (i) Let x ∈ X. Then, x ∈ S(σ), that is, σ(x) = x, if and only ifσ(σ(x)) = σ(x), that is, σ(x) ∈ S(σ).

(ii) This follows from the fact that σ(x) = x if and only if x = σ−1(x).(iii) Suppose that x ∈ X \ (S(σ) ∪ S(τ)). Then σ(x) = x = τ(x) and

therefore (στ)(x) = σ(τ(x)) = σ(x) = x. It follows that x /∈ S(στ).(iv) Let x ∈ X. Suppose first that x ∈ X \ (S(σ) ∪ S(τ)). Then, by (iii),

x /∈ S(στ) and x /∈ S(τσ), so that σ(τ(x)) = x = τ(σ(x)). Suppose now thatx is in the support of one of the two permutations, say x ∈ S(σ). It thenfollows from our assumptions that x /∈ S(τ) and therefore τ(x) = x. Also,by (i), σ(x) ∈ S(σ) and therefore, again by our assumptions, σ(x) /∈ S(τ), sothat τ(σ(x)) = σ(x). We thus have τ(σ(x)) = σ(x) = σ(τ(x)). It follows thatστ = τσ.

(v) Let x ∈ X. Then, x ∈ S(σ), that is σ(x) = x, if and only if,(τστ−1)(τ(x)) = τ(σ(x)) is not equal to τ(x). Thus, x ∈ S(σ) if and only ifτ(x) ∈ S(τστ−1). ��

Let Sym0(X) denote the subset of Sym(X) consisting of all permutationsof X with finite support.

Proposition C.2.2. Let X be a set. Then the set Sym0(X) is a normalsubgroup of Sym(X).

Proof. The support of the identity map IdX is the empty set and thereforeIdX ∈ Sym0(X). By Proposition C.2.1(ii), if σ ∈ Sym0(X) then σ−1 ∈Sym0(X). On the other hand, by Proposition C.2.1(iii), the set Sym0(X) isclosed under multiplication. Thus Sym0(X) is a subgroup of Sym(X). Finally,from Proposition C.2.1(v) we deduce that if σ ∈ Sym0(X) then τστ−1 ∈Sym0(X) for all τ ∈ Sym(X). It follows that Sym0(X) is a normal subgroupof Sym(X). ��

Let r ≥ 2 be an integer and let x1, x2, . . . , xr be distinct elements in X.We denote by

γ = (x1 x2 · · · xr)

C.2 Permutations with Finite Support 361

the permutation of X that maps x1 to x2, x2 to x3, . . . , xr−1 to xr, xr to x1,and maps each element of X \ {x1, x2, . . . , xr} to itself. Thus, the supportof γ is the set {x1, x2, . . . , xr}. One says that γ is a cycle of length r, or anr-cycle. A 2-cycle is called a transposition. Observe that

γ = (x γ(x) γ2(x) · · · γr−1(x))

for all x ∈ {x1, x2, . . . , xr} and that the inverse of γ is the r-cycle

γ−1 = (xr xr−1 · · · x2 x1).

Proposition C.2.3. Let X be a set and let σ ∈ Sym0(X). Then there existsan integer n ≥ 0 and cycles γ1, γ2, . . . , γn with pairwise disjoint supports suchthat

σ = γ1γ2 · · · γn. (C.1)

Moreover, such a factorization is unique up to a permutation of the factors.

Proof. If σ = IdX then n = 0 and there is nothing to prove.Suppose now that σ = IdX . Let S = S(σ) ⊂ X be the support of σ.

We introduce an equivalence relation on S by setting x ∼ y if and only ifthere exists k ∈ Z such that y = σk(x). Let S = X1

∐X2

∐· · ·

∐Xn be the

partition of S into the equivalence classes of ∼ and let us set ri = |Xi| for1 ≤ i ≤ n. Note that ri ≥ 2 for all i. For each i = 1, 2, . . . , n, choose a repre-sentative xi ∈ Xi. Observe that the elements xi, σ(xi), σ2(xi), . . . , σri−1(xi)are all distinct since otherwise the class of xi would have less that ri elements.Consider the cycle

γi = (xi σ(xi) σ2(xi) · · · σri−1(xi)).

The support of γi is Xi. Thus, the cycles γ1, γ2, . . . , γn have pairwise disjointsupports. Clearly σ = γ1γ2 · · · γn.

Suppose now that σ = δ1δ2 · · · δs, where δ1, δ2, . . . , δs are cycles with pair-wise disjoint supports. The supports of the cycles δi are the equivalence classesof ∼. We deduce that s = n. Moreover, up to a permutation of the fac-tors, we may suppose that the support of γi equals the support of δi for alli = 1, 2, . . . , n. We have

γi = (xi γi(xi) γ2i (xi) · · · γri−1

i (xi))

= (xi σ(xi) σ2(xi) · · · σri−1(xi))

= (xi δi(xi) δ2i (xi) · · · δri−1

i (xi))= δi

and this completes the proof. ��

Corollary C.2.4. Every permutation in Sym0(X) can be expressed as aproduct of transpositions.


Proof. First observe that every cycle is a product of transpositions. Indeed,for all distinct x1, x2, . . . , xr ∈ X, we have

(x1 x2 · · · xr) = (x1 xr)(x1 xr−1) · · · (x1 x3)(x1 x2). (C.2)

By applying Proposition C.2.3 we deduce that every σ ∈ Sym0(X) is a prod-uct of transpositions. ��

C.3 Conjugacy Classes in Sym0(X)

Let G be a group and let H ⊂ G be a subgroup. We recall that two elementsh and h′ in H are said to be conjugate in G (resp. in H) if there exists anelement g ∈ G (resp. g ∈ H) such that h′ = ghg−1. Clearly conjugacy in G(resp. in H) defines an equivalence relation on H.

Proposition C.3.1. Let X be a set. Let γ ∈ Sym0(X) be a cycle of length rand let σ ∈ Sym(X). Then σγσ−1 is also a cycle of length r. More precisely,if γ = (x1 x2 · · · xr), then σγσ−1 equals the cycle

(σ(x1) σ(x2) · · · σ(xr)). (C.3)

Proof. First observe that, by Proposition C.2.1(v) the support of σγσ−1 isthe set {σ(x1), σ(x2), . . . , σ(xn)}. Given 1 ≤ i ≤ r, we have

(σγσ−1)(σ(xi)) = (σγ)(xi) = σ(xi+1),

where r + 1 = 1. It follows that σγσ−1 = (σ(x1) σ(x2) · · · σ(xr)). ��

Let σ ∈ Sym0(X). The type of σ is the sequence t(σ) = (tr)r≥2 where tris the number of cycles of length r in the factorization of σ as a product ofcycles with pairwise disjoint supports (cf. Proposition C.2.3).

Proposition C.3.2. Let X be a set and let σ and σ′ in Sym0(X). Then thefollowing conditions are equivalent:

(a) σ and σ′ are conjugate in Sym0(X);(b) σ and σ′ are conjugate in Sym(X);(c) σ and σ′ have the same type.

Proof. The implication (a) ⇒ (b) is obvious. Suppose (b). let

σ = γ1γ2 · · · γr (C.4)

be the factorization of σ as a product of cycles with disjoint supports and letα ∈ Sym(X) be a permutation such that σ′ = ασα−1. From (C.4) we deducethat σ′ = (αγ1α

−1)(αγ2α−1) · · · (αγrα

−1). It follows from Proposition C.3.1that t(σ′) = t(σ). This shows (b) ⇒ (c).

C.4 The Alternating Group 363

Finally, suppose that

σ = (x1 x2 · · · xr1)(y1 y2 · · · yr2) · · · (z1 z2 · · · zr�)

andσ′ = (x′

1 x′2 · · · x′

r1)(y′

1 y′2 · · · y′

r2) · · · (z′1 z′2 · · · z′r�

)

are two permutations of the same type. Consider a permutation α, withsupport the union of the supports of σ and σ′, which maps xi to x′

i for alli = 1, 2, . . . , r1, yj to y′

j , for all j = 1, 2, . . . , r2, . . ., and zk to z′k for all k =1, 2, . . . , r�. Note that α ∈ Sym0(X). Then (cf. the proof of Proposition C.3.1)ασα−1 = α′. It follows that σ and σ′ are conjugate in Sym0(X). ��

C.4 The Alternating Group

Proposition C.4.1. Let X be a set and let σ ∈ Sym0(X). Suppose that σcan be expressed as a product of n transpositions. Then the parity of n onlydepends on σ.

Proof. Suppose that σ can be expressed both as a product of an even and asa product of an odd number of transpositions. Then, the same holds for σ−1.It follows that choosing an even writing for σ and an odd one for σ−1, wecan write the identity element IdX = σσ−1 as a product of an odd numberof transpositions, say

IdX = τ1τ2 · · · τ2m+1. (C.5)

Let x be an element in X appearing in the support of one of the transpo-sitions τi in (C.5).

As transpositions with disjoint support commute (cf. Proposition C.2.1(iv))and (y z)(x z) = (x y)(y z) for all distinct elements y, z in X \ {x}, we canmove all transpositions of the form (x y) to the left in (C.5). In other words,we can write the identity IdX as a product of 2m + 1 transpositions

IdX = τ ′1τ

′2 · · · τ ′

rτ′r+1 · · · τ ′

2m+1 (C.6)

where 1 ≤ r ≤ 2m + 1 and x belongs to the support of τ ′i if and only if

1 ≤ i ≤ r. Let τ ′r = (x y) and observe that it cannot appear only once

in the product (C.6), otherwise the element x would be mapped onto y,while it has to remain fixed, since that product is the identity. It followsthat there exists 1 ≤ j ≤ r − 1 such that τ ′

j = τ ′r and τ ′

i = τ ′r for all

i = j + 1, j + 2, . . . , r − 1. Now, for all j + 1 ≤ i ≤ r − 1, if τ ′i = (x z), we

have τ ′iτ

′r = (x z)(x y) = (x y)(y z) = τ ′

r(y z). Thus, we can move to the leftthe transposition τ ′

r next to τ ′j and cancel them out, without changing the

value of the product. Repeating this operation, we reduce every time by 2 the


number of transpositions in (C.6). Eventually, we reach a single transposition.This clearly yields a contradiction. ��

Consider the mapε : Sym0(X) → {−1, 1} (C.7)

defined by setting ε(σ) = 1 if σ is a product of an even number of permu-tations and ε(σ) = −1 otherwise. Note that ε is well defined by virtue ofProposition C.4.1. It is obvious that ε is a group homomorphism. Moreover,ε is surjective if X has at least two elements.

The normal subgroup ker(ε) ⊂ Sym0(X) is called the alternating group onX and it is denoted by Sym+

0 (X).

Proposition C.4.2. Let X be a set. An r-cycle is in Sym+0 (X) if and only

if r is odd. In particular, every 3-cycle belongs to Sym+0 (X).

Proof. We have seen (cf. (C.2)) that and r-cycle is a product of r − 1 trans-positions. ��

Recall that a nontrivial group G is simple if the only normal subgroups ofG are the trivial subgroup {1G} and G itself.

Theorem C.4.3. Let X be a set having at least five distinct elements. Thenthe group Sym+

0 (X) is simple.

Proof. Every element of Sym+0 (X) is a product of permutations of the form

(s t)(u v) or (s t)(s u), where s, t, u and v are distinct elements of X. Since(s t)(u v) = (s u t)(s u v) and (s t)(s u) = (s u t), it follows that Sym+

0 (X)is generated by the set of all 3-cycles.

Let x and y be two distinct elements in X. Clearly, any 3-cycle is of one ofthe forms (x y s), (x s y), (x s t), (y t u), or (s t u), where s, t, u are distinctelements in X \ {x, y}. We have

(x s y) = (x y s)2,

(x s t) = (x y t)(x s y) = (x y t)(x y s)2,

(y s t) = (x t y)(x y s) = (x y t)2(x y s),

(s t u) = (x s y)(x y u)(x t y)(x y s) = (x y s)2(x y u)(x y t)2(x y s).

This shows that Sym+0 (X) is generated by the 3-cycles (x y z), where z ∈

X \ {x, y}.Let now N ⊂ Sym+

0 (X) be a nontrivial normal subgroup. Let us show thatN = Sym+

0 (X). We distinguish a few cases (corresponding to the differentpossible cycle structures of a nontrivial element in N).

Case 1. N contains a 3-cycle (x y s). Then, for any z ∈ X \{x, y, s} we havethat the 3-cycle

C.4 The Alternating Group 365

(x y z) = (x y)(s z)(x y s)2(s z)(x y) = [(x y)(s z)]−1(x y z)2[(s z)(x y)]

belongs to N . From the preceding part of the proof, we deduce that N =Sym+

0 (x).Case 2. N contains an element σ whose factorization as product of cycleswith disjoint supports contains a cycle (x1 x2 · · · xr) of length r ≥ 4. Writeσ = (x1 x2 · · · xr)ρ, where ρ ∈ Sym0(X) is the product of the remainingcycles. Consider the cycle γ = (x1 x2 x3). Then, σ(γσ−1γ−1) ∈ N , as N is anormal subgroup. By Proposition C.3.1, we have

σ(γσ−1γ−1) = (σγσ−1)γ−1 = (x2 x3 x4)(x3 x2 x1) = (x1 x4 x2).

Thus, N contains a 3-cycle and, by Case 1, N = Sym+0 (X).

Case 3. N contains an element σ whose factorization as a product of cycleswith disjoint supports contains at least two cycles (x1 x2 x3) and (x4 x5 x6)of length 3. Write σ = (x1 x2 x3)(x4 x5 x6)ρ, where ρ ∈ Sym0(X) is theproduct of the remaining cycles. Consider the cycle γ = (x1 x2 x4). Then, asabove, σγσ−1γ−1 ∈ N . But, again by Proposition C.3.1, we have

σγσ−1γ−1 = (x2 x3 x5)(x4 x2 x1) = (x1 x4 x3 x5 x2).

Thus, N contains a 5-cycle and by Case 2, N = Sym+0 (X).

Case 4. N contains an element σ which factorizes as a product of onesingle 3-cycle (x1 x2 x3) and transpositions with disjoint supports. Writeσ = (x1 x2 x3)ρ, where ρ ∈ Sym0(X) is the product of the transpositions.Note that ρ2 = IdX . Then σ2 ∈ N and

σ2 = (x1 x2 x3)ρ(x1 x2 x3)ρ = (x1 x2 x3)2ρ2 = (x1 x3 x2).

Thus, again as in Case 1, N = Sym+0 (X).

Case 5. N contains an element σ which is the product of an (even) numberof transpositions with disjoint supports. We can write σ = (x1 x2)(x3 x4)ρwhere ρ ∈ Sym0(X) satisfies ρ2 = IdX . Consider the cycle γ = (x1 x2 x3).Then, the element π = σγσ−1γ−1 belongs to N . By Proposition C.3.1,

π = σγσ−1γ−1 = (x2 x1 x4)(x1 x3 x2) = (x1 x3)(x2 x4).

Since X has at least five distinct elements, there exists an element y ∈ X \{x1, x2, x3, x4}. Set δ = (x1 x3 y). Then, π(δπ−1δ−1) ∈ N . But, one moretime by Proposition C.3.1,

π(δπ−1δ−1) = (πδπ−1)δ−1 = (x3 x1 y)(y x3 x1) = (x1 x3 y).

We are again in Case 1, and therefore N = Sym+0 (X).

In all cases, N = Sym+0 (X), and this shows that Sym+

0 (X) is simple. ��


Note that if X is a finite set, then Sym0(X) = Sym(X). Given an integern ≥ 1, we denote by Symn the symmetric group of the set {1, 2, . . . , n}. Thegroup Symn is called the symmetric group of degree n. By Remark C.1.1, ifX is a finite set with |X| = n, we have Symn

∼= Sym(X).The subgroup Sym+

0 ({1, 2, . . . , n}) is called the alternating group of degreen and it is denoted by Sym+

n .

Remark C.4.4. The groups Sym+1 (= Sym1) and Sym+

2 are trivial groups.The group Sym+

3 = {Id{1,2,3}, (1 2 3), (1 3 2)} is cyclic of order 3 andtherefore it is a simple group.

On the other hand, the subgroup

K = {Id{1,2,3,4}, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ⊂ Sym+4

has index two in Sym+4 . It follows that K is a normal subgroup of Sym+

4 .Therefore, Sym+

4 is not a simple group.From Theorem C.4.3 we deduce that the group Sym+

n is simple for alln ≥ 5.

Appendix D

Free Groups

D.1 Concatenation of Words

Let A be a set.A word on the alphabet set A is an element of the set

A∗ =⋃

n∈N

An,

where An is the Cartesian product of A with itself n times, that is, the setconsisting of all n-tuples (a1, a2, . . . , an) with ak ∈ A for 1 ≤ k ≤ n. Theunique element of A0 is denoted by ε and is called the empty word .

The concatenation of two words w = (a1, a2, . . . , am) ∈ Am and w′ =(a′

1, a′2, . . . , a

′n) ∈ An is the word ww′ ∈ Am+n defined by

ww′ = (a1, a2, . . . , am, a′1, a

′2, . . . , a

′n).

We have εw = wε = w and (ww′)w′′ = w(w′w′′) for all w, w′, w′′ ∈ A∗. Thus,A∗ is a monoid for the concatenation product whose identity element is theempty word ε.

Observe that each word w = (a1, a2, . . . , an) ∈ An may be uniquely writtenas a product of elements of A = A1, namely w = a1a2 · · · an.

D.2 Definition and Construction of Free Groups

Definition D.2.1. A based free group is a triple (F, X, i), where F is a group,X is a set, and i : X → F is a map from X to F satisfying the followinguniversal property: for every group G and any map f : X → G, there existsa unique homomorphism φ : F → G such that f = φ ◦ i.


367

http://dx.doi.org/10.1007/978-3-642-14034-1_12

368 D Free Groups

F

φ

X

i

fG

A group F is called free if there exist a set X and a map i : X → F such thatthe triple (F, X, i) is a based free group. One then says that (X, i) is a freebase for F and that F is a free group based on (X, i).

Remarks D.2.2. (a) If (F, X, i) is a based free group and α : Y → X is abijective map from a set Y onto X, then the triple (F, Y, i◦α) is also a basedfree group. Indeed, if f : Y → G is a map from Y into a group G, then thereis a unique homomorphism φ : F → G such that f = φ ◦ i ◦ α, namely theunique homomorphism φ : F → G satisfying f ◦ α−1 = φ ◦ i.

(b) If (F, X, i) is a based free group and ψ : F → F ′ is an isomorphism fromF onto a group F ′, then the triple (F ′, X, ψ ◦ i) is a based free group. Indeed,if f : X → G is a map from X into a group G, then there exists a uniquehomomorphism φ′ : F ′ → G such that f = φ′ ◦ ψ ◦ i, namely the homomor-phism given by φ′ = φ ◦ ψ−1, where φ : F → G is the unique homomorphismsatisfying f = φ ◦ i.

Proposition D.2.3. Let (F, X, i) be a based free group. Then the followinghold:

(i) the map i is injective;(ii) the set i(X) generates the group F ;(iii) the triple (F,X ′, i′), where X ′ = i(X) and i′ : X ′ → F is the inclusion

map, is also a based free group.

Proof. (i) Let x1 and x2 be two distinct elements in X. Consider the mapf : X → Z/2Z defined by f(x) = 0 if x �= x2 and f(x2) = 1. Since (F, X, i)is a based free group, there exists a homomorphism φ : F → Z/2Z such thatf = φ ◦ i. As f(x1) �= f(x2), this implies i(x1) �= i(x2). Therefore, the map iis injective.

(ii) Denote by H the subgroup of F generated by i(X). Consider the mapi∗ : X → H defined by i∗(x) = i(x) for all x ∈ X. Since (F, X, i) is a basedfree group, there exists a homomorphism φ : F → H such that i∗ = φ ◦ i.Consider now the inclusion map ρ : H → F . The homomorphisms IdF andρ◦φ satisfy IdF ◦i = ρ◦φ◦ i. By uniqueness, we get IdF = ρ◦φ. This impliesthat ρ is surjective, that is, H = F . Therefore, i(X) generates F .

(iii) The fact that (F,X ′, i′) is a based free group immediately follows fromRemark D.2.2(a) since i′ = i ◦ j−1 where j : X → X ′ is the bijective mapdefined by j(x) = i(x) for all x ∈ X. ��

From Proposition D.2.3(iii), we deduce that if F is a free group then thereexists a subset X ⊂ F such that the triple (F, X, i), where i : X → F is theinclusion map, is a based free group. Such a subset X ⊂ F is then called afree base subset , or simply a base, for F .

D.2 Definition and Construction of Free Groups 369

Proposition D.2.4. Let (F, X, i) be a based free group and let Y ⊂ X. LetK denote the subgroup of F generated by Y and let j : Y → K be the mapdefined by j(y) = i(y) for all y ∈ Y . Then (K, Y, j) is a based free group.

Proof. Let G be a group and let f : Y → G be a map. Let us show that thereexists a unique homomorphism φ : K → G satisfying f = φ ◦ j. Uniquenessfollows from the fact that j(Y ) = i(Y ) generates K. Choose a map f ′ : X → Gextending f . As (F, X, i) is a based free group, there exists a homomorphismφ′ : F → G such thatf ′ = φ′ ◦ i. Then φ = φ′|K : K → G satisfies f = φ ◦ j.This proves that (K, Y, j) is a based free group. ��

In the case when i is an inclusion map, this gives us the following:

Corollary D.2.5. Let F be a free group with base X ⊂ F . Let Y ⊂ X andlet K denote the subgroup of F generated by Y . Then K is a free group withbase Y . ��

Proposition D.2.6. Let (F1, X1, i1) and (F2, X2, i2) be two based free groups.Suppose that there is a bijective map u : X1 → X2. Then there exists a uniquegroup isomorphism ϕ : F1 → F2 satisfying ϕ ◦ i1 = i2 ◦ u.

F1ϕ−−−−→ F2

i1

�⏐⏐�⏐⏐i2

X1 −−−−→u

X2

Proof. Since (F1, X1, i1) is a based free group, there exists a unique homo-morphism ϕ : F1 → F2 such that i2 ◦ u = ϕ ◦ i1. It suffices to show that ϕis bijective. To see this, we now use the fact that (F2, X2, i2) is a based freegroup. This implies that there exists a homomorphism ϕ′ : F2 → F1 such thati1◦u−1 = ϕ′◦i2. The maps IdF1 and ϕ′◦ϕ are endomorphisms of F1 satisfyingIdF1 ◦i1 = i1 and (ϕ′ ◦ ϕ) ◦ i1 = i1. Since (F1, X1, i1) is a based free group,it follows that IdF1 = ϕ′ ◦ ϕ by uniqueness. Similarly, we get ϕ ◦ ϕ′ = IdF2 .This shows that ϕ is bijective. ��

Theorem D.2.7. Let X be a set. Then there exist a group F and a mapi : X → F such that the triple (F, X, i) is a based free group.

Proof. Let X ′ be a disjoint copy of X, that is, a set X ′ such that X∩X ′ = ∅

together with a bijective map γ : X → X ′. Let A = X ∪ X ′. For each a ∈ A,define the element a ∈ A by setting

a =

{γ(a) if a ∈ X,

γ−1(a) if a ∈ X ′.

Observe that the map a �→ a is an involution of A, that is, it satisfies ˜a = afor all a ∈ A.

370 D Free Groups

Consider now the set A∗ consisting of all words on the alphabet set A (seeSect. D.1). Recall that A∗ is a monoid for the concatenation product whoseidentity element is the empty word ε.

We say that a word w ∈ A∗ may be obtained from a word w′ ∈ A∗ by anelementary reduction if there exist an element a ∈ A and words u, v ∈ A∗

such that w = uv and w′ = uaav. Given words w, w′ ∈ A∗, we write w ∼ w′

if either w may be obtained from w′ by an elementary reduction or w′ maybe obtained from w by an elementary reduction. Finally, we define a relation≡ on A∗ by writing w ≡ w′ for w, w′ ∈ A∗ if and only if there exist aninteger n ≥ 1 and a sequence of words w1, w2, . . . , wn ∈ A∗ with w1 = w andwn = w′ such that wi ∼ wi+1 for each i = 1, 2, . . . , n − 1. It is immediate tocheck that ≡ is an equivalence relation on A∗. Denote by [w] the equivalenceclass of an element w ∈ A∗ and consider the quotient set F = A∗/ ≡, thatis, the set consisting of all equivalence classes [w], w ∈ A∗. Observe that ifu, u′, v ∈ A∗ and u ∼ u′ then uv ∼ u′v and vu ∼ vu′. It follows from thisobservation that if the words u, v, u′, v′ ∈ A∗ satisfy u ≡ u′ and v ≡ v′ thenuv ≡ u′v′, so that we can define the product of [u] and [v] in F by setting

[u][v] = [uv].

Let us show that this product gives a group structure on F . The associativ-ity immediately follows from the associativity of the concatenation productin A∗. Indeed, for all u, v, w ∈ A∗, we have

([u][v])[w] = [uv][w] = [(uv)w] = [u(vw)] = [u][vw] = [u]([v][w]).

On the other hand, for all w ∈ A∗, we have

[ε][w] = [εw] = [w] and [w][ε] = [wε] = [w]

which shows that [ε] is an identity element. Finally, let us show that everyelement in F admits an inverse. For w = a1a2 · · · an ∈ A∗, where n ≥ 0 andak ∈ A for 1 ≤ k ≤ n, define the word w ∈ A∗ by

w = anan−1 · · · a1.

Observe that

ww = a1a2 · · · anan · · · a2a1

∼ a1 · · · an−1an−1 · · · a1

...∼ a1a2a2a1

∼ a1a1

∼ ε.

D.2 Definition and Construction of Free Groups 371

Thus, we have ww ≡ ε. This gives us [w][w] = [ww] = [ε]. Similarly, we get[w][w] = [ε]. It follows that [w] is an inverse of [w] for the product operationin F . This proves that F is a group.

Consider the map i : X → F defined by i(x) = [x] for all x ∈ X. Ifw = a1a2 · · · an ∈ A∗, where ak ∈ A for 1 ≤ k ≤ n, then

[w] = [a1a2 · · · an] = [a1][a2] · · · [an].

Since [a] = i(a) if a ∈ X and [a] = [a]−1 = i(a)−1 if a ∈ X ′, we deduce thati(X) generates the group F .

Let us show that the triple (F, X, i) is a based free group. Suppose thatf : X → G is a map from X into a group G. We have to prove that thereexists a unique homomorphism φ : F → G such that f = φ ◦ i. Uniquenessfollows from the fact that i(X) generates F . To construct φ, we first extendf to a map g : A → G by setting

g(a) =

{f(a) if a ∈ X,

f(a)−1 if a ∈ X ′.

Observe that g(a) = g(a)−1 for all a ∈ A. Define now a map Φ : A∗ → G bysetting

Φ(w) = g(a1)g(a2) · · · g(an)

for all w = a1a2 · · · an ∈ A∗. Note that we have

Φ(ww′) = Φ(w)Φ(w′) (D.1)

for all w, w′ ∈ A∗. Moreover, if w may be obtained from w′ by an elementaryreduction, that is, w = uv and w′ = uaav for some a ∈ A and u, v ∈ A∗, then

Φ(w′) = Φ(u)g(a)g(a)Φ(v) = Φ(u)g(a)g(a)−1Φ(v) = Φ(u)Φ(v) = Φ(w).

It follows that Φ(w) = Φ(w′) whenever w, w′ ∈ A∗ satisfy w ∼ w′. Byinduction, we deduce that Φ(w) = Φ(w′) for all w, w′ ∈ A∗ such that w ≡ w′.Thus, we can define a map φ : F → G by setting φ([w]) = Φ(w) for all w ∈ A∗.By using (D.1), we get

φ([w][w′]) = φ([ww′]) = Φ(ww′) = Φ(w)Φ(w′) = φ([w])φ([w′]).

Therefore, φ is a group homomorphism. On the other hand, for all x ∈ X,we have

φ ◦ i(x) = φ([x]) = g(x) = f(x),

which shows that φ ◦ i = f . Consequently, the triple (F, X, i) is a based freegroup. ��

Given an arbitrary set X, it follows from Theorem D.2.7 that there exista based free group (F, X, i). Then one often says that F = F (X) is the free

372 D Free Groups

group based on X (this is a minor abuse of language since if (F ′, X, i′) isanother based free group then there is a unique isomorphism ϕ : F → F ′

such that ϕ ◦ i = i′, by Proposition D.2.6). For k ∈ N, the free group basedon {1, 2, . . . , k} is denoted by Fk.

We recall that one says that two sets X1 and X2 are equipotent , or thatthey have the same cardinality, if there exists a bijective map u : X1 → X2.

Theorem D.2.8. Let F1 and F2 be two free groups based on X1 ⊂ F1 andX2 ⊂ F2. Then the following conditions are equivalent:

(a) the groups F1 and F2 are isomorphic;(b) the sets X1 and X2 are equipotent.

For the proof, we shall need a few preliminary results. We use the follow-ing notation. If G1 and G2 are groups, we denote by Hom(G1, G2) the setconsisting of all homomorphisms φ : G1 → G2. We denote by P(X) the setof all subsets of a set X.

Lemma D.2.9. Let F be a free group based on X ⊂ F . Then the setsHom(F, Z/2Z) and P(X) are equipotent.

Proof. Since F is free with base X, each map f : X → Z/2Z can be uniquelyextended to a homomorphism φ : F → Z/2Z. Therefore, the restriction mapyields a bijection from Hom(F, Z/2Z) onto the set of maps from X to Z/2Z.Consequently, the sets Hom(F, Z/2Z) and P(X) are equipotent. ��

Lemma D.2.10. Let F be a free group based on X ⊂ F . Suppose that theset X is infinite. Then the sets X and F are equipotent.

Proof. Let A = X ∪X−1. Since X generates F , the map ρ : A∗ → F definedby ρ(a1, a2, . . . , an) = a1a2 . . . an is surjective. As X is infinite, the sets X,A, and A∗ are all equipotent. It follows that there is a surjective map fromX onto F and therefore an injective map from F into X. By applying theCantor-Bernstein theorem (cf. Corollary H.3.5), we deduce that X and F areequipotent. ��

Proof of Theorem D.2.8. The fact that (b) implies (a) immediately followsfrom Proposition D.2.6.

To prove the converse implication, suppose that there exists an isomor-phism α : F1 → F2. Then α induces a bijective map between the setsHom(F1, Z/2Z) and Hom(F2, Z/2Z). It follows that P(X1) and P(X2) areequipotent by Lemma D.2.9. If X1 is infinite, this implies that X2 is alsoinfinite, and we conclude that X1 and X2 are equipotent by applyingLemma D.2.10. On the other hand, if X1 is finite, the fact that P(X1) andP(X2) are equipotent implies that X2 is also finite and that 2|X1| = 2|X2|.This gives us |X1| = |X2| and we conclude that X1 and X2 are also equipo-tent in this case. ��

D.3 Reduced Forms 373

Corollary D.2.11. Let F be a free group and let X1, X2 ⊂ F be two basesof F . Then X1 and X2 are equipotent. ��

Let F be a free group. The cardinality of a base X ⊂ F depends only on Fby Corollary D.2.11. This cardinality is called the rank of F . Two free groupsare isomorphic if and only if they have the same rank by Theorem D.2.8.A group is free of finite rank k ∈ N if and only if it is isomorphic to Fk.

D.3 Reduced Forms

In order to state the first result of this section, we use the notation introducedin the proof of Theorem D.2.7. A word w ∈ A∗ is said to be reduced if itcontains no subword of the form aa with a ∈ A, that is, if there is no wordw′ ∈ A∗ which can be obtained from w by applying an elementary reduction.Note that the empty word ε is reduced and that every subword of a reducedword is itself reduced.

Theorem D.3.1. Every equivalence class for ≡ contains a unique reducedword.

Proof. It is clear that any word w ∈ A∗ can be transformed into a reducedword by applying a suitable finite sequence of elementary reductions. Thisshows the existence of a reduced word in any equivalence class for ≡.

Let us prove uniqueness. Consider the subset R ⊂ A∗ consisting of allreduced words. For a ∈ A and r ∈ R, define the word αa(r) by

αa(r) =

{w if r = aw for some w ∈ A∗,

ar otherwise.

Note that the word αa(r) is always reduced. Thus, we get a map αa : R → Rdefined for each a ∈ A. Let us show that

αa ◦ αea = α

ea ◦ αa = IdR . (D.2)

Let a ∈ A and consider an arbitrary element r ∈ R. If r = aw for somew ∈ A∗, then α

ea(r) = w and hence αa(αea(r)) = αa(w) = aw = r (observe

that w cannot start by a since r is reduced). Otherwise, we have αea(r) = ar

and therefore αa(αea(r)) = αa(ar) = r. It follows that αa ◦ α

eA = IdR. Byreplacing a by a in this equality, we get α

ea ◦αa = IdR since · is an involutionon A. This proves (D.2).

From (D.2), we deduce that αa ∈ Sym(R) for all a ∈ A. By setting

ρ(w) = αa1 ◦ αa2 ◦ · · · ◦ αan

374 D Free Groups

for every word w = a1a2 · · · an ∈ A∗, we get a monoid homomorphismρ : A∗ → Sym(R). Note that

ρ(r)(ε) = r for all r ∈ R. (D.3)

On the other hand, it immediately follows from (D.2) that ρ(w) = ρ(w′)whenever w, w′ ∈ A∗ satisfy w ≡ w′. Thus, if r1 an r2 are reduced words inthe same equivalence class for ≡, we have ρ(r1) = ρ(r2) and therefore r1 = r2

by applying (D.3). ��

Corollary D.3.2. Let (F, X, i) be a based free group. Then every elementf ∈ F can be uniquely written in the form

f = i(x1)h1i(x2)h2 · · · i(xn)hn (D.4)

with n ≥ 0, xi ∈ X and hi ∈ Z \ {0} for 1 ≤ i ≤ n, and xi �= xi+1 for1 ≤ i ≤ n − 1.

Definition D.3.3. The expression (D.4) is called the reduced form of theelement f in the based free group (F, X, i).

Proof of Corollary D.3.2. We can assume that (F, X, i) is the based free groupconstructed in the proof of Theorem D.2.7. By construction, the elementf ∈ F is an equivalence class for ≡. This equivalence class contains a uniquereduced word r by Theorem D.3.1. The word r can be uniquely written inthe form

r = ak11 ak2

2 · · · aknn ,

where n ≥ 0, ai ∈ A and ki ∈ N \ {0} for 1 ≤ i ≤ n, and ai �= ai+1 for1 ≤ i ≤ n−1. This gives us an expression of the form (D.4) with xi = ai andhi = ki if ai ∈ X, and xi = ai and hi = −ki otherwise. Uniqueness of such anexpression for f follows from the uniqueness of the reduced word r ∈ f . ��

Corollary D.3.4. Let G be a group. Let U ⊂ G be a subset such that

uk11 uk2

2 · · ·uknn �= 1G, (D.5)

for all u1, u2, . . . , un ∈ U and k1, k2, . . . , kn ∈ Z \ {0} with ui �= ui+1 for1 ≤ i ≤ n− 1 and n ≥ 1. Then the subgroup of G generated by U is free withbase U .

Proof. Denote by H the subgroup of G generated by U and by ι : U → Hthe inclusion map. Let (F,U, i) be a based free group (cf. Theorem D.2.7).Then there exists a unique homomorphism φ : F → H such that ι = φ ◦ i.Note that φ is surjective since φ(i(U)) = ι(U) = U generates H. Let us showthat φ is also injective. Consider an element f ∈ F written in reduced form,that is, in the form f = i(u1)h1i(u2)h2 · · · i(un)hn with n ≥ 0, ui ∈ U andhi ∈ Z \ {0} for 1 ≤ i ≤ n, and ui �= ui+1 for 1 ≤ i ≤ n − 1. Then we

D.4 Presentations of Groups 375

have φ(f) = uh11 uh2

2 · · ·uhnn since φ(i(ui)) = ι(ui) = ui for 1 ≤ i ≤ n. It

follows from (D.5) that φ(f) = 1H if and only if n = 0, that is, if and only iff = 1F . This shows that φ is an isomorphism. It follows from Remark D.2.2that (H,U, ι) is a based free group. ��

D.4 Presentations of Groups

Proposition D.4.1. Let G be a group and let S be a generating subset of G.Let F denote the free group based on S. Then, the group G is isomorphic toa quotient of F .

Proof. Let i : S → F and f : S → G denote the inclusion maps. Since F is freewith base S, there exists a homomorphism φ : F → G such that f = φ ◦ i.This implies that S is contained in the image of φ. Since S generates G,we deduce that φ is surjective. Therefore, the group G is isomorphic to thequotient group F/ Ker(φ). ��

As every group admits a generating subset (e.g., the group itself), wededuce the following:

Corollary D.4.2. Every group is isomorphic to a quotient of a free group.��

A group is said to be finitely generated if it admits a finite generatingsubset. Proposition D.4.1 gives us:

Corollary D.4.3. Every finitely generated group is isomorphic to a quotientof a free group of finite rank. ��

Let A be a subset of a group G. The intersection of all normal subgroups ofG containing A is a normal subgroup of G which is called the normal closureof A in G.

Let G be a group. By Corollary D.4.2, there exist a free group F and anepimorphism φ : F → G. Let X be a free base for F . If R is a subset of Fwhose normal closure is the kernel of φ, then one says that G admits thepresentation

G = 〈X; R〉. (D.6)

Note that as X generates F (cf. Proposition D.2.3(ii)), we have that φ(X)generates G. The elements x ∈ X (rather than the φ(x) ∈ G, x ∈ X) arecalled, by abuse of language, the generators of the presentation (D.6). Theelements r ∈ R are called the relators of the presentation (D.6). For everyr ∈ R let ur, vr be elements in F such that r = ur(vr)−1. Then (D.6) is oftenexpressed as G = 〈X : r = 1, r ∈ R〉 or G = 〈X : ur = vr, r ∈ R〉.

Note that G admits a presentation (D.6) with X finite if and only if G isfinitely generated (cf. Corollary D.4.3).

376 D Free Groups

If a group G admits a presentation (D.6) with both X and R finite, thenG is said to be finitely presented.

D.5 The Klein Ping-Pong Theorem

The following theorem is often used to prove that a group is free:

Theorem D.5.1. Let G be a group. Let X be a generating subset of G havingat least two distinct elements. Suppose that G acts on a set E and that thereis a family (Ax)x∈X of nonempty pairwise disjoint subsets of E such that

xk

⎛

⎝⋃

y∈X\{x}Ay

⎞

⎠ ⊂ Ax for all x ∈ X and k ∈ Z \ {0}. (D.7)

Then G is a free group with base X.

Proof. Consider an element g ∈ G written as a nontrivial reduced word onthe generating subset X, that is, in the form

g = xk11 xk2

2 . . . xknn ,

where n ≥ 1, xi ∈ X and ki ∈ Z \ {0} for 1 ≤ i ≤ n, and xi �= xi+1 for1 ≤ i ≤ n − 1. By Corollary D.3.4, we have to show that g �= 1G.

Suppose first that either X contains at least three distinct elements or Xhas exactly two elements and x1 = xn. In this case, we can find an elementy ∈ X such that y �= x1 and y �= xn. By successive applications of D.7, weget

gAy = xk11 xk2

2 . . . xkn−2n−2 x

kn−1n−1 xkn

n Ay

⊂ xk11 xk2

2 . . . xkn−2n−2 x

kn−1n−1 Axn

⊂ xk11 xk2

2 . . . xkn−2n−2 Axn−1

⊂ xk11 xk2

2 . . . Axn−2

. . .

⊂ xk11 xk2

2 Ax3

⊂ xk11 Ax2

⊂ Ax1 .

As the sets Ay and Ax1 are disjoint and Ay �= ∅ by our hypotheses, wededuce that g �= 1G.

D.5 The Klein Ping-Pong Theorem 377

It remains to treat the case when X has exactly two elements and x1 �= xn.Then

xk11 gx−k1

1 = x2k11 xk2

2 xk33 . . . xkn

n x−k11

is a reduced form of xk11 gx−k1

1 . We have xk11 gx−k1

1 �= 1G by the first case. Wededuce that we also have g �= 1G in this case. ��

Remark D.5.2. The proof of Theorem D.5.1 shows that the hypotheses on thefamily (Ax)x∈X may be relaxed: in fact, it suffices that the subsets Ax ⊂ X,x ∈ X, satisfy D.7 and Ay �⊂ Ax for all distinct elements x, y ∈ X.

Corollary D.5.3. Let F be a free group of rank 2 and let n ≥ 2 be an integer.Then F contains a free subgroup of rank n.

Proof. Suppose that F is based on the elements a and b. Let G be the sub-group of F generated by the subset

X = {aiba−i : 0 ≤ i ≤ n − 1}.

Consider the action of G on F given by left multiplication. Define, for eachx = aiba−i ∈ X, the subset Ax ⊂ F as being the set of elements of F whosereduced form starts by aibk for some k ∈ Z \ {0}. The subsets Ax, x ∈ X,clearly satisfy the hypotheses of Theorem D.5.1. Therefore G is free withbase X. ��

Appendix E

Inductive Limits and Projective Limitsof Groups

E.1 Inductive Limits of Groups

Let I be a directed set. An inductive system of groups over I consists ofthe following data: (1) a family of groups (Gi)i∈I indexed by I, (2) for eachpair i, j ∈ I such that i ≤ j, a homomorphism ψji : Gi → Gj satisfying thefollowing conditions:

ψii = IdGi (identity map on Gi) for all i ∈ I,

ψkj ◦ ψji = ψki for all i, j, k ∈ I such that i ≤ j ≤ k.

Then one speaks of the inductive system (Gi, ψji) or simply of the inductivesystem (Gi) if the homomorphisms ψji are understood.

Let (Gi, ψji) be an inductive system of groups over I. Consider the relation∼ on the disjoint union E =

∐i∈I Gi defined as follows. If xi ∈ Gi and

xj ∈ Gj are elements of E, then xi ∼ xj if and only if there is an elementk ∈ I such that i ≤ k, j ≤ k and ψki(xi) = ψkj(xj). It easy to check that ∼ isa equivalence relation on the set E. Let G = E/ ∼ be the set of equivalenceclasses of ∼. For xi ∈ Gi, let [xi] ∈ G denote the class of xi. One defines abinary operation on G in the following way. Given xi ∈ Gi and xj ∈ Gj , onedefines the class [xi][xj ] by setting

[xi][xj ] = [ψki(xi)ψkj(xj)],

where k ∈ I is such that i ≤ k and j ≤ k. One checks that [xi][xj ] dependsneither on the choice of the representatives xi and xj , nor on the choice of k.Moreover, this operation gives a group structure on G. The group G is calledthe inductive limit (or the direct limit) of the inductive system (Gi) andone writes G = lim−→Gi. For each i ∈ I, there is a canonical homomorphismhi : Gi → G defined by hi(xi) = [xi]. Note that it immediately follows fromthe construction of G that G =

⋃i∈I hi(Gi). Moreover, one has


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380 E Inductive Limits and Projective Limits of Groups

hj ◦ ψji = hi

for all i, j ∈ I such that i ≤ j.

Examples E.1.1. (a) Let (Gi)i∈I be a family of groups and, for i, j ∈ I suchthat i ≤ j, let ψji : Gi → Gj be the trivial homomorphism, namely, ψji(g) =1Gj for all g ∈ Gi. Then, (Gi, ψji) is an inductive system whose inductivelimit is a trivial group.

(b) Let G be a group and denote by H the set of all finitely generatedsubgroups of G. Then (H,⊂) is a directed set and

⋃H∈H H = G. Moreover,

the set H together with the inclusion maps ψK,H : H → K, for all H,K ∈ Hwith H ⊂ K, is an inductive system whose limit is canonically isomorphicto G.

E.2 Projective Limits of Groups

Let I be a directed set. A projective system of groups over I consists of thefollowing data: (1) a family of groups (Gi)i∈I indexed by I, (2) for eachpair i, j ∈ I such that i ≤ j, a homomorphism ϕij : Gj → Gi satisfying thefollowing conditions:

ϕii = IdGi (identity map on Gi) for all i ∈ I,

ϕij ◦ ϕjk = ϕik for all i, j, k ∈ I such that i ≤ j ≤ k.

Then one speaks of the projective system (Gi, ϕij) or simply of the projectivesystem (Gi) if the homomorphisms ϕij are understood.

Let (Gi, ϕij) be a projective system of groups over I. Let P =∏

i∈I Gi

denote the direct product of the groups Gi. One immediately checks that

G = {(xi) ∈ P : ϕij(xj) = xi for all i, j ∈ I such that i ≤ j}

is a subgroup of P . The group G is called the projective limit of the projectivesystem (Gi), and one writes G = lim←−Gi. For each i ∈ I, there is a canonicalhomomorphism fi : G → Gi obtained by restriction of the projection mapπi : P → Gi. One has

ϕij ◦ fj = fi

for all i, j ∈ I such that i ≤ j.

Examples E.2.1. (a) Let (Gi)i∈I be a family of groups and, for i, j ∈ I suchthat i ≤ j, let ϕij : Gj → Gi be the trivial homomorphism. Then, (Gi, ϕij)is a projective system whose projective limit is a trivial group.

(b) Let G be a group. Denote by N the set of all normal subgroupsof G. Then (N ,⊃) is a directed set. The family of groups (G/H)H∈N , when

E.2 Projective Limits of Groups 381

equipped with the canonical quotient homomorphisms ϕH,K : G/K → G/Hfor all H,K ∈ N with H ⊃ K, gives rise to a projective system whoseprojective limit is canonically isomorphic to G.

Appendix F

The Banach-Alaoglu Theorem

All vector spaces considered in this appendix are vector spaces over the fieldR of real numbers.

F.1 Topological Vector Spaces

A topological vector space is a real vector space X endowed with a topologysuch that the maps

X × X → X

(x, y) �→ x + y

and

R × X → X

(λ, x) �→ λx

are continuous.

Example F.1.1. Let ‖ · ‖ be a norm on a real vector space X and let d denotethe metric on X defined by d(x, y) = ‖x − y‖ for all x, y ∈ X. Then thetopology defined by d yields a structure of topological vector space on X.

Recall that a subset C of a real vector space X is said to be convex if

(1 − λ)x + λy ∈ C

for all λ ∈ [0, 1] and x, y ∈ C.A topological real vector space X is said to be locally convex if there is a

base of neighborhoods of 0 consisting of convex subsets of X.


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384 F The Banach-Alaoglu Theorem

F.2 The Weak-∗ Topology

Let X be a real vector space equipped with a norm ‖ · ‖. Recall that a linearmap u : X → R is continuous if and only if u is bounded on the unit ballB(X) = {x ∈ X : ‖x‖ ≤ 1}, that is, if and only if there is a constant C ≥ 0such that |u(x)| ≤ C for all x ∈ B(X). The topological dual of X is the vectorspace X∗ consisting of all continuous linear maps u : X → R. The operatornorm on X∗ is the norm defined by

‖u‖ = supx∈B(X)

|u(x)|.

The topology defined by the operator norm is called the strong topologyon X∗.

Given x ∈ X, let ψx : X∗ → R denote the evaluation map u �→ u(x) at x.The weak-∗ topology on X∗ is the initial topology associated with the familyof all evaluation maps ψx : X∗ → R, x ∈ X. Thus, the weak-∗ topology isthe smallest topology on X∗ for which all evaluation maps ψx are continuous.Observe that every subset of X∗ which is open for the weak-∗ topology is alsoopen for the strong topology since all the evaluation maps are continuous forthe strong topology. It follows that convergence with respect to the strongtopology implies convergence with respect to the weak-∗ topology.

The weak-∗ topology provides a topological vector space structure on X∗.A base of open neighborhoods of 0 for the weak-∗ topology is given by allsubsets of the form

V (F, ε) = {u ∈ X∗ : |u(x)| < ε for all x ∈ F},

where F is a finite subset of X and ε > 0. Since all the sets V (F, ε) are convex,the topological vector space structure associated with the weak-∗ topologyon X∗ is locally convex.

The weak-∗ topology on X∗ is Hausdorff. Indeed, if u1 and u2 are twodistinct elements in X∗, then there exists x ∈ X such that u1(x) = u2(x). IfU1 and U2 are disjoint open subsets of R containing u1(x) and u2(x) respec-tively, then ψ−1

x (U1) and ψ−1x (U2) are disjoint open subsets of X∗ containing

u1 and u2 respectively.

F.3 The Banach-Alaoglu Theorem

In general, the unit ball in X∗ is not compact for the strong topology (thisfollows from the fact that the unit ball in a normed space is never compactunless the space is finite-dimensional). However, this ball is always compactfor the weak-∗ topology. This result, which is known as the Banach-Alaoglu

F.3 The Banach-Alaoglu Theorem 385

theorem is one of the central results of classical functional analysis and maybe easily deduced from the Tychonoff theorem.

Theorem F.3.1 (Banach-Alaoglu theorem). Let X be a real normedvector space and let ‖ · ‖ denote the operator norm on its topological dual X∗.Then the unit ball B(X∗) = {u ∈ X∗ : ‖u‖ ≤ 1} is compact for the weak-∗topology on X∗.

Proof. Observe first that X∗ is a vector subspace of the vector space RX

consisting of all real-valued functions on X. On the other hand, setting Ix =[−‖x‖, ‖x‖] ⊂ R for each x ∈ X. We have B(X∗) ⊂

∏x∈X Ix by definition of

the operator norm. Let us equip RX =

∏x∈X R with the product topology.

Then it is clear that the topology induced on∏

x∈X Ix is the product topologyand that the topology induced on X∗ is the weak-∗ topology. Let f be anelement of R

X which is the limit of a net (ui) of elements of B(x∗). For allx, y ∈ X and λ ∈ R, we have ui(λx) = λui(x), ui(x + y) = ui(x) + ui(y)and |ui(x)| ≤ ‖x‖ since ui ∈ B(X∗). By taking limits, we get f(λx) = λf(x),f(x + y) = f(x) + f(y) and |f(x)| ≤ ‖x‖. This shows that f ∈ B(X∗).Consequently, B(X∗) is closed in R

X . Since∏

x∈X Ix is compact by Tychonofftheorem (Theorem A.5.2), we deduce that B(X∗) is compact. ��

Appendix G

The Markov-Kakutani Fixed PointTheorem


G.1 Statement of the Theorem

Let C be a convex subset of a real vector space X. A map f : C → C is calledaffine if

f((1 − λ)x + λy) = (1 − λ)f(x) + λf(y)

for all λ ∈ [0, 1] and x, y ∈ C.

Theorem G.1.1 (Markov-Kakutani). Let K be a nonempty convex com-pact subset of a Hausdorff topological vector space X. Let F be a set of con-tinuous affine maps f : K → K. Suppose that all elements of F commute,that is, f1 ◦ f2 = f2 ◦ f1 for all f1, f2 ∈ F . Then there exists a point in Kwhich is fixed by all the elements of F .

G.2 Proof of the Theorem

In the proof of the Markov-Kakutani theorem, we shall use the followinglemmas.

Lemma G.2.1. Let K be a compact subset of a topological vector space Xand let V be a neighborhood of 0 in X. Then there exists a real number α > 0such that λK ⊂ V for every real number λ such that |λ| < α.

Proof. Since the multiplication by a scalar R×X → X is continuous, we canfind, for each x ∈ X, a real number αx and an open neighborhood Ωx ⊂ Xof x such that


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388 G The Markov-Kakutani Fixed Point Theorem

|λ| < αx ⇒ λxΩx ⊂ V. (G.1)

The sets Ωx, x ∈ K, form an open cover of K. As K is compact, there is afinite subset F ⊂ K such that

K ⊂⋃

x∈F

Ωx. (G.2)

If we take α = minx∈F αx, then α > 0 and

|λ| < α ⇒ λK ⊂ V

by (G.1) and (G.2). ��

Lemma G.2.2. Let K be a compact subset of a topological vector space X.Let (xi)i∈I be a net of points in K and let (λi)i∈I be a net of real numbersconverging to 0 in R. Then the net (λixi)i∈I converges to 0 in X.

Proof. Let V be a neighborhood of 0 in X. By Lemma G.2.1, we can findα > 0 such that λK ⊂ V for every λ such that |λ| < α. As the net (λi)converges to 0, there exists i0 ∈ I such that i ≥ i0 implies |λi| < α. Thuswe have λixi ∈ V for all i ≥ i0. This shows that the net (λixi) convergesto 0. ��

The following lemma is the theorem of Markov-Kakutani in the particularcase when the set F is reduced to a single element.

Lemma G.2.3. Let K be a nonempty convex compact subset of a Hausdorfftopological vector space X and let f : K → K be an affine continuous map.Then f has a fixed point in K.

Proof. Let us set C = {y − f(y) : y ∈ K}. The fact that f admits a fixedpoint in K is equivalent to the fact that 0 ∈ C. Choose an arbitrary pointx ∈ K and consider the sequence (xn)n≥1 of points of X defined by

xn =1n

n−1∑

k=0

(fk(x) − fk+1(x)).

We have fk(x) − fk+1(x) = fk(x) − f(fk(x)) ∈ C for 0 ≤ k ≤ n − 1. Onthe other hand, the set C is convex, since K is convex and f is affine. Thusxn ∈ C for every n ≥ 1.

Asxn =

1n

x − 1n

fn(x).

and fn(x) ∈ K for every n ≥ 1, it follows from Lemma G.2.2 that thesequence (xn)n≥1 converges to 0. The set C is compact since it is the imageof the compact set K by the continuous map y �→ y−f(y). As every compactsubset of a Hausdorff space is closed, we deduce that C is closed in X. Thus0 ∈ C. ��

Notes 389

Proof of Theorem G.1.1. Let f ∈ F and consider the set

Fix(f) = {x ∈ K : f(x) = x}

of its fixed points. The set Fix(f) is not empty by Lemma G.2.3 and it iscompact since it is a closed subset of the compact set K. On the other hand,Fix(f) is convex since K is convex and f is affine. If g ∈ F and x ∈ Fix(f),then the fact that f and g commute implies that g(x) ∈ Fix(f) since

f(g(x)) = g(f(x)) = g(x).

Therefore we can apply Lemma G.2.3 to the restriction of g to Fix(f). Itfollows that g fixes a point in Fix(f), that is,

Fix(f) ∩ Fix(g) = ∅.

By induction on n, we get

Fix(f1) ∩ Fix(f2) ∩ · · · ∩ Fix(fn) = ∅

for all f1, f2, . . . , fn ∈ F . Since K is compact, from the finite intersectionproperty (see Sect. A.5) we deduce that

⋂

f∈FFix(f) = ∅.

This shows that there is a point in K which is fixed by all elements of F . ��

Notes

The proof presented here is due to S. Kakutani [Kak] (see [Jac]). In the proofof A. Markov [Mar], the local convexity of X is needed.

Appendix H

The Hall Harem Theorem

H.1 Bipartite Graphs

A bipartite graph is a triple G = (X,Y,E), where X and Y are arbitrary sets,and E is a subset of the Cartesian product X × Y . The set X (resp. Y ) iscalled the set of left (resp. right) vertices and E is called the set of edges ofthe bipartite graph G (see Fig. H.1).

Fig. H.1 The bipartite graph G = (X, Y, E) with X = {x1, x2, x3, x4, x5}, Y = {y1, y2,y3, y4} and E = {(x1, y1), (x2, y1), (x2, y2), (x2, y3), (x3, y4), (x5, y3), (x5, y4)}

A bipartite subgraph of a bipartite graph G = (X,Y,E) is a bipartite graphG′ = (X ′, Y ′, E′) with X ′ ⊂ X, Y ′ ⊂ Y and E′ ⊂ E (see Fig. H.2).

Let G = (X,Y,E) be a bipartite graph.Two edges (x, y), (x′, y′) ∈ E are said to be adjacent if x = x′ or y = y′.Given a vertex x ∈ X (resp. y ∈ Y ) the right-neighborhood of x (resp. the

left-neighborhood of y) is the subset NR(x) ⊂ Y (resp. NL(y) ⊂ X) definedby (see Figs. H.3–H.4):


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392 H The Hall Harem Theorem

Fig. H.2 The bipartite subgraph G′ = (X′, Y ′, E′) of the bipartite graph G in Fig. H.1with X′ = {x1, x2, x4, x5}, Y ′ = {y1, y2, y3} and E′ = {(x2, y1), (x2, y2), (x2, y3), (x5, y3)}

Fig. H.3 The right-neighborhood NR(x2) ⊂ Y of the vertex x2 ∈ X in the bipartite

graph G of Fig. H.1

NR(x) = N GR(x) = {y ∈ Y : (x, y) ∈ E}

(resp. NL(y) = N GL (y) = {x ∈ X : (x, y) ∈ E}).

For subsets A ⊂ X and B ⊂ Y , we define the right-neighborhood NR(A)of A and the left-neighborhood NL(B) of B by

NR(A) = N GR(A) =

⋃

a∈A

NR(a) and NL(B) = N GL (B) =

⋃

b∈B

NL(b).

One says that the bipartite graph G = (X,Y,E) is finite if the sets X andY are finite. One says that G is locally finite if the sets NR(x) and NR(y) arefinite for all x ∈ X and y ∈ Y . Note that if G is locally finite then the setsNR(A) and NL(B) are finite for all finite subsets A ⊂ X and B ⊂ Y .

H.2 Matchings 393

Fig. H.4 The left-neighborhood NL(y1) ⊂ X of the vertex y1 ∈ Y in the bipartite graph Gof Fig. H.1

H.2 Matchings

Let G = (X,Y,E) be a bipartite graph.A matching in G is a subset M ⊂ E of pairwise nonadjacent edges. In

other words, a subset M ⊂ E is a matching if and only if both projectionmaps p : M → X and q : M → Y are injective.

A matching M is called left-perfect (resp. right-perfect) if for each x ∈ X(resp. y ∈ Y ), there exists y ∈ Y (resp. x ∈ X) such that (x, y) ∈ M(see Fig. H.5). Thus, a matching M is left-perfect (resp. right-perfect) if andonly if the projection map p : M → X (resp. q : M → Y ) is surjective (andtherefore bijective).

A matching M is called perfect if it is both left-perfect and right-perfect.

Fig. H.5 A right-perfect matching M ⊂ E in the bipartite graph G of Fig. H.1. Note thatthere is no left-perfect matching (and therefore no perfect matching) in G since |X| > |Y |

Remarks H.2.1. (a) A subset M ⊂ E is a left-perfect (resp. right-perfect)matching if and only if there is an injective map ϕ : X → Y (resp. an injective


map ψ : Y → X) such that M = {(x, ϕ(x)) : x ∈ X} (resp. M = {(ψ(y), y) :y ∈ Y }).

(b) Similarly, a subset M ⊂ E is a perfect matching if and only if there isa bijective map ϕ : X → Y such that M = {(x, ϕ(x)) : x ∈ X}.

Given a bipartite graph G = (X,Y,E), one often regards the set X as a setof boys and Y as a set of girls. One interprets (x, y) ∈ E as the condition thatx and y know each other. In this context, a matching M ⊂ E is a process ofgetting boys and girls that know each other married (no polygamy is allowedhere). The matching is left-perfect (resp. right-perfect) if and only if everyboy (resp. girl) gets married. Finally, M is perfect if and only if there remainno singles.

H.3 The Hall Marriage Theorem

We use the notation | · | to denote cardinality of sets.

Definition H.3.1 (Hall conditions). Let G = (X,Y,E) be a bipartitegraph. One says that G satisfies the left (resp. right) Hall condition if

|NR(A)| ≥ |A| (H.1)

(resp. |NL(B)| ≥ |B|) for every finite subset A ⊂ X (resp. for every finitesubset B ⊂ Y ).

One says that G satisfies the Hall marriage conditions if G satisfies boththe left and the right Hall conditions.

Theorem H.3.2. Let G = (X,Y,E) be a locally finite bipartite graph. Thenthe following conditions are equivalent.

(a) G satisfies the left (resp. right) Hall condition;(b) G admits a left (resp. right) perfect matching.

Proof. It is obvious that (b) implies (a). Let us prove that (a) implies (b). Bysymmetry, it suffices to show that if G satisfies the left Hall condition then itadmits a left perfect matching.

We fist treat the case when the set X is finite by induction on n = |X|. Inthe case |X| = 1, the statement is trivially satisfied. Suppose that we haveproved the statement whenever |X| ≤ n − 1 and let us prove it for |X| = n.We distinguish two cases.

Case (i): Suppose that

|NR(A)| ≥ |A| + 1 (H.2)

for all nonempty proper subsets A ⊂ X. Then fix x0 ∈ X and y0 ∈ Ysuch that (x0, y0) ∈ E. Consider the bipartite subgraph G′ = (X ′, Y ′, E′),

H.3 The Hall Marriage Theorem 395

where X ′ = X \ {x0}, Y ′ = Y \ {y0}, and E′ = E ∩ (X ′ × Y ′). Then, forevery nonempty subset A′ ⊂ X ′, we have |NR(A′)| ≥ |A′| + 1 by (H.2), andhence |N G′

R (A′)| ≥ |A′| since N G′

R (A′) = NR(A′) \ {y0}. As |X ′| = n − 1, itfollows from our induction hypothesis that G′ admits a left-perfect matchingM ′ ⊂ E′. Then M = M ′ ∪ {(x0, y0)} is a left-perfect matching for G.

Case (ii): Suppose that we are not in Case (i). This means that there existsa nonempty proper subset X ′ ⊂ X such that

|NR(X ′)| = |X ′|. (H.3)

Then the bipartite subgraph G′ = (X ′, Y ′, E′), where Y ′ = NR(X ′) andE′ = E ∩ (X ′ × Y ′) clearly satisfies the left Hall condition. As |X ′| ≤ n − 1,there exists, by our induction hypothesis, a left-perfect matching M ′ ⊂ E′

for G′. Consider now the bipartite subgraph G′′ = (X ′′, Y ′′, E′′), where X ′′ =X \ X ′, Y ′′ = Y \ Y ′, and E′′ = E ∩ (X ′′ × Y ′′). We claim that the left Hallcondition also holds for G′′. Otherwise, there would be some subset A′′ ⊂ X ′′

such that,

|N G′′

R (A′′)| < |A′′|, (H.4)

and then the left Hall condition for G would be violated by A = X ′∪A′′ ⊂ Xsince

|NR(A)| = |NR(X ′ ∪ A′′)|

= |NR(X ′) ∪NR(A′′)|

= |NR(X ′) ∪N G′′

R (A′′)|

≤ |NR(X ′)| + |N G′′

R (A′′)|

< |X ′| + |A′′| (by (H.3) and (H.4))

= |X ′ ∪ A′′|

= |A|.

Therefore, as |X ′′| < |X| = n, induction applies again yielding a left-perfectmatching M ′′ ⊂ E′′ for G′′. It then follows that M = M ′ ∪ M ′′ is a left-perfect matching for G. This completes the proof that (a) implies (b) in thecase when X is finite.

To treat the general case, we shall apply the Tychonoff product theorem.Suppose that G = (X,Y,E) is a (possibly infinite) locally finite bipartitegraph satisfying the left Hall condition, that is, |NR(A)| ≥ |A| for every fi-nite subset A ⊂ X. Let us equip the Cartesian product K =

∏x∈X NR(x)

with its prodiscrete topology, that is, with the product topology obtained bytaking the discrete topology on each factor NR(x). As each set NR(x) is finite,the space K is compact by the Tychonoff product theorem (Corollary A.5.3).


For each x ∈ X, let πx : K → NR(x) denote the projection map. Let F be theset consisting of all nonempty finite subsets of X. Consider, for each F ∈ F ,the set C(F ) ⊂ K consisting of all z ∈ K which satisfy πx1(z) = πx2(z) forall distinct elements x1 and x2 in F . It follows from this definition and thecontinuity of the projection maps πx, that C(F ) is an intersection of closedsubsets of X and hence closed in K. On the other hand, C(F ) is not empty.Indeed, choose, for each x ∈ X, an element ψ(x) ∈ NR(x) (observe thatNR(x) is not empty by the left Hall condition). Also set G′ = (X ′, Y ′, E′)where X ′ = F , Y ′ = NR(F ) and E′ = E ∩ (X ′ × Y ′). Then, the left Hallcondition for G implies the left Hall condition for the finite bipartite graph G′

and therefore, by Theorem H.3.2, there exists a perfect matching for G′. ByRemark H.2.1, there exists an injective mapping ϕ : F = X ′ → Y ′ = NR(F ).Then, the element (Φ(x))x∈X ∈ K, defined by Φ(x) = ϕ(x) if x ∈ F andΦ(x) = ψ(x) if x ∈ X \ F , clearly belongs to C(F ).

Since

C(F1) ∩ C(F2) ∩ · · · ∩ C(Fn) ⊃ C(F1 ∪ F2 ∪ · · · ∪ Fn)

for all F1, F2, . . . , Fn ∈ F , we deduce that the family {C(F ) : F ∈ F} ofsubsets of K has the finite intersection property. By compactness of K, thereis a point z0 ∈

⋂F∈F F . Then M = {(x, πx(z0)) : x ∈ X} is a left-perfect

matching for G. ��

Remark H.3.3. The hypothesis of local finiteness for the bipartite graph cannot be removed from the statement of the previous theorem. To see this,consider the bipartite graph G = (X,Y,E), where X = Y = N and E ={(0, n+1) : n ∈ N}∪ {(n+1, n) : n ∈ N}. Observe that G is not locally finiteas |NR(0)| = ∞. Also, it satisfies the left Hall condition. Indeed, given a finitesubset A ⊂ X, the set NR(A) is infinite if 0 ∈ A, while |NR(A)| = |{n − 1 :n ∈ A}| = |A| if 0 /∈ A. However, G admits no left perfect matching. Indeed,for any matching M ⊂ E, either 0 ∈ X remains unmatched, or, if (0, n) ∈ Mfor some n ∈ N, then n + 1 ∈ X remains unmatched (see Fig. H.6).

Theorem H.3.4. Let G = (X,Y,E) be a bipartite graph. Suppose that Gadmits both a left perfect matching and a right perfect matching. Then Gadmits a perfect matching.

Proof. Let MX (resp. MY ) be a left perfect (resp. right perfect) matchingfor G. Consider the equivalence relation in M = MX ∪ MY defined bydeclaring two edges e and e′ in relation if there exists a finite sequencee = e0, e1, . . . , en = e′ in M such that ei and ei+1 are adjacent for all0 ≤ i ≤ n − 1. Then, each equivalence class consists of either (see Fig. H.7):

– a single edge, or– a cycle of even length 2n ≥ 4, or– an infinite chain which can be bi-infinite, or infinite only in one direction.

H.3 The Hall Marriage Theorem 397

Fig. H.6 The bipartite graph G = (X, Y, E), where X = Y = N and E = {(0, n + 1) : n ∈N} ∪ {(n + 1, n) : n ∈ N}

Note that an equivalence class is reduced to a single edge (x, y) if and onlyif (x, y) ∈ MX ∩ MY .

On the other hand, an equivalence class is a cycle of length 2n if and onlyif it is of the form

C ={(x1, y1), (x2, y2), . . . , (xn, yn)} ∪ {(x1, y2), (x2, y3),. . . , (xn−1, yn), (xn, y1)}

with xi ∈ X, all distinct, and yj ∈ Y , all distinct, 1 ≤ i, j ≤ n. In this case,we then set

M(C) = {(x1, y1), (x2, y2), . . . , (xn, yn)}.

Also, a bi-infinite chain is an equivalence class of the form

C = {(xn, yn) : n ∈ Z} ∪ {(xn, yn+1) : n ∈ Z}

with xi ∈ X, all distinct, and yj ∈ Y , all distinct, i, j ∈ Z. In this case, wethen set

M(C) = {(xn, yn) : n ∈ Z}.


Fig. H.7 In the bipartite graph G = (X, Y, E), where X = {x0, x1, x2, x3, x4, x5}, Y =

{y0, y1, y2, y3, y4, y5} and E = {(xi, yi) : i = 0, 1, . . . , 5} ∪ {(xi, yi+1) : i = 1, 2, . . . , 4} ∪{(x5, y1)}, we have a left-perfect matching MX = {(xi, yi) : i = 0, 1, . . . , 5} (which is indeed

perfect) and a right-perfect matching MY = {(x0, y0)} ∪ {(xi, yi+1) : i = 1, 2, . . . , 4} ∪{(x5, y1)} (which is also perfect). There are two equivalence classes in M = MX ∪ MY ,namely, MX ∩ MY = {(x0, y0)}, which consists of a single edge, and C = {(xi, yi) : i =

1, 2 . . . , 5} ∪ {(xi, yi+1) : i = 1, 2, . . . , 4} ∪ {(x5, y1)}, which is a cycle of length 10

Finally, if the equivalence class is an infinite chain, which is infinite onlyin one direction, then it is either of the form

C = {(xn, yn) : n ∈ N} ∪ {(xn, yn+1) : n ∈ N}

orC = {(xn, yn) : n ∈ N} ∪ {(xn+1, yn) : n ∈ N}

with xi ∈ X, all distinct, and yj ∈ Y , all distinct, i, j ∈ N. In both cases, wethen set

M(C) = {(xn, yn) : n ∈ N}.

It is then clear that the set M ⊂ E defined by

M =⋃

CM(C),

where C runs over all equivalence classes in M , is the required perfect match-ing for G. ��

Corollary H.3.5 (Cantor–Bernstein Theorem). Let X and Y be twosets. Suppose that there exist injective maps f : X → Y and g : Y → X. Thenthere exists a bijective map h : X → Y .

Proof. Consider the bipartite graph G = (X,Y,E), where E = {(x, f(x)) :x ∈ X} ∪ {(g(y), y) : y ∈ Y }. Now, MX = {(x, f(x)) : x ∈ X} and

H.4 The Hall Harem Theorem 399

MY = {(g(y), y) : y ∈ Y } are left perfect and right perfect matchings inG, respectively (cf. Remark H.2.1(a)). By the previous theorem, there existsa perfect matching M ⊂ E for G. Then M = {(x, h(x)) : x ∈ X}, whereh : X → Y is bijective (cf. Remark H.2.1(b)). ��

Theorem H.3.6 (The Hall marriage Theorem). Let G = (X,Y,E) bea locally finite bipartite graph. Then the following conditions are equivalent:

(a) G satisfies the Hall-marriage conditions;(b) G admits a perfect matching.

Proof. The left (resp. right) Hall condition implies, by Theorem H.3.2, theexistence of a left- (resp. right-) perfect matching for G. Then, Theorem H.3.4guarantees the existence of a perfect matching for G. The converse implicationis trivial. ��

H.4 The Hall Harem Theorem

Let G = (X,Y,E) be a bipartite graph and let k ≥ 1 be an integer.A subset M ⊂ E is called a perfect (1, k)-matching if it satisfies the fol-

lowing conditions: (1) for each x ∈ X, there are exactly k elements in y ∈ Ysuch that (x, y) ∈ M , (2) for each y ∈ Y , there is a unique element x ∈ Xsuch that (x, y) ∈ M (see Fig. H.8). Thus, a subset M ⊂ E is a perfect (1, k)-matching if and only if there exists a k-to-one surjective map ψ : Y → X suchthat M = {(ψ(y), y) : y ∈ Y } (recall that a surjective map f : S → T from aset S onto a set T is said to be k-to-one if each element in T has exactly kpreimages in S). Note that when k = 1, a perfect (1, k)-matching is the samething as a perfect matching.

In the language of boys, girls, and marriages, a perfect (1, k)-matching is aprocess for marrying each boy with exactly k girls (among the girls he knows)in such a way that each girl is married with exactly one boy (among the boysshe knows). The girls that are married with a given boy constitute his harem.

Definition H.4.1. Let G = (X,Y,E) be a locally finite bipartite graph andlet k ≥ 1 be an integer. One says that G satisfies the Hall k-harem conditionsif

|NR(A)| ≥ k|A| and|NL(B)| ≥ 1

k |B| for all finite subsets A ⊂ X and B ⊂ Y .(H.5)

Theorem H.4.2 (The Hall harem Theorem). Let G = (X,Y,E) be alocally finite bipartite graph and let k ≥ 1 be an integer. Then, the followingconditions are equivalent.

(a) G satisfies the Hall k-harem conditions;(b) G admits a perfect (1, k)-matching.


Fig. H.8 The bipartite graph G = (X, Y, E), where X = {x1, x2}, Y = {y1, y2, y3, y4} and

E = {(x1, yi) : i = 1, 2, 3, 5} ∪ {(x2, yj) : j = 1, 2, 4, 5, 6}, with a perfect (1, 3)-matchingM = {(x1, yi) : i = 1, 2, 3}∪ {(x2, yj) : j = 4, 5, 6} and the harems H(x1) and H(x2) of x1

and x2 respectively

Proof. The fact that (b) implies (a) is trivial.Let us prove that (a) implies (b). Suppose that the Hall k-harem conditions

are satisfied by G. Let X1, X2, . . . , Xk be disjoint copies of X and let φi : X →Xi, i = 1, 2, . . . , k, denote the copy maps. It will be helpful to think of φi(x)as the clone of x ∈ X in Xi.

Consider the new bipartite graph G′ = (X ′, Y ′, E′), where X ′ =∐k

i=1 Xi,Y ′ = Y , and

E′ = {(φi(x), y) : (x, y) ∈ E, i = 1, 2, . . . , k} ⊂ X ′ × Y ′.

Let A′ be a finite subset of X ′ and denote by A′ the set of x ∈ X such thatsome clone of x belongs to A′. Observe that |A′| ≤ k|A′| and N G′

R (A′) =N G

R(A′). Thus, using (a), we get

|N G′

R (A′)| = |N GR(A′)| ≥ k|A′| ≥ |A′|. (H.6)

On the other hand, if B′ is a finite subset of Y ′ = Y , then N G′

L (B′) is the setconsisting of all clones of elements of N G

L (B′) so that

|N G′

L (B′)| = k|N GL (B′)| ≥ k

(1k|B′|

)= |B′| (H.7)

by (a). Inequalities H.6 and H.7 say that G′ satisfies the Hall marriage condi-tions. Therefore, G′ admits a perfect matching M ′ ⊂ E′ by Theorem H.3.6.Then the set M , consisting of all pairs (x, y) ∈ E such that (x′, y) ∈ M ′ forsome clone x′ of x, is clearly a perfect (1, k)-matching for G. ��

Notes 401

Notes

The Hall marriage theorem was first established for finite bipartite graphsby P. Hall [Hall1] and then extended to infinite locally finite bipartite graphsby M. Hall [Hall-M-1]. The proof of Theorem H.3.2 which is given in thisappendix is based on [Halm].

Appendix I

Complements of Functional Analysis


I.1 The Baire Theorem

Let (X, d) be a metric space. Given x ∈ X and r > 0, we denote by

BX(x, r) = {y ∈ X : d(x, y) ≤ r}

the closed ball of radius r centered at x and by

OBX(x, r) = {y ∈ X : d(x, y) < r}

the open ball of radius r centered at x. For a subset A ⊂ X we denote by A(resp. Int A) the closure (resp. the interior) of A.

Theorem I.1.1 (Baire’s Theorem). Let (X, d) be a complete metric space.Let (Xn)n≥1 be a sequence of closed subsets such that

IntXn = ∅ (I.1)

for all n ≥ 1. Then

Int

⎛

⎝⋃

n≥1

Xn

⎞

⎠ = ∅. (I.2)

Proof. Let us set An = X \ Xn, so that An is an open dense subset of X.Let us show that

⋂n≥1 An is dense in X. Let x0 ∈ X and r0 > 0. We have

to show that

BX(x0, r0)⋂

⎛

⎝⋂

n≥1

An

⎞

⎠ �= ∅. (I.3)


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404 I Complements of Functional Analysis

As A1 is dense and open we can find x1 ∈ BX(x0, r0)⋂

A1 and r1 > 0 suchthat {

BX(x1, r1) ⊂ OBX(x0, r0)⋂

A1

0 < r1 < r02 .

Continuing this way, we produce by induction a sequence (xn)n≥1 in X anda sequence (rn)n≥1 in R such that

{BX(xn+1, rn+1) ⊂ OBX(xn, rn)

⋂An+1

0 < rn+1 < rn

2 .(I.4)

Given 0 ≤ n ≤ m we have

xm ∈ BX(xn, rn) (I.5)

so that d(xn, xm) < r02n . We deduce that (xn)n≥1 is a Cauchy sequence.

As X is complete, (xn)n≥1 is convergent so that there exists x ∈ X suchthat limn→∞ xn = x. Passing to the limit (for m → ∞) in (I.5) we obtainx ∈ BX(xn, rn) ⊂ BX(x0, r0) and, by (I.4), x ∈ BX(xn, rn) ⊂ An for alln ≥ 1. As a consequence, x ∈ Bx0,r0

⋂(⋂

n≥1 An) and (I.3) follows. Thisshows that

⋂n≥1 An is dense in X. Taking complements, we deduce that⋃

n≥1 Xn =⋃

n≥1(X \ An) = X \⋂

n≥1 An has empty interior, and (I.2)follows.

Remark I.1.2. We can restate Baire’s theorem in the following two ways.(i) Let (X, d) be a nonempty complete metric space. Let (Xn)n≥1 be a

sequence of closed subsets such that⋃

n≥1 Xn = X. Then there exists n0

such that IntXn0 �= ∅.(ii) (By passing to complements) Let (X, d) be a nonempty complete metric

space. Let (Ωn)n≥1 be a sequence of open dense subsets in X. Then⋂

n≥1 Ωn

is dense in X.

I.2 The Open Mapping Theorem

Recall that if X and Y are two topological spaces, then a map T : X → Y issaid to be open if for any open set A in X the image T (A) is open in Y .

Theorem I.2.1 (Open mapping theorem). Let X and Y be two Banachspaces. Then every surjective continuous linear map T : X → Y is open.

Proof. Let T : X → Y be a surjective continuous linear map. We need to showthat for every x ∈ X and every neighborhood N of x, the image T (N) is aneighborhood of T (x). Since, by linearity, T (x + N) = T (x) + T (N), we canreduce to the case when x = 0. Moreover, since neighborhoods contain openballs, it is sufficient to show that for every r′ > 0 there exists r′′ > 0 such that

I.2 The Open Mapping Theorem 405

T (OBX(0, r′)) ⊃ OBY (0, r′′). Finally, since T (OBX(0, r′)) = r′T (OBX(0, 1))and OBY (0, r′′) = r′′OBY (0, 1), we are only left to show that there existsr > 0 such that

T (OBX(0, 1)) ⊃ OBY (0, r). (I.6)

For all n ≥ 1 set Yn = nT (OBX(0, 1)) = T (OBX(0, n)). As T is surjective wehave

⋃n≥1 Yn = Y . It follows from Theorem I.1.1 that there exists n0 ≥ 1

such that IntYn0 �= ∅. Since Int(n0T (OBX(0, 1))) = n0 Int(T (OBX(0, 1))),we deduce that Int(T (OBX(0, 1))) �= ∅. Thus we can find r > 0 and y ∈ Ysuch that

OBY (y, 4r) ⊂ T (OBX(0, 1)). (I.7)

It follows that y ∈ T (OBX(0, 1)) and, by symmetry,

− y ∈ T (OBX(0, 1)). (I.8)

Summing (I.7) and (I.8) we obtain

2OBY (0, 2r) = OBY (0, 4r)= OBY (y, 4r) − y

⊂ T (OBX(0, 1)) + T (OBX(0, 1))

⊂ 2T (OBX(0, 1)).

We deduce that OBY (0, 2r) ⊂ T (OBX(0, 1)) so that, for all n ≥ 0,

OBY

(0,

r

2n

)⊂ T (OBX(0, 1/2n+1)). (I.9)

Let y0 ∈ OBY (0, r) and let us show that there exists x′ ∈ OBX(0, 1) suchthat y0 = T (x′). We deduce from (I.9) (with n = 0) that there exists x0 ∈OBX(0, 1

2 ) such that ‖y0 −T (x0)‖ < r2 . Set y1 = y0 −T (x0) and observe that

y1 ∈ OBY (0, r2 ). Continuing this way, we find a sequence (yn)n≥1 in Y and a

sequence (xn)n≥1 in X such that

yn+1 = yn − T (xn) ∈ OBY

(0,

r

2n+1

), (I.10)

xn ∈ OBX

(0,

12n+1

)(I.11)

and‖yn − T (xn)‖ <

r

2n+1. (I.12)

Now, (I.11) is equivalent to ‖xn‖ < 12n+1 so that setting x′

n = x0+x1+· · ·+xn,the sequence (x′

n)n≥1 is convergent in X. It follows that there exists x′ ∈ Xsuch that limn→∞ x′

n = x′ and, moreover ‖x′‖ < 1. On the other hand,


(I.10) and (I.12) give ‖y0 − T (x′n)‖ = ‖y0 − T (x0 + x1 + · · · + xn)‖ = ‖y1 −

T (x1 + x2 + · · · + xn)‖ = · · · = ‖yn − T (xn)‖ ≤ r2n+1 so that, passing to

the limit, we deduce that T (x′) = y0, since T is continuous. This shows thatOBY (0, r) ⊂ T (OBX(0, 1)). It follows that T is an open map.

Corollary I.2.2. Let X and Y be two Banach spaces. Let T : X → Y be abijective continuous linear map. Then the inverse linear map T−1 : Y → Xis continuous.

Proof. This follows immediately from Theorem I.2.1 since the inverse mapT−1 : Y → X is continuous if and only if T : X → Y is an open map.

Corollary I.2.3. Let X and Y be two Banach spaces. Let T : X → Y be acontinuous linear map. Suppose that for every ε > 0 there exists x ∈ X suchthat ‖x‖ = 1 and ‖T (x)‖ < ε. Then T is not bijective.

Proof. Suppose by contradiction that T is bijective. Then, by Corollary I.2.2,the inverse linear map T−1 : Y → X is continuous. Thus we can find a con-stant M > 0 such that ‖T−1(y)‖ ≤ M‖y‖ for all y ∈ Y . Since T is bijective,this is equivalent to ‖x‖ ≤ M‖T (x)‖ for all x ∈ X. This clearly contradictsthe hypotheses. Thus, T is not bijective.

I.3 Spectra of Linear Maps

Let (X, ‖·‖) be a Banach space. We denote by L(X) the space of all continuouslinear maps T : X → X endowed with the norm

‖T‖ = supx∈Xx�=0

‖T (x)‖‖x‖ .

We denote by IdX : X → X the identity map.

Definition I.3.1. Let T ∈ L(X). The set

σ(T ) = {λ ∈ R : (T − λ IdX) is not bijective} (I.13)

is called the real spectrum of T .

Proposition I.3.2. Let T ∈ L(X). Then the spectrum σ(T ) is a compact setand

σ(T ) ⊂ [−‖T‖, ‖T‖]. (I.14)

Proof. Let λ ∈ R and suppose that |λ| > ‖T‖. For y ∈ X the equation

(T − λ IdX)(x) = y (I.15)

I.4 Uniform Convexity 407

admits a unique solution x ∈ X. Indeed (I.15) is equivalent to

x =1λ

(T (x) − y). (I.16)

Moreover, ‖ 1λ (T (x1)−y)− 1

λ (T (x2)−y)‖ ≤ 1|λ|‖T‖·‖x1−x2‖ for all x1, x2 ∈ X

and 1|λ|‖T‖ < 1. It then follows from the Banach fixed point theorem that

there exists a unique x ∈ X satisfying (I.16). It follows that T − λ IdX isbijective and therefore λ /∈ σ(T ). This shows (I.14).

Let us show that R \ σ(T ) is open. Let λ0 ∈ R \ σ(T ) so that T − λ0 IdX

is bijective. By Corollary I.2.2, we have that the inverse map (T −λ0 IdX)−1

is also continuous so that 0 �= ‖(T − λ0 IdX)−1‖ < ∞. Let λ ∈ R such that|λ − λ0| < 1

‖(T−λ0 IdX)−1‖ . For y ∈ X we have that the linear equation (I.15)can be written as T (x) − λ0x = y + (λ − λ0)x, that is,

x = (T − λ0 IdX)−1[y + (λ − λ0)x]. (I.17)

By applying again the Banach fixed point theorem, we deduce that thereexists a unique x ∈ X satisfying (I.17). This shows that T −λ IdX is bijective,that is, λ ∈ R\σ(T ). It follows that R\σ(T ) is open. Therefore σ(T ) is closed.

I.4 Uniform Convexity

Definition I.4.1. A normed space (X, ‖ · ‖) is said to be uniformly convexif, for every ε > 0, there exists δ > 0 such that

‖x − y‖ > ε implies∥∥∥∥

x + y

2

∥∥∥∥ < 1 − δ

for all x, y ∈ X with ‖x‖, ‖y‖ ≤ 1.

Let Z be a nonempty set not reduced to a single point. Then the Banachspace �1(Z) is not uniformly convex. For instance, if z, z′ ∈ Z are distinct,setting x = δz and y = δz′ , one has ‖x‖1 = ‖y‖1 = 1, ‖x − y‖1 = 2, but‖x+y

2 ‖1 = 1.On the other hand, we have the following:

Proposition I.4.2. Let X be a vector space equipped with a scalar product〈·, ·〉 and denote by ‖ ·‖ the associated norm. Then the normed space (X, ‖ ·‖)is uniformly convex. In particular, every Hilbert space is uniformly convex.

Proof. Let x, y ∈ X such that ‖x‖, ‖y‖ ≤ 1. Then, we have


∥∥∥∥x+ y

2

∥∥∥∥2

+14‖x− y‖2 =

14

(〈x + y, x + y〉 + 〈x − y, x − y〉)

=14

(〈x, x〉+ 〈y, y〉+ 2〈x, y〉+ 〈x, x〉+ 〈y, y〉− 2〈x, y〉)

=14

(2‖x‖2 + 2‖y‖2

)

≤ 1.

Therefore, ∥∥∥∥x + y

2

∥∥∥∥2

≤ 1 − 14‖x − y‖2. (I.18)

Let now ε > 0 and set δ = 1 − 12

√4 − ε2. Suppose that ‖x − y‖ > ε. This

implies 1 − 14‖x − y‖2 < 1 − ε2

4 = (1 − δ)2. From (I.18) we then deduce that‖x+y

2 ‖ < 1 − δ. This shows that X is uniformly convex.

Appendix J

Ultrafilters

J.1 Filters and Ultrafilters

Let X be a set. We denote by P(X) the set of all subsets of X.

Definition J.1.1. A filter on X is a nonempty set F ⊂ P(X) satisfying thefollowing conditions:

(F-1) ∅ /∈ F ;(F-2) if A ∈ F and A ⊂ B ⊂ X, then B ∈ F ;(F-3) if A ∈ F and B ∈ F , then A ∩ B ∈ F .

Examples J.1.2. (a) Let X be a set and let A0 be a nonempty subset of X.Then the set {A ∈ P(X) : A0 ⊂ A} is a filter on X. A filter F on X is saidto be a principal filter if there exists a nonempty subset A0 of X such thatF = {A ∈ P(X) : A0 ⊂ A}. One then says that F is the principal filter basedon A0.

(b) Let X be a set and Ω ⊂ P(X). Suppose that ∅ /∈ Ω and that, givenA′

1, A′2 ∈ Ω, there exists A′ ∈ Ω such that A′ ⊂ A′

1 ∩ A′2. Then the set

F(Ω) = {A ∈ P(X) : there exists A′ ∈ Ω such that A′ ⊂ A} (J.1)

is a filter on X and one has Ω ⊂ F(Ω). The filter F(Ω) is called the filtergenerated by Ω.

(c) Let (I,≤) be a nonempty directed set. A subset A ⊂ I is called residualin I if there exists i ∈ I such that A ⊃ {j ∈ I : i ≤ j}. Clearly, the set Fr(I)of all residual subsets of I is a filter on I. It is the filter generated by the sets{j ∈ I : i ≤ j}, i ∈ I. The filter Fr(I) is called the residual filter on I.

(d) Let X be an infinite set. Then the set F = {A ∈ P(X) : X\A is finite}is a filter. It is called the Frechet filter on X. If we consider the directedset (N,≤), a subset A ⊂ N is residual if and only if its complement N \ A isfinite. Therefore, the residual filter on N equals the Frechet filter on N.


409

http://dx.doi.org/10.1007/978-3-642-14034-1_18

410 J Ultrafilters

(e) Let X be a topological space and let x be a point in X. Then, the setFx of all neighborhoods of x is a filter on X.

(f) Let X be a nonempty uniform space (cf. Appendix B). Then the setU ⊂ P(X × X) of all entourages of X is a filter on X × X.

Note that for a filter F ⊂ P(X) the following also holds true:

(F-4) if A ⊂ X, then A and X \ A cannot both belong to F ;(F-5) if A1, A2, . . . , An ∈ F , n ≥ 1, then

⋂ni=1 Ai ∈ F ;

(F-6) X ∈ F .

Indeed, (F-4) follows from (F-1) and (F-3) with B = X \ A. From (F-3)we deduce by induction condition (F-5). Finally, (F-6) follows from the factthat F �= ∅ and (F-2).

Proposition J.1.3. Let X be a set and Ω0 ⊂ P(X). Then, there exists afilter on X containing Ω0 if an only if Ω0 has the finite intersection property,that is, every finite family of elements in Ω0 has nonempty intersection.

Proof. Let Ω0 ⊂ P(X) be a set and suppose that there exists a filter F suchthat Ω0 ⊂ F . By (F-1) and (F-5) we have that Ω0 has the finite intersectionproperty. Conversely, suppose that Ω0 has the finite intersection property.Consider the set Ω consisting of all finite intersections of elements of Ω0.Then, Ω satisfies the conditions in Example J.1.2(b), and therefore the filtergenerated by Ω is a filter containing Ω0.

Definition J.1.4. A filter ω ⊂ P(X) is called an ultrafilter on X if it satis-fies the condition

(UF) if A ⊂ X, then A ∈ ω or (X \ A) ∈ ω.

Example J.1.5. Let X be a set and x ∈ X. Then the principal filter based on{x} is an ultrafilter. An ultrafilter ω on X is called principal if there existsx ∈ X such that ω is the principal filter based on {x}. Note that a principalfilter is an ultrafilter if and only if it is based on a singleton. If X is finite,then every ultrafilter is, clearly, principal.

An ultrafilter which is not principal is called non-principal (or free).

Theorem J.1.6. Let X be a nonempty set. Let F0 be a filter on X. Thenthere exists an ultrafilter ω on X containing F0.

Let us first prove the following:

Lemma J.1.7. Let X be a set. Let F be a filter on X. Suppose there existsa set A0 ∈ P(X) such that A0 /∈ F and (X \ A0) /∈ F . Then there exists afilter F ′ on X such that

J.1 Filters and Ultrafilters 411

(1) F ⊂ F ′;(2) A0 ∈ F ′.

Proof. Consider the principal filter F0 based on A0 and set

F ′ = {B ∈ P(X) : B ⊃ A ∩ A′ for some A ∈ F and A′ ∈ F0}. (J.2)

Let us show that F ′ satisfies the required conditions. First of all, takingA′ = X we have A = A ∩ X ∈ F ′ for all A ∈ F . This shows (1). On theother hand, taking A = X and A′ = A0 we have A0 = X ∩ A0 ∈ F , and thisshows (2).

We are only left to show that F ′ is a filter. Let B ∈ F ′ and denote byA ∈ F and A′ ∈ F0 two sets such that A ∩ A′ ⊂ B.

Let us first show that A ∩ A′ �= ∅. Suppose the contrary. Then A ⊂(X \A′) ⊂ (X \ A0). As A belongs to the filter F , it follows from (F-2) that(X \A0) ∈ F , contradicting our assumptions. It follows that A∩A′ �= ∅ andtherefore B �= ∅. This shows (F-1). Suppose now that B′ ∈ P(X) containsB. Then (A ∩ A′) ⊂ B′ so that B′ ∈ F ′. This shows (F-2). Finally, letB1, B2 ∈ F ′. For i = 1, 2 denote by Ai ∈ F and A′

i ∈ F0 two sets such thatAi ∩ A′

i ⊂ Bi. We have

B1 ∩ B2 ⊃ (A1 ∩ A′1)

⋂(A2 ∩ A′

2) = (A1 ∩ A2)⋂

(A′1 ∩ A′

2).

But A1 ∩ A2 belongs to the filter F , and A′1 ∩ A′

2 ∈ F0 as both A′1 and A′

2

contain A0. It follows that B1 ∩ B2 ∈ F ′. This shows (F-3). It follows thatF ′ is a filter.

Proof of Theorem J.1.6. Consider the set Φ0 consisting of all filters on Xcontaining F0. This is a nonempty set, partially ordered by inclusion. Let Φbe a totally ordered subset of Φ0. We claim that F =

⋃F∈Φ F is an upper

bound for Φ. We only have to show that F belongs to Φ0. As ∅ /∈ F for allF ∈ Φ we also have ∅ /∈ F . This shows that F satisfies condition (F-1). Letnow A ∈ F and B ⊂ X be such that A ⊂ B. Then there exists F ∈ Φ suchthat A ∈ F . As F is a filter, we have B ∈ F , by (F-2). It then follows thatB ∈ F . This shows that F satisfies condition (F-2). Finally, suppose thatA, B ∈ F . Then there exist FA,FB ∈ Φ such that A ∈ FA and B ∈ FB . As Φis totally ordered, up to exchanging A and B we can suppose that FA ⊂ FB .We then have A, B ∈ FB and therefore A∩B ∈ FB , by (F-3). It follows thatA∩B ∈ F . Therefore F satisfies condition (F-3) as well. This shows that Φ0

is inductive.By Zorn’s lemma, Φ0 contains a maximal element ω. Let us show that ω

is an ultrafilter on X. Let A0 ⊂ X. Suppose that A0, (X \ A0) /∈ ω. Then,by Lemma J.1.7, there exists a filter ω′ containing A0 and ω. As ω′ is in Φ0

and properly contains ω, this contradicts the maximality of ω. It follows thateither A0 or X \A0 belongs to ω. This shows that ω satisfies condition (UF),and therefore it is an ultrafilter.

412 J Ultrafilters

Corollary J.1.8. Let X be a set. Let F be a filter on X. Then the followingconditions are equivalent.

(a) F is an ultrafilter;(b) F is a maximal filter, that is, if F ′ is a filter containing F , then F ′ = F .

Proof. Suppose (a) and let F ′ be a filter properly containing F . Let A ∈F ′ \ F . As F is an ultrafilter, (X \ A) ∈ F ⊂ F ′. Thus, both A and (X \ A)belong to F ′ contradicting (F-4). This shows that F ′ = F and the implication(a) ⇒ (b) follows. Suppose now that F is a maximal filter. By Theorem J.1.6there exists an ultrafilter ω containing F . By maximality we have F = ω,that is, F is an ultrafilter. This shows (b) ⇒ (a).

J.2 Limits Along Filters

Definition J.2.1. Let X be a topological space. A filter F on X is said tobe convergent if there exists a point x0 ∈ X such that all neighborhoods ofx0 belong to F . One then says that x0 is a limit of F and that F convergesto x0.

Remark J.2.2. A topological space X is Hausdorff if and only if every con-vergent filter on X has a unique limit point in X. Indeed, suppose that Xis Hausdorff and let F be a filter converging to two distinct points x, y ∈ X.Let U ∈ Fx ⊂ F and V ∈ Fy ⊂ F be such that U ∩ V = ∅. As U, V ∈ Fwe also have U ∩V ∈ F and this contradicts (F-1). Conversely, suppose thatX is not Hausdorff. Then there exist two distinct points x, y ∈ X such thatfor all U ∈ Fx and V ∈ Fy one has U ∩ V �= ∅. It follows that the setΩ = Fx ∪Fy ⊂ P(X) has the finite intersection property. Thus, by Proposi-tion J.1.3, there exists a filter F containing both Fx and Fy. It follows thatF converges to both x and y.

Theorem J.2.3. Let X be a topological space. Then the following conditionsare equivalent:

(a) X is compact;(b) every ultrafilter on X is convergent.

Proof. Suppose (a) and let ω be an ultrafilter on X. Let A1, A2, . . . , An,n ≥ 1, be elements in ω and denote by A1, A2, . . . , An their closures. As⋂n

i=1 Ai ⊃⋂n

i=1 Ai �= ∅ (by (F-1) and (F-5)), we have that the family (A)A∈ω

in X has the finite intersection property and therefore, by compactness of X,it has a nonempty intersection. Let x ∈

⋂A∈ω A. As x ∈ A for all A ∈

ω, it follows that for every V ∈ Fx we have A ∩ V �= ∅. Consider theset Ω = {A ∩ V : A ∈ ω, V ∈ Fx} ⊂ P(X). Observe that Ω has thefinite intersection property and denote by F(Ω) the filter generated by Ω

J.2 Limits Along Filters 413

(cf. Example J.1.2(b)). We have ω ⊂ Ω ⊂ F(Ω) and, as ω is an ultrafilter,it follows from Corollary J.1.8 that ω = F(Ω). As Fx ⊂ Ω, we deduce thatFx ⊂ ω, that is, ω converges to x.

Conversely, suppose (b). Let (Ci)i∈I be a family of closed subsets of Xwith the finite intersection property. To prove that X is compact we have toshow that ⋂

i∈I

Ci �= ∅. (J.3)

By Proposition J.1.3 there exists a filter F such that Ci ∈ F for all i ∈ I.By Theorem J.1.6 there exists an ultrafilter ω such that F ⊂ ω. By ourassumptions, there exists x ∈ X such that ω converges to x, equivalently,Fx ⊂ ω. Let i ∈ I. by (F-1) and (F-3), we have Ci ∩ V �= ∅ for all V ∈ Fx.As Ci is closed, we have x ∈ Ci. Thus x ∈

⋂i∈I Ci, and (J.3) follows.

Definition J.2.4. Let X be a set, Y a topological space, y0 a point of Y ,f : X → Y a map and F a filter on X. One says that y0 is a limit of f alongF (or that f(x) converges to y0 along F), and one writes

f(x) −→x→F

y0,

if f−1(V ) = {x ∈ X : f(x) ∈ V } belongs to F for all neighborhoods V of y0.If such a limit point y0 is unique one writes

limx→F

f(x) = y0.

Examples J.2.5. (a) Let X and Y be two topological spaces, f : X → Y amap, x0 ∈ X and y0 ∈ Y . One has that f(x) converges to y0 in Y for xtending to x0 if and only if f(x) −→

x→Fx0

y0.

(b) Let X be a topological space and f : N → X a map. The sequence(f(n))n∈N converges to x ∈ X, if and only if x is a limit of f along theFrechet filter on N.

(c) Let (I,≤) be a directed set and let F be the residual filter on I. LetY be a topological space and (yi)i∈I a net in Y . Then (yi)i∈I converges to apoint y0 ∈ Y if and only if yi −→

i→Fy0. Note that (b) is a particular case of

the present example.

Corollary J.2.6. Let X be a set, Y a compact topological space, f : X →Y a map, and ω an ultrafilter on X. Then there exists y0 ∈ Y such thatf(x) −→

x→Fy0. Moreover, if X is Hausdorff such an y0 is unique.

Proof. The set f(ω) = {f(A) : A ∈ ω} ⊂ P(Y ) has the finite intersectionproperty since f(A)∩f(B) ⊃ f(A∩B) �= ∅ for all A, B ∈ ω. Let F denote thefilter generated by f(ω). By Theorem J.1.6, there exists an ultrafilter ω′ on Ywhich contains F . As Y is compact, by Theorem J.2.3 there exists y0 ∈ Y suchthat ω′ converges to y0. Let us show that if V is a neighborhood of y0, then

414 J Ultrafilters

f−1(V ) belongs to ω. Suppose the contrary. As ω is an ultrafilter, by (UF)we have X \ f−1(V ) = {x ∈ X : f(x) /∈ V } ∈ ω. Setting U = f(X \ f−1(V )),we have U ∈ f(ω) ⊂ F ⊂ ω′. But V ∈ Fy0 ⊂ ω′ and, by construction,U ∩ V = ∅. As U ∩ V ∈ ω′, this contradicts (F-1). This shows that y0 isa limit of f along ω. If X is Hausdorff, then uniqueness of y0 follows fromRemark J.2.2.

Proposition J.2.7. Let X be a set and let F be a filter on X. Let Y , Z betwo topological spaces and f : X → Y and g : X → Z be two maps. EquipY × Z with the product topology and let F : X → Y × Z be the map definedby F (x) = (f(x), g(x)) for all x ∈ X. Suppose that there exist y0 ∈ Y andz0 ∈ Z such that f(x) −→

x→Fy0 and g(x) −→

x→Fz0. Then F (x) −→

x→F(y0, z0).

Proof. Let W ⊂ Y ×Z be a neighborhood of the point (y0, z0). By definitionof the product topology, there exist neighborhoods U ⊂ Y of y0 and V ⊂ Z ofz0 such that U ×V ⊂ W . As f(x) −→

x→Fy0, we have f−1(U) ∈ F . Similarly, as

g(x) −→x→F

z0, we have g−1(V ) ∈ F . It follows that f−1(U) ∩ g−1(V ) belongs

to F and as F−1(W ) ⊃ F−1(U × V ) = f−1(U) ∩ g−1(V ), we deduce from(F-2) that F−1(W ) ∈ F . This shows that F (x) −→

x→F(y0, z0).

Proposition J.2.8. Let X be a set and let F be a filter on X. Let Y , Z betwo topological spaces and f : X → Y and g : Y → Z be two maps. Supposethat there exists y0 ∈ Y such that f(x) −→

x→Fy0 and that g is continuous at

y0. Then (g ◦ f)(x) −→x→F

g(y0).

Proof. Let W ⊂ Z be a neighborhood of g(y0). By continuity of g there existsa neighborhood V ⊂ Y of y0 such that g−1(W ) ⊃ V . By definition of limit,we have f−1(V ) ∈ F . As (g ◦ f)−1(W ) = f−1(g−1(W )) ⊃ f−1(V ), it followsfrom (F-2) that (g ◦ f)−1(W ) ∈ F . This shows that (g ◦ f)(x) −→

x→Fg(y0).

Proposition J.2.9. Let X be a set and let F be a filter on X. Let f1, f2 : X →R be two maps such that f1 ≤ f2 (i.e. f1(x) ≤ f2(x) for all x ∈ X). Supposethat there exists y1 ∈ R (resp. y2 ∈ R) such that y1 = limx→F f1(x) (resp.y2 = limx→F f2(x)). Then y1 ≤ y2.

Proof. Suppose, by contradiction, that y1 > y2. Set r = y1−y23 and let V1 =

(y1−r, y1+r) and V2 = (y2−r, y2+r). Note that V1∩V2 = ∅. By definition oflimit we have f−1

1 (V1) ∈ F and f−12 (V2) ∈ F . As F is a filter we deduce that

f−11 (V1) ∩ f−1

2 (V2) ∈ F . On the other hand we have f−11 (V1) ∩ f−1

2 (V2) = ∅

since f1 ≤ f2. This contradicts (F-2). It follows that y1 ≤ y2.

Recall that given a set X, we denote by �∞(X) the Banach space con-sisting of all bounded real maps f : X → R equipped with the norm‖f‖∞ = sup{|f(x)| : x ∈ X}. Moreover, a linear map m : �∞(X) → R

satisfying m(1) = 1 and m(x) ≥ 0 for all x ∈ �∞(E) such that x ≥ 0, iscalled a mean on X (cf. Definition 4.1.4).

Notes 415

Corollary J.2.10. Let X be a set and let ω be an ultrafilter on X. For everyf ∈ �∞(X) there exists a unique y0 ∈ R such that f(x) −→

x→ωy0. Moreover,

the map mω : �∞(X) → R defined by mω(f) = limx→ω f(x) is a mean on X,in particular mω is continuous and ‖mω‖ = 1.

Proof. Let f ∈ �∞(X). Using the fact that f(x) ∈ Y = [−‖f‖∞, ‖f‖∞] forall x ∈ X and that Y is compact Hausdorff, we deduce from Corollary J.2.6that there exists a unique y0 ∈ R such that f(x) −→

x→ωy0.

Let us show that the map mω is linear. Let a ∈ R. Consider the continuousmap g : R → R defined by g(y) = ay for all y ∈ R. Then af = g ◦ f and fromProposition J.2.8 we deduce

mω(af) = limx→ω

(af)(x) = a limi→ω

f(x) = amω(f).

Let now f, g ∈ �∞(X). Consider the map F : X → R2 defined by setting

F (x) = (f(x), g(x)) for all x ∈ X and the continuous map G : R2 → R defined

by G(y1, y2) = y1 + y2. Then f + g = G ◦ F and from Proposition J.2.7 andProposition J.2.8 we deduce

mω(f + g) = limx→ω

(f + g)(x) =(

limx→ω

f(x))

+(

limx→ω

g(x))

= mω(f) + mω(g).

This shows that mω is linear.Let now f(x) = 1 for all x ∈ X. For every neighborhood V of 1 ∈ R we

have f−1(V ) = {x ∈ X : f(x) ∈ V } = X ∈ ω, by (F-6). This shows thatmω(1) = mω(f) = limx→ω f(x) = 1.

Finally, it follows immediately from Proposition J.2.9 that if f ≥ 0 thenmω(f) = limx→ω f(x) ≥ 0.

We have shown that mω is a mean. The last properties of mω follow fromProposition 4.1.7.

Notes

The definition of filter is due to H. Cartan (1937). The full treatment ofconvergence along filters is given in Bourbaki [Bou] as an alternative to thesimilar notion of a net developed in 1922 by E. H. Moore and H. L. Smith.Given a Hausdorff topological space X the set βX of all ultrafilters on Xcan be given the structure of a compact Hausdorff space, called the Stone-Cech compactification of X. This is the largest compact Hausdorff space“generated” by X, in the sense that any map from X to a compact Hausdorffspace factors through βX in a unique way. The elements x of X correspondto the principal ultrafilters ω({x}) on X.

Let now X be any set. With every ultrafilter ω on X one associatesthe {0, 1}-valued finitely additive probability measure μω on X defined

416 J Ultrafilters

by μω(A) = 1 if A ∈ ω and μω(A) = 0 if A ∈ P(X) \ ω. Conversely,given a {0, 1}-valued finitely additive probability measure μ on X, the setωμ = {A ∈ P(X) : μ(A) = 1} is an ultrafilter on X. This establishes a one-to-one correspondence between the ultrafilters ω on X and the {0, 1}-valuedfinitely additive probability measures on X. In Sect. 4.1 we considered theset MP(X) (resp. M(X)) of all finitely additive probability measures (resp.of all means) on X and we showed that there exists a natural bijective mapΦ : MP(X) → M(X). Then one has Φ−1(μω) = mω for all ultrafilters ωon X.

Open Problems

In the list below we collect some open problems related to the topics treatedin this book.

(OP-1) Let G be an amenable periodic group which is not locally finite.Does there exist a finite set A and a cellular automaton τ : AG → AG

which is surjective but not injective?(OP-2) Let G be a periodic group which is not locally finite and let A

be an infinite set. Does there exist a bijective cellular automatonτ : AG → AG which is not invertible?

(OP-3) Let G be a periodic group which is not locally finite and let V bean infinite-dimensional vector space over a field K. Does there exista bijective linear cellular automaton τ : V G → V G which is notinvertible?

(OP-4) Is every Gromov-hyperbolic group residually finite (resp. residuallyamenable, resp. sofic, resp. surjunctive)?

(OP-5) (Gottschalk’s conjecture) Is every group surjunctive?(OP-6) Let G be a periodic group which is not locally finite and let A

be an infinite set. Does there exist a cellular automaton τ : AG →AG whose image τ(AG) is not closed in AG with respect to theprodiscrete topology?

(OP-7) Let G be a periodic group which is not locally finite and let V bean infinite-dimensional vector space over a field K. Does there exista linear cellular automaton τ : V G → V G whose image τ(V G) is notclosed in V G with respect to the prodiscrete topology?

(OP-8) Let G and H be two quasi-isometric groups. Suppose that G issurjunctive. Is it true that H is surjunctive?

(OP-9) Let G be a non-amenable group. Does there exist a finite set A anda cellular automaton τ : AG → AG which is pre-injective but notsurjective?

(OP-10) Does there exist a non-sofic group?(OP-11) Does there exist a surjunctive group which is non-sofic?

T. Ceccherini-Silberstein, M. Coornaert, Cellular Automata and Groups,Springer Monographs in Mathematics,DOI 10.1007/978-3-642-14034-1, © Springer-Verlag Berlin Heidelberg 2010

417

http://dx.doi.org/10.1007/978-3-642-14034-1

418 Open Problems

(OP-12) Let G and H be two quasi-isometric groups. Suppose that G is sofic.Is it true that H is sofic?

(OP-13) Let G be a non-amenable group and let K be a field. Does thereexist a finite-dimensional K-vector space V and a linear cellularautomaton τ : V G → V G which is pre-injective but not surjective?

(OP-14) Let G be a non-amenable group and let K be a field. Does thereexist a finite-dimensional K-vector space V and a linear cellularautomaton τ : V G → V G which is surjective but not pre-injective?

(OP-15) (Kaplanski’s stable finiteness conjecture) Is the group algebra K[G]stably finite for any group G and any field K? Equivalently, is everygroup L-surjunctive, that is, is it true that, for any group G, anyfield K, and any finite-dimensional K-vector space V , every injectivelinear cellular a automaton τ : V G → V G is surjective?

(OP-16) (Kaplanski’s zero-divisors conjecture) Is it true that the group alge-bra K[G] has no zero-divisors for any torsion-free group G and anyfield K? Equivalently, is it true that, for any torsion-free group Gand any field K, every non-identically-zero linear cellular automatonτ : K

G → KG is pre-injective?

(OP-17) Is every unique-product group orderable?

Comments

(OP-1) The answer to this question is affirmative if G is non-periodic, i.e.,it contains an element of infinite order (see Exercise 3.23), or if G is non-amenable (Theorem 5.12.1). On the other hand, if G is a locally finite groupand A is a finite set, then every surjective cellular automaton τ : AG → AG

is injective (see Exercise 3.21). An example of an amenable periodic groupwhich is not locally finite is provided by the Grigorchuck group described inSect. 6.9.

(OP-2) The answer is affirmative if G is not periodic (cf. [CeC11, Corol-lary 1.2]). On the other hand, if G is locally finite and A is an arbitraryset, then every bijective cellular automaton τ : AG → AG is invertible (cf.Exercise 3.20 or [CeC11, Proposition 4.1]).

(OP-3) The answer is affirmative if G is not periodic (cf. [CeC11, The-orem 1.1]). On the other hand, if G is locally finite and V is an arbitraryvector space, then every bijective linear cellular automaton τ : V G → V G isinvertible (cf. [CeC11, Proposition 4.1]).

(OP-5) Every sofic group is surjunctive (cf. Theorem 7.8.1).(OP-6) When A is a finite set and G is an arbitrary group, it follows

from Lemma 3.3.2 that the image of every cellular automaton τ : AG → AG

is closed in AG. When A is an infinite set and G is a non-periodic group,it is shown in [CeC11, Corollary 1.4] that there exists a cellular automatonτ : AG → AG whose image is not closed in AG. On the other hand, when G

Comments 419

is locally finite, then, for any set A, the image of every cellular automatonτ : AG → AG is closed in AG (cf. Exercise 3.22 or [CeC11, Proposition 4.1]).

(OP-6) When V is a finite-dimensional vector space over a field K andG is an arbitrary group, it follows from Theorem 8.8.1 that the image ofevery linear cellular automaton τ : V G → V G is closed in V G. When Vis an infinite-dimensional vector space and G is a non-periodic group, it isshown in [CeC11, Theorem 1.3] that there exists a linear cellular automatonτ : V G → V G whose image is not closed in V G. On the other hand, when G islocally finite, then, for any vector space V , the image of every linear cellularautomaton τ : V G → V G is closed in V G (cf. [CeC11, Proposition 4.1]).

(OP-9) The answer is affirmative if G contains a nonabelian free subgroup(cf. Proposition 5.11.1).

(OP-13) The answer is affirmative if G contains a nonabelian free subgroup(cf. Corollary 8.10.2).

(OP-14) The answer is affirmative if G contains a nonabelian free subgroup(cf. Corollary 8.11.2).

(OP-15) See the discussion in the notes at the end of Chap. 8.(OP-16) See the discussion in the notes at the end of Chap. 8.

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List of Symbols

Symbol Definition Pageε the empty word 367� the dominance relation in the set of growth

functions γ : N → [0, +∞) 162∼ the equivalence relation in the set of growth

functions γ : N → [0, +∞) 162‖T‖p→p the �p-norm of a linear map

T : �p(E) → �p(E) 1940R, 0 the zero element of the ring R 2911G the identity element of the group G 21R, 1 the unity element of the ring R 291γG

S , γS the growth function of the group G relativeto the finite symmetric generating subsetS ⊂ G 160

ΔGS , ΔS the discrete laplacian on the group G

associated with the subset S ⊂ G 9Δ

(2)S the restriction of ΔS to the Hilbert space

�2(G) 201∂E(Ω) the E-boundary of the subset Ω ⊂ G 116ιS(G) the isoperimetric constant of the group G

with respect to the finite symmetricgenerating subset S ⊂ G 191

λGS , λS the growth rate of the group G with respect

to the finite symmetric generating subsetS ⊂ G 169

λ(e) the label of the edge e ∈ E in a labeledgraph G = (Q, E) 153

λ(π) the label of the path π in a labeled graphG = (Q, E) 154, 223

π− the initial vertex of the path π in anS-labeled graph 154


429

http://dx.doi.org/10.1007/978-3-642-14034-1

430 List of Symbols

Symbol Definition Pageπ+ the terminal vertex of the path π in an

S-labeled graph 154σ(T ) the real spectrum of T ∈ L(X) 406ψq,r the S-labeled graph isomorphism from

BS(r) onto B(q, r) such that ψq,r(1G) = q 265Ω−E the E-interior of the subset Ω ⊂ G 115Ω+E the E-closure of the subset Ω ⊂ G 115A∗ the monoid consisting of all words on the

alphabet A 367AG the set of all configurations x : G → A 2B(q, n) the ball of radius n in an S-labeled graph

Q = (Q, E) centered at the vertex q ∈ Q 265BG

S (g, n), BS(g, n) the ball of radius n in G centered at theelement g ∈ G with respect to the wordmetric 153

BGS (n), BS(n) the ball of radius n in G centered at the

identity element 1G ∈ G with respect to theword metric 153

CA(G;A) the monoid consisting of all cellularautomata τ : AG → AG 13

CA(G, H; A) the submonoid of CA(G;A) consisting of allcellular automata τ : AG → AG admitting amemory set S such that S ⊂ H 16

(Ci(G))i≥0 the lower central series of the group G 93CS(G) the Cayley graph of the group G with

respect to the finite symmetric generatingsubset S ⊂ G 156

dF the normalized Hamming distance onSym(F ) 251

dGS , dS the word metric on G with respect to the

finite symmetric generating subset S ⊂ G 152dQ the graph metric in the edge-symmetric

S-labeled graph Q 155D(G) the derived subgroup of the group G 92(Di(G))i≥0 the derived series of the group G 92entF (X) the entropy of the subset X ⊂ AG with

respect to the right Følner net F 125F (X) the free group based on the set X 371Fn the free group of rank n 372Fix(α) the set of fixed points of the permutation α 251Fix(H) the set of configurations x ∈ AG fixed by H 4G = 〈X; R〉 the presentation of the group G given by the

generating subset X and the set ofrelators R 375

List of Symbols 431

Symbol Definition Pagegx the configuration defined by

gx(h) = x(g−1h) 2G = (X,Y,E) the bipartite graph with left (resp. right)

vertex set X (resp. Y ) and set of edges E 391HR the Heisenberg group with coefficients in the

ring R 94ICA(G;A) the group consisting of all invertible cellular

automata τ : AG → AG 24IdX the identity map on the set X 2�(w) the length of the word w ∈ A∗ 35�p(E) the Banach space of all p-summable

functions x : E → R 193�∞(E) the Banach space of all bounded functions

x : E → R 78�GS (g), �S(g) the word-length of the element g ∈ G with

respect to the finite symmetric generatingsubset S ⊂ G 152

LCA(G; V ) the algebra of all linear cellular automataτ : V G → V G 287

LCA(G, H; V ) the subalgebra of LCA(G;V ) consisting ofall linear cellular automata τ : V G → V G

admitting a memory set S such that S ⊂ H 289L(X) the language associated with the subshift X 35Ln(X) the set of admissible words of length n of the

subshift X 35L(X) the space of all continous endomorphisms of

the Banach space X 406M

(p)S the �p-Markov operator associated with the

finite subset S ⊂ G 195Matd(R) the ring consisting of all d × d matrices with

entries in the ring R 305mdimF (X) the mean dimension of the vector subspace

X ⊂ V G with respect to the right Følnernet F 308

M(E) the set of all means on the set E 79PM(E) the set of all finitely additive probability

measures on the set E 79N (Γ ) the space of all normal subgroups of the

group Γ or, equivalently, the space of allΓ -marked groups 61

NL(B) ⊂ X the left-neighborhood of the subset B ⊂ Yin the bipartite graph (X,Y,E) 392

NL(y) ⊂ X the left-neighborhood of the vertex y ∈ Y inthe bipartite graph (X,Y,E) 391

432 List of Symbols

Symbol Definition PageNR(A) ⊂ Y the right-neighborhood of the subset A ⊂ X

in the bipartite graph (X,Y,E) 392NR(x) ⊂ Y the right-neighborhood of the vertex x ∈ X

in the bipartite graph (X,Y,E) 391P(E) the set of all subsets of the set E 77per(B) the period of the matrix B 227per(G) the period of the labeled graph G 227per(X) the period of the irreducible sofic subshift X 227pern(X) the number of nZ-periodic configurations in

the subshift X 227Q(r) the set of all vertices of the S-labeled graph

Q = (Q, E) for which there exists anS-labeled graph isomorphismψq,r : BS(r) → B(q, r) satisfyingψq,r(1G) = q 265

Q = (Q, E) the S-labeled graph with vertex set Q andedge set E ⊂ Q × S × Q 153

R[G] the group ring of the group G withcoefficients in the ring R 292

Rop the opposite ring of the ring R 293Sym(X) the symmetric group of the set X 359Sym0(X) the subgroup of Sym(X) consisting of all

permutations with finite support 360Sym+

0 (X) the alternating group on X 364Symn the symmetric group of degree n 366Sym+

n the alternating group of degree n 366Sym(X,�) the subgroup of Sym(X) that preserve the

partial order � of the set X 179, 333U(R) the multiplicative group consisting of all

invertible elements in the ring R, 292V [G] the vector subspace of V G consisting of all

configurations x : V → G with finite support 288x|Ω the restriction of the configuration x ∈ AG

to the subset Ω ⊂ G 3X∗ the topological dual of the real normed

space X 384X(A) the subshift of finite type defined by the set

of admissible patterns A 32Xf the set of all configurations in the subshift

X whose G-orbit is finite 71XP the subshift defined by the set of forbidden

patterns P 32XG the subshift defined by the labeled graph G 223Z(G) the center of the group G 94

Index

Δ-irreducible subshift, 34

action

continuous —, 3

equivariantly approximable —, 279

expansive —, 65

faithful —, 50

topologically mixing —, 29

topologically transitive —, 32

uniformly continuous —, 64

additive cellular automaton, 335

adjacency matrix of a labeled graph, 227

admissible pattern, 32

admissible word, 35

affine

— group, 93

— map, 387

algebra, 286

— homomorphism, 289

— isomorphism, 290

almost

— -homomorphism, 234, 254

— equal configurations, 112

— perfect group, 55

— periodic configuration, 72

alphabet, 2

alternating group, 364

— of rank n, 366

amenable

— group, 87

elementary — group, 215, 338

Artinian module, 69

automaton

additive cellular —, 335

cellular —, 6

linear cellular —, 284

automorphism group, 45

back-tracking, 156

Baire theorem, 403

Banach-Alaoglu theorem, 385

base

— of a uniform structure, 353

free —, 368

based free group, 367

bi-invariant metric, 251

bi-orderable group, 341

bipartite

— graph, 391

— subgraph, 391

finite — graph, 392

locally finite — graph, 392

Boolean ring, 340

boundary, 116

Burnside problem, 214

Cantor-Bernstein theorem, 398

Cayley graph, 156

cellular automaton, 6

additive —, 335

induced —, 17

invertible —, 24

linear —, 284, 335

reversible —, 24

characteristic map, 79

closed path, 155

closure, 115

cluster point of a net, 345

color, 2

commensurable groups, 171

commutative-transitive group, 279

commutator

— of two group elements, 92

— subgroup, 92

simple —, 176


433

http://dx.doi.org/10.1007/978-3-642-14034-1

434 Index

compact topological space, 347

completion

proamenable —, 108

pronilpotent —, 108

prosolvable —, 108

complexity of a paradoxical decomposition,

106

composition of paths, 154

concatenation, 367

configuration, 2

H-periodic —, 3

almost periodic —, 72

Garden of Eden —, 111

language of a —, 73

Toeplitz —, 74

conjugate elements, 362

connected labeled graph, 155

Connes embedding conjecture, 277

context-free subshift, 225

convergent net, 344

convex subset, 383

convolution product, 291

convolutional encoders, 335

Curtis-Hedlund theorem, 20

cycle, 361

Day’s problem, 105

Dedekind finite ring, 336

degree

— of a graph, 155

— of a symmetric group, 366

— of a vertex, 155

— of an alternating group, 366

derived

— series, 92

— subgroup, 92

directed set, 343

directly finite ring, 327

discrete uniform structure, 352

divisible group, 38

dominance of growth functions, 162

edge

— -symmetric labeled graph, 155

— of a bipartite graph, 391

inverse —, 155

of a labeled graph, 153

elementary

— amenable group, 215, 338

— reduction, 370

empty

— path, 154

— word, 367

entourage, 352

entropy

topological —, 142

equipotent sets, 372

equivalence of growth functions, 162

equivariant map, 5

equivariantly approximable action, 279

even subshift, 35

expansive action, 65

expansivity

constant, 65

entourage, 65

exponential growth, 164

Følner

— conditions, 96

— theorem, 99

left — net, 96

left — sequence, 96

right — net, 96

right — sequence, 96

faithful action, 50

Fibonacci sequence, 219

field, 284

filter, 409

— generated, 409

convergent —, 412

Frechet —, 409

limit of a —, 412

principal —, 409

residual —, 409

ultra—, 410

finite intersection property, 347

finitely

— additive probability measure, 77

— generated group, 152, 375

— presented group, 376

bi-invariant — additive probabilitymeasure, 85

left-invariant — additive probabilitymeasure, 85

right-invariant — additive probabilitymeasure, 85

forbidden

— pattern, 32

— word, 35

Frechet filter, 409

free

— base, 368

— base subset, 368

— group, 368

— group of rank k, 372

— ultrafilter, 410

based — group, 367

Index 435

rank of a — group, 373

fully residually free group, 279

Garden of Eden

— configuration, 111

— pattern, 112

— theorem, 114, 128

— theorem for linear cellular automata,

312

generating subset, 151

generator of a presentation, 375

golden mean subshift, 35

graph

— metric, 155

bipartite —, 391

Cayley —, 156

degree of a regular labeled —, 155

finite labeled —, 154

labeled —, 153

loop in a labeled —, 154

regular labeled —, 155

tree, 156

Grigorchuk group, 179

abelianization of the —, 222

group

— algebra, 294

— of p-adic integers, 41

— of intermediate growth, 190

— ring, 292

affine —, 93

almost perfect —, 55

alternating —, 364

alternating — of rank n, 366

amenable —, 87

automorphism —, 45

bi-orderable —, 341

Cayley graph of a —, 156

commutative-transitive —, 279

divisible —, 38

elementary amenable —, 215, 338

finitely generated —, 152, 375

finitely presented —, 376

free —, 368

free — of rank k, 372

fully residually free —, 279

Grigorchuk —, 179

Heisenberg —, 94

Hopfian —, 44

hyperlinear —, 277

Kaloujnine —, 221

Klein bottle —, 341

L-surjunctive —, 324

lamplighter —, 108

LEA —, 247

LEF —, 247

linear —, 51

locally P —, 58

locally indicable —, 337

marked —, 61

metabelian —, 92

nilpotent —, 93

orderable —, 331

periodic —, 29, 105

polycyclic —, 106, 108

presentation of a —, 375

profinite —, 41

residually C —, 238

residually P —, 62, 131

residually amenable —, 132

residually finite —, 37

simple —, 44

sofic —, 254

solvable —, 93

surjunctive —, 57

symmetric —, 359

symmetric — of rank n, 366

unique-product —, 331

virtually P —, 41

growth

— function, 160, 162

— rate, 169

— type of a group, 163

equivalence class of — functions, 163

equivalence of — functions, 162

exponential —, 164

intermediate —, 190

polynomial —, 164

subexponential —, 164

Hall

— k-harem conditions, 399

— condition, 394

— harem theorem, 399

— marriage theorem, 399

Hamming metric, 252

Hausdorff metric, 357

Hausdorff-Bourbaki

— topology, 356

uniform structure, 356

Heisenberg group, 94

homomorphism

almost- —, 234, 254

labeled graph —, 154

Hopfian

— group, 44

— module, 339

hyperlinear group, 277

436 Index

ICC-property, 338

idempotent, 331

proper —, 331

induced

— cellular automaton, 17

— labeled subgraph, 154

inductive

— limit, 379

— system of groups, 379

initial

— topology, 346

— uniform structure, 355

interior, 115

intermediate growth, 190

inverse

— edge, 155

— path, 155

invertible cellular automaton, 24

irreducible

— matrix, 227

— subshift, 32

isomorphism

labeled graph —, 154

isoperimetric constant, 191

Kaloujnine group, 221

abelianization of the —, 222

Kesten-Day theorem, 201

Klein bottle group, 341

Klein Ping-Pong theorem, 376

L-surjunctive group, 324

labeled graph, 153

— homomorphism, 154

— isomorphism, 154

adjacency matrix of a —, 227

connected —, 155

edge-symmetric —, 155

finite —, 154

locally finite —, 155

path in a —, 154

subgraph, 154

subshift defined by a —, 223

labelling map, 153

lamplighter group, 108

language

— of a configuration, 73

— of a subshift over Z, 35

Laplacian, 10

lattice, 95

LEA-group, 247

LEF-group, 247

left-invariant

— finitely additive probability measure,85

— metric, 251

length

— of a cycle, 361

— of a word, 35, 152

letter, 2

limit

— along an ultrafilter, 413

— of a filter, 412

— point of a net, 344

inductive —, 379

projective —, 380

linear

— cellular automaton, 284, 335

— group, 51

Lipschitz-equivalence, 162

local defining map, 6

locally

— P group, 58

— convex topological vector space, 383

— embeddable, 235

— finite bipartite graph, 392

— finite labeled graph, 155

— indicable group, 337

loop, 154

lower central series, 93

majority action, 10

marked group, 61

Markov operator, 195

Markov-Kakutani theorem, 387

matching, 393

left-perfect —, 393

perfect —, 393

right-perfect—, 393

mean, 78

— dimension, 308

bi-invariant —, 86

left-invariant —, 86

right-invariant —, 86

memory set, 6

minimal —, 15

metabelian group, 92

metric

bi-invariant —, 251

graph —, 155

Hamming —, 252

word —, 153

metrizable uniform structure, 352

Milnor problem, 215

minimal

— memory set, 15

— set, 72

Index 437

— subshift, 72

module

Artinian —, 69

Hopfian —, 339

Noetherian —, 339

projective —, 339

monoid, 13

Moore neighborhood, 166

Morse subshift, 74

entropy of the —, 144

neighbor, 155

net, 343

nilpotency degree, 93

nilpotent group, 93

Noetherian

— module, 339

— ring, 340

non-principal ultrafilter, 410

normal closure, 375

Open mapping theorem, 404

operator norm, 384

opposite ring, 293

orderable group, 331

Ore ring, 340

paradoxical decomposition

left —, 98

right —, 98

partially ordered set, 343

path

— in a labeled graph, 154

closed —, 155

closed simple —, 156

composition, 154

empty —, 154

inverse —, 155

label of a —, 154

proper —, 156

simple —, 156

pattern, 2

admissible —, 32

forbidden —, 32

periodic group, 29, 105

permutation, 359

support of a —, 360

polycyclic group, 106, 108

polynomial, 164

pre-injective map, 112

presentation

— of a group, 375

generator of a —, 375

relator of a —, 375

principal— filter, 409

— ultrafilter, 410proamenable completion, 108

prodiscrete— topology, 3, 346

— uniform structure, 22, 355product

— topology, 346— uniform structure, 355

profinite— completion, 55

— group, 41— kernel, 39

— topology, 53projective

— limit, 380— module, 339

— system of groups, 380pronilpotent completion, 108

proper— idempotent, 331

— path, 156prosolvable completion, 108

quasi-isometric— embedding, 204

— groups, 206quasi-isometry, 204

rank of a free group, 373reduced

— form, 374— product, 244

— word, 373regular labeled graph, 155

relator of a presentation, 375residual

— filter, 409— set, 409

— subgroup, 39residually

— C group, 238

— P group, 62, 131— amenable group, 132

— finite group, 37restriction, 17

reversible cellular automaton, 24right-invariant

— finitely additive probability measure,85

— metric, 251ring

Boolean —, 340Dedekind finite —, 336

438 Index

directly finite —, 327

group —, 292

Noetherian —, 340

opposite —, 293

Ore —, 340

stably finite —, 328

unit-regular—, 340

von Neumann finite —, 336

set

directed —, 343

partially ordered —, 343

shift, 2

simple

— group, 44

— path, 156

closed — path, 156

sofic

— subshift, 225

— group, 254

solvable group, 93

spectrum

real —, 406

stably finite ring, 328

state, 2

strong topology, 82, 384

strongly irreducible subshift, 34

subalgebra, 287

subexponential growth, 164

subgraph

induced labeled —, 154

labeled —, 154

submonoid, 17

subnet, 344

subsemigroup, 322

subshift, 31

N-power —, 228

Nth higher block —, 227

Δ-irreducible —, 34

— defined by a labeled graph, 223

— of finite type, 32

context-free —, 225

even —, 35

golden mean —, 35

irreducible —, 32

language of a — over Z, 35

minimal —, 72

Morse —, 74, 144

sofic —, 225

strongly irreducible —, 34

surjunctive —, 71

Toeplitz —, 74

topologically mixing —, 33

subword, 35

support

— of a configuration, 288

— of a pattern, 2

— of a permutation, 360

surjunctive

— group, 57

—subshift, 71

symbol, 2

symmetric

— group, 359

— subset, 152

syndetic subset, 72

Tarski

— alternative, 99

— number of a group, 106

Tarski-Følner theorem, 99

theorem

Baire —, 403

Banach-Alaoglu —, 385

Cantor-Bernstein —, 398

Curtis-Hedlund —, 20

Garden of Eden —, 114, 128

Garden of Eden — for linear cellular

automata, 312

Gromov-Weiss —, 272

Hall harem —, 399

Hall marriage —, 399

Kesten-Day —, 201

Klein Ping-Pong —, 376

Markov-Kakutani —, 387

open mapping —, 404

Tarski-Følner —, 99

Tychonoff —, 348

Thue-Morse sequence, 73

tiling, 122

Toeplitz

— configuration, 74

— subshift, 74

topological

— dual, 384

— entropy, 142

— manifold, 69

— vector space, 383

topologically mixing

— action, 29

— subshift, 33

topologically transitive action, 32

topology

Hausdorff-Bourbaki —, 356

initial —, 346

prodiscrete —, 3, 346

product —, 346

profinite —, 53

Index 439

strong —, 384weak-∗ —, 384

total ordering, 331totally disconnected topological space, 346

transposition, 361tree, 156

trivial uniform structure, 352Tychonoff theorem, 348

ultrafilter, 410free —, 410

limit along an — , 413non-principal —, 410

principal —, 410ultraproduct, 244

uniform— convexity, 407

— embedding, 355— isomorphism, 355

— structure, 351Hausdorff-Bourbaki — structure, 356

induced — structure, 353prodiscrete — structure, 22, 355

uniformly continuous— action, 64

— map, 353unique

— -product group, 331— rank property, 340

unit-regular ring, 340universe, 2

valence of a vertex, 155

vertex

— of a bipartite graph, 391

— of a labeled graph, 153

degree of a —, 155

neighbor, 155

valence of a —, 155

virtually P group, 41

von Neumann

— conjecture, 105

— finite ring, 336

— neighborhood, 165

weak-∗ topology, 384

word, 367

— length, 152

— metric, 153

admissible —, 35

empty —, 367

forbidden —, 35

length of a —, 35

reduced —, 373

subword of a —, 35

wreath product, 52

zero-divisor, 330

— conjecture, 337

left —, 330

right —, 330

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Springer onographs in Mathematics - Université Lorrainemath.univ-metz.fr/~gnc/bibliographie/Ceccherini... · 2013-03-22 · models for self-reproducing machines. About twenty years