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Representations and Characters of Groups
Now in its second edition, this text provides a modern introduction to
the representation theory of ®nite groups. The authors have revised the
popular ®rst edition and added a considerable amount of new material.
The theory is developed in terms of modules, since this is appropriate
for more advanced work, but considerable emphasis is placed upon
constructing characters. The character tables of many groups are given,
including all groups of order less than 32, and all simple groups of
order less than 1000.
Among the applications covered are Burnside's paqb theorem, the
use of character theory in studying subgroup structure and permutation
groups, and a description of how to use representation theory to
investigate molecular vibration.
Each chapter is accompanied by a variety of exercises, and full
solutions to all the exercises are provided at the end of the book. This
will be ideal as a text for a course in representation theory, and in
view of the applications of the subject, will be of interest to mathema-
ticians, chemists and physicists alike.
REPRESENTATIONS AND
CHARACTERS OF GROUPS
GORDON JAMES and MARTIN LIEBECKDepartment of Mathematics,
Imperial College, London
Second Edition
PUBLISHED BY CAMBRIDGE UNIVERSITY PRESS (VIRTUAL PUBLISHING) FOR AND ON BEHALF OF THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street, Cambridge CB2 IRP 40 West 20th Street, New York, NY 10011-4211, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia http://www.cambridge.org © Cambridge University Press 1993, 2001 This edition © Cambridge University Press (Virtual Publishing) 2003 First published in printed format 1993 Second edition 2001 A catalogue record for the original printed book is available from the British Library and from the Library of Congress Original ISBN 0 521 81205 4 hardback Original ISBN 0 521 00392 X paperback ISBN 0 511 01700 6 virtual (netLibrary Edition)
Contents
Preface page vii
1 Groups and homomorphisms 1
2 Vector spaces and linear transformations 14
3 Group representations 30
4 FG-modules 38
5 FG-submodules and reducibility 49
6 Group algebras 53
7 FG-homomorphisms 61
8 Maschke's Theorem 70
9 Schur's Lemma 78
10 Irreducible modules and the group algebra 89
11 More on the group algebra 95
12 Conjugacy classes 104
13 Characters 117
14 Inner products of characters 133
15 The number of irreducible characters 152
16 Character tables and orthogonality relations 159
17 Normal subgroups and lifted characters 168
18 Some elementary character tables 179
19 Tensor products 188
20 Restriction to a subgroup 210
21 Induced modules and characters 224
22 Algebraic integers 244
23 Real representations 263
24 Summary of properties of character tables 283
25 Characters of groups of order pq 288
26 Characters of some p-groups 298
27 Character table of the simple group of order 168 311
v
28 Character table of GL(2, q) 322
29 Permutations and characters 337
30 Applications to group theory 348
31 Burnside's Theorem 361
32 An application of representation theory to molecular vibration 367
Solutions to exercises 397
Bibliography 454
Index 455
vi Representations and characters of groups
Preface
We have attempted in this book to provide a leisurely introduction to
the representation theory of groups. But why should this subject
interest you?
Representation theory is concerned with the ways of writing a group
as a group of matrices. Not only is the theory beautiful in its own right,
but it also provides one of the keys to a proper understanding of ®nite
groups. For example, it is often vital to have a concrete description of a
particular group; this is achieved by ®nding a representation of the
group as a group of matrices. Moreover, by studying the different
representations of the group, it is possible to prove results which lie
outside the framework of representation theory. One simple example: all
groups of order p2 (where p is a prime number) are abelian; this can be
shown quickly using only group theory, but it is also a consequence of
basic results about representations. More generally, all groups of order
paqb ( p and q primes) are soluble; this again is a statement purely about
groups, but the best proof, due to Burnside, is an outstanding example
of the use of representation theory. In fact, the range of applications of
the theory extends far beyond the boundaries of pure mathematics, and
includes theoretical physics and chemistry ± we describe one such
application in the last chapter.
The book is suitable for students who have taken ®rst undergraduate
courses involving group theory and linear algebra. We have included two
preliminary chapters which cover the necessary background material.
The basic theory of representations is developed in Chapters 3±23, and
our methods concentrate upon the use of modules; although this accords
with the more modern style of algebra, in several instances our proofs
differ from those found in other textbooks. The main results are elegant
and surprising, but at ®rst sight they sometimes have an air of mystery
vii
about them; we have chosen the approach which we believe to be the
most transparent.
We also emphasize the practical aspects of the subject, and the text is
illustrated with a wealth of examples. A feature of the book is the wide
variety of groups which we investigate in detail. By the end of Chapter
28, we have presented the character tables of all groups of order less
than 32, of all p-groups of order at most p4, and of all the simple
groups of order less than 1000.
Every chapter is accompanied by a set of Exercises, and the solutions
to all of these are provided at the end of the book.
We would like to thank Dr Hans Liebeck for his careful reading of
our manuscript and the many helpful suggestions which he made.
Preface to Second EditionIn this second edition, we have included two new chapters; one
(Chapter 28) deals with the character tables of an in®nite series of groups,
and the other (Chapter 29) covers aspects of the representation theory of
permutation groups. We have also added a considerable amount of new
material to Chapters 20, 23 and 30, and made minor amendments else-
where.
viii Representations and characters of groups
1
1
Groups and homomorphisms
This book is devoted to the study of an aspect of group theory, so we
begin with a reÂsume of facts about groups, most of which you should
know already. In addition, we introduce several examples, such as
dihedral groups and symmetric groups, which we shall use extensively
to illustrate the later theory. An elementary course on abstract algebra
would normally cover all the material in the chapter, and any book on
basic group theory will supply you with further details. One or two
results which we shall use only infrequently are demoted to the
exercises at the end of the chapter ± you can refer to the solutions if
necessary.
Groups
A group consists of a set G, together with a rule for combining any
two elements g, h of G to form another element of G, written gh; this
rule must satisfy the following axioms:
(1) for all g, h, k in G,
(gh)k � g(hk);
(2) there exists an element e in G such that for all g in G,
eg � ge � g;
(3) for all g in G, there exists an element gÿ1 in G such that
ggÿ1 � gÿ1 g � e:
We refer to the rule for combining elements of G as the product
operation on G.
Axiom (1) states that the product operation is associative; the
element e in axiom (2) is an identity element of G; and gÿ1 is an
inverse of g in axiom (3).
It is elementary to see that G has just one identity element, and that
every g in G has just one inverse. Usually we write 1, rather than e,
for the identity element of G.
The product of an element g with itself, gg, is written g2; similarly
g3 � g2 g, gÿ2 � (gÿ1)2, and so on. Also, g0 � 1.
If the number of elements in G is ®nite, then we call G a ®nite
group; the number of elements in G is called the order of G, and is
written |G|.
1.1 Examples
(1) Let n be a positive integer, and denote by C the set of all complex
numbers. The set of nth roots of unity in C, with the usual multi-
plication of complex numbers, is a group of order n. It is written as
Cn and is called the cyclic group of order n. If a � e2ði=n, then
Cn � f1, a, a2, : : : , anÿ1g,and an � 1.
(2) The set Z of all integers, under addition, is a group.
(3) Let n be an integer with n > 3, and consider the rotation and
re¯ection symmetries of a regular n-sided polygon.
There are n rotation symmetries: these are r0, r1, . . . , rnÿ1 where rk
is the (clockwise) rotation about the centre O through an angle 2ðk=n.
There are also n re¯ection symmetries: these are re¯ections in the n
lines passing through O and a corner or the mid-point of a side of the
polygon.
These 2n rotations and re¯ections form a group under the product
operation of composition (that is, for two symmetries f and g, the
product fg means `®rst do f, then do g'). This group is called the
dihedral group of order 2n, and is written D2n.
Let A be a corner of the polygon. Write b for the re¯ection in the
2 Representations and characters of groups
line through O and A, and write a for the rotation r1. Then the n
rotations are
1, a, a2, : : : , anÿ1
(where 1 denotes the identity, which leaves the polygon ®xed); and the
n re¯ections are
b, ab, a2b, : : : , anÿ1b:
Thus all elements of D2n are products of powers of a and b ± that is,
D2n is generated by a and b.
Check that
an � 1, b2 � 1 and bÿ1ab � aÿ1:
These relations determine the product of any two elements of the
group. For example, we have ba j � aÿ jb (using the relation ba �aÿ1b), and hence
(aib)(a jb) � aiba jb � aiaÿ jbb � aiÿ j:
We summarize all this in the presentation
D2n � ha, b: an � 1, b2 � 1, bÿ1ab � aÿ1i:(4) For n a positive integer, the set of all permutations of
{1, 2, . . . , n}, under the product operation of composition, is a group.
It is called the symmetric group of degree n, and is written Sn. The
order of Sn is n!.
(5) Let F be either R (the set of real numbers) or C (the set of
complex numbers). The set of all invertible n 3 n matrices with entries
in F, under matrix multiplication, forms a group. This group is called
the general linear group of degree n over F, and is denoted by
GL(n, F). It is an in®nite group. The identity of GL(n, F) is of course
the identity matrix, which we denote by In or just I.
A group G is said to be abelian if gh � hg for all g and h in G.
While Cn and Z are abelian, most of the other examples given above
are non-abelian groups.
Subgroups
Let G be a group. A subset H of G is said to be a subgroup if H is
itself a group under the product operation inherited from G. We use
the notation H < G to indicate that H is a subgroup of G.
Groups and homomorphisms 3
It is easy to see that a subset H of a group G is a subgroup if and
only if the following two conditions hold:
(1) 1 2 H, and
(2) if h, k 2 H then hkÿ1 2 H.
1.2 Examples
(1) For every group G, both {1} and G are subgroups of G.
(2) Let G be a group and g 2 G. The subset
hgi � fgn: n 2 Zgis a subgroup of G, called the cyclic subgroup generated by g. If
gn � 1 for some n > 1, then kgl is ®nite. In this case, let r be the
least positive integer such that g r � 1; then r is equal to the number
of elements in kgl ± indeed,
hgi � f1, g, g 2, : : : , grÿ1g:We call r the order of the element g.
If G � kgl for some g 2 G then we call G a cyclic group. The
groups Cn and Z in Examples 1.1 are cyclic.
(3) Let G be a group and let a, b 2 G. De®ne H to be the subset of G
consisting of all elements which are products of powers of a and b ±
that is, all elements of the form
ai1 b j1 ai2 b j2 : : :ain b jn
for some n, where ik , jk 2 Z for 1 < k < n. Then H is a subgroup of
G; we call H the subgroup generated by a and b, and write
H � ha, bi:Given any ®nite set S of elements of G, we can similarly de®ne hSi,the subgroup of G generated by S.
This construction gives a powerful method of ®nding new groups as
subgroups of given groups, such as general linear or symmetric groups.
We illustrate the construction in the next example, and again in
Example 1.5 below.
(4) Let G � GL(2, C), the group of invertible 2 3 2 matrices with
entries in C, and let
A � i 0
0 ÿi
� �, B � 0 1
ÿ1 0
� �:
4 Representations and characters of groups
Put H � kA, Bl, the subgroup of G generated by A and B. Check that
A4 � I , A2 � B2, Bÿ1 AB � Aÿ1:
Using the third relation, we see that every element of H has the form
Ai Bj for some integers i, j; and using the ®rst two relations, we can
take 0 < i < 3 and 0 < j < 1. Hence H has at most eight elements.
Since the matrices
Ai Bj (0 < i < 3, 0 < j < 1)
are all distinct, in fact jH j � 8.
The group H is called the quaternion group of order 8, and is
written Q8. The above three relations determine the product of any two
elements of Q8, so we have the presentation
Q8 � hA, B: A4 � I , A2 � B2, Bÿ1 AB � Aÿ1i:(5) A transposition in the symmetric group Sn is a permutation which
interchanges two of the numbers 1, 2, . . . , n and ®xes the other n ÿ 2
numbers. Every permutation g in Sn can be expressed as a product of
transpositions. It can be shown that either all such expressions for g
have an even number of transpositions, or they all have an odd number
of transpositions; we call g an even or an odd permutation, accord-
ingly. The subset
An � fg 2 Sn: g is an even permutationgis a subgroup of Sn, called the alternating group of degree n.
Direct products
We describe a construction which produces a new group from given
ones.
Let G and H be groups, and consider
G 3 H � f(g, h): g 2 G and h 2 Hg:De®ne a product operation on G 3 H by
(g, h)(g9, h9) � (gg9, hh9)
for all g, g9 2 G and all h, h9 2 H. With this product operation, G 3 H
is a group, called the direct product of G and H.
Groups and homomorphisms 5
More generally, if G1, . . . , Gr are groups, then the direct product
G1 3 : : : 3 Gr is
f(g1, : : : , gr): gi 2 Gi for 1 < i < rg,with product operation de®ned by
(g1, : : : , gr)(g91, : : : , g9r) � (g1 g91, : : : , grg9r):
If all the groups Gi are ®nite, then G1 3 . . . 3 Gr is also ®nite, of
order |G1| . . . |Gr |.
1.3 Example
The group C2 3 . . . 3 C2 (r factors) has order 2r and all its non-
identity elements have order 2.
Functions
A function from one set G to another set H is a rule which assigns a
unique element of H to each element of G. In this book, we generally
apply functions on the right ± that is, the image of g under a function
W is written as gW, not as Wg. We often indicate that W is a function
from G to H by the notation W: G! H. By an expression W: g! h,
where g 2 G and h 2 H, we mean that h � gW.
A function W: G! H is invertible if there is a function ö: H! G
such that for all g 2 G, h 2 H,
(gW)ö � g and (hö)W � h:
Then ö is called the inverse of W, and is written as Wÿ1. A function Wfrom G to H is invertible if and only if it is both injective (that is,
g1W � g2W for g1, g2 2 G implies that g1 � g2) and surjective (that is,
for every h 2 H there exists g 2 G such that gW � h). An invertible
function is also called a bijection.
Homomorphisms
Given groups G and H, those functions from G to H which `preserve
the group structure' ± the so-called homomorphisms ± are of particular
importance.
If G and H are groups, then a homomorphism from G to H is a
function W: G! H which satis®es
(g1 g2)W � (g1W)(g2W) for all g1, g2 2 G:
6 Representations and characters of groups
An invertible homomorphism is called an isomorphism. If there is an
isomorphism W from G to H, then G and H are said to be isomorphic,
and we write G � H; also, Wÿ1 is an isomorphism from H to G, so
H � G.
The following example displays a technique which can often be used
to prove that certain functions are homomorphisms.
1.4 Example
Let G � D2n � ka, b: an � b2 � 1, bÿ1ab � aÿ1l, and write the 2n
elements of G in the form aib j with 0 < i < n ÿ 1, 0 < j < 1. Let H
be any group, and suppose that H contains elements x and y which
satisfy
xn � y2 � 1, yÿ1xy � xÿ1:
We shall prove that the function W: G! H de®ned by
W: aib j ! xi y j (0 < i < nÿ 1, 0 < j < 1)
is a homomorphism.
Suppose that 0 < r < n ÿ 1, 0 < s < 1, 0 < t < n ÿ 1, 0 < u < 1.
Then
arbsatbu � aib j
for some i, j with 0 < i < n ÿ 1, 0 < j < 1. Moreover, i and j are
determined by repeatedly using the relations
an � b2 � 1, bÿ1ab � aÿ1:
Since we have xn � y2 � 1, yÿ1xy � xÿ1, we can also deduce that
xrysxtyu � xi y j:
Therefore,
(arbsatbu)W � (aib j)W � xi y j � xrysxtyu
� (arbs)W . (atbu)W,
and so W is a homomorphism.
We now demonstrate the technique of Example 1.4 in action.
1.5 Example
Let G � S5 and let x, y be the following permutations in G:
x � (1 2 3 4 5), y � (2 5)(3 4):
Groups and homomorphisms 7
(Here we adopt the usual cycle notation ± thus, (1 2 3 4 5) denotes the
permutation 1! 2! 3! 4! 5! 1, and so on.) Check that
x5 � y2 � 1, yÿ1xy � xÿ1:
Let H be the subgroup kx, yl of G. Using the above relations, we see
that
H � fxi y j: 0 < i < 4, 0 < j < 1g,a group of order 10.
Now recall that
D10 � ha, b: a5 � b2 � 1, bÿ1ab � aÿ1i:By Example 1.4, the function W: D10 ! H de®ned by
W: aib j ! xi y j (0 < i < 4, 0 < j < 1)
is a homomorphism. Since W is invertible, it is an isomorphism. Thus,
H � kx, yl � D10.
Cosets
Let G be a group and let H be a subgroup of G. For x in G, the
subset
Hx � fhx: h 2 Hgof G is called a right coset of H in G. The distinct right cosets of H
in G form a partition of G (that is, every element of G is in precisely
one of the cosets).
Suppose now that G is ®nite, and let Hx1, . . . , Hxr be all the
distinct right cosets of H in G. For all i, the function
h! hxi (h 2 H)
is a bijection from H to Hxi, and so jHxij � jH j. Since
G � Hx1 [ : : : [ Hxr, and
Hxi \ Hxj is empty if i 6� j,
we deduce that
jGj � rjH j:In particular, we have
8 Representations and characters of groups
1.6 Lagrange's Theorem
If G is a ®nite group and H is a subgroup of G, then jH j divides |G|.
The number r of distinct right cosets of H in G is called the index
of H in G, and is written as jG: H j. Thus
jG: H j � jGj=jH jwhen G is ®nite.
Normal subgroups
A subgroup N of a group G is said to be a normal subgroup of G
if gÿ1 Ng � N for all g 2 G (where gÿ1 Ng � fgÿ1 ng: n 2 Ng); we
write N v G to indicate that N is a normal subgroup of G.
Suppose that N v G and let G=N be the set of right cosets of N in
G. The importance of the condition gÿ1 Ng � N (for all g 2 G) is that
it can be used to show that for all g, h 2 G, we have
fxy: x 2 Ng and y 2 Nhg � Ngh:
Hence we can de®ne a product operation on G=N by
(Ng)(Nh) � Ngh for all g, h 2 G:
This makes G=N into a group, called the factor group of G by N.
1.7 Examples
(1) For every group G, the sub-groups {1} and G are normal sub-
groups of G.
(2) For n > 1, we have An v Sn. If n > 2 then there are just two right
cosets of An in Sn, namely
An � fg 2 Sn: g eveng, and
An(1 2) � fg 2 Sn: g oddg:Thus |Sn:An| � 2, and so Sn=An � C2.
(3) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l and let N �ka2l � {1, a2}. Then N v G and
G=N � fN , Na, Nb, Nabg:Since (Na)2 � (Nb)2 � (Nab)2 � N, we see that G=N � C2 3 C2.
The subgroup kal is also normal in G, but the subgroup H � kbl is
not normal in G, since b 2 H while aÿ1ba � a2b =2 H.
Groups and homomorphisms 9
Simple groups
A group G is said to be simple if G 6� {1} and the only normal
subgroups of G are {1} and G. For example, the cyclic group Cp, with
p a prime number, is simple. We shall give examples of non-abelian
simple groups in later chapters ± the smallest one is A5.
If G is a ®nite group which is not simple, then G has a normal
subgroup N such that both N and G/N have smaller order than G; and
in a sense, G is `built' out of these two smaller groups. Continuing
this process with the smaller groups, we eventually see that G is `built'
out of a collection of simple groups. (This is analogous to the fact that
every positive integer is built out of its prime factors.) Thus, simple
groups are fundamental to the study of ®nite groups.
Kernels and images
To conclude the chapter, we relate normal subgroups and factor groups
to homomorphisms. Let G and H be groups and suppose that
W: G! H is a homomorphism. We de®ne the kernel of W by
Ker W � fg 2 G: gW � 1g:(1:8)
Then Ker W is a normal subgroup of G. Also, the image of W is
Im W � fgW: g 2 Gg,(1:9)
and Im W is a subgroup of H.
The following result describes the way in which the kernel and
image of W are related.
1.10 Theorem
Suppose that G and H are groups and let W: G! H be a homomorph-
ism. Then
G=Ker W � Im W:
An isomorphism is given by the function
Kg ! gW (g 2 G)
where K � Ker W.
10 Representations and characters of groups
1.11 Example
The function W: Sn ! C2 given by
W: g ! 1, if g is an even permutation,
ÿ1, if g is an odd permutation,
�is a homomorphism. We have Ker W � An, and for n > 2, Im W � C2.
We know from Example 1.7(2) that Sn/An � C2, illustrating Theorem
1.10.
Summary of Chapter 1
1. Examples of groups are
Cn � ka: an � 1l,D2n � ka, b: an � b2 � 1, bÿ1ab � aÿ1l,Q8 � ka, b: a4 � 1, a2 � b2, bÿ1ab � aÿ1l,Sn � the symmetric group of degree n,
An � the alternating group of degree n,
GL(n, C) � the group of invertible n 3 n matrices over C,
G1 3 . . . 3 Gr, the direct product of the groups G1, . . . , Gr.
2. A normal subgroup N of G is a subgroup such that gÿ1 Ng � N for
all g in G. The factor group G=N consists of the right cosets Ng
(g 2 G), with multiplication
(Ng)(Nh) � Ngh:
3. A homomorphism W: G! H is a function such that
(g1 g2)W � (g1W)(g2W)
for all g1, g2 in G. The kernel, Ker W, is a normal subgroup of G,
and the image, Im W, is a subgroup of H. The factor group G=Ker Wis isomorphic to Im W.
Exercises for Chapter 1
1. Show that if G is an abelian group which is simple, then G is
cyclic of prime order.
2. Suppose that G and H are groups, with G simple, and that
W: G! H is a surjective homomorphism. Show that either W is an
isomorphism or H � {1}.
Groups and homomorphisms 11
3. Suppose that G is a subgroup of Sn, and that G is not contained
in An. Prove that G \ An is a normal subgroup of G, and
G= (G \ An) � C2:
4. Let
G � D8 � ha, b: a4 � b2 � 1, bÿ1ab � aÿ1i, and
H � Q8 � hc, d: c4 � 1, c2 � d2, dÿ1cd � cÿ1i:
(a) Let x, y be the permutations in S4 which are given by
x � (1 2), y � (3 4),
and let K be the subgroup kx, yl of S4. Show that both the
functions ö: G! K and ø: H! K, de®ned by
ö: arbs ! xrys,
ø: crds ! xrys (0 < r < 3, 0 < s < 1),
are homomorphisms. Find Kerö and Kerø.
(b) Let X, Y be the 2 3 2 matrices which are given by
X � 0 i
i 0
� �, Y � 0 ÿ1
1 0
� �,
and let L be the subgroup hX , Y i of GL(2, C). Show that just
one of the functions ë: G! L and ì: H! L, de®ned by
ë: arbs ! X rY s,
ì: crds ! X rY s (0 < r < 3, 0 < s < 1),
is a homomorphism. Prove that this homomorphism is an
isomorphism.
5. Prove that D4m � D2m 3 C2 if m is odd.
6. (a) Show that every subgroup of a cyclic group is cyclic.
(b) Let G be a ®nite cyclic group, and let n be a positive integer
which divides |G|. Prove that
fg 2 G: gn � 1g
is a cylic subgroup of G of order n.
(c) If G is a ®nite cyclic group and x, y are elements of G with
the same order, show that x is a power of y.
12 Representations and characters of groups
7. Show that the set of non-zero complex numbers, under the usual
multiplication, is a group. Prove that every ®nite subgroup of this
group is cyclic.
8. Show that every group of even order contains an element of
order 2.
9. Find elements A and B of GL(2, C) such that A has order 8, B
has order 4, and
B2 � A4 and Bÿ1 AB � Aÿ1:
Show that the group kA, Bl has order 16.
(Hint: compare Q8 in Example 1.2(4).)
10. Suppose that H is a subgroup of G with |G:H| � 2. Prove that
H v G.
Groups and homomorphisms 13
14
2
Vector spaces and linear transformations
An attractive feature of representation theory is that it combines two
strands of mainstream mathematics, namely group theory and linear
algebra. For reference purposes, we gather the results from linear
algebra concerning vector spaces, linear transformations and matrices
which we shall use later. Most of the material will be familiar to you
if you have taken a ®rst course on linear algebra, so we omit the
proofs. An exception occurs in the last section, where we deal with
projections; here, we explain in detail how the results work, in case
you have not come across projections before.
Vector spaces
Let F be either R (the set of real numbers) or C (the set of complex
numbers). A vector space over F is a set V, together with a rule for
adding any two elements u, v of V to form an element u � v of V, and
a rule for multiplying any element v of V by any element ë of F to
form an element ëv of V. (The latter rule is called scalar multi-
plication.) Moreover, these rules must satisfy:
(2.1) (a) V is an abelian group under addition;
(b) for all u, v in V and all ë, ì in F,
(1) ë(u � v) � ëu � ëv,
(2) (ë � ì)v � ëv � ìv,
(3) (ëì)v � ë(ìv),
(4) 1v � v.
The elements of V are called vectors, and those of F are called scalars.
We write 0 for the identity element of the abelian group V under
addition.
2.2 Examples
(1) Let R2 denote the set of all ordered pairs (x, y) where x and y are
real numbers. De®ne addition and scalar multiplication on R2 by
(x, y)� (x9, y9) � (x� x9, y� y9),
ë(x, y) � (ëx, ëy):
Then R2 is a vector space over R.
(2) More generally, for each positive integer n, we consider row
vectors
(x1, x2, : : : , xn)
where x1, x2, . . . , xn belong to F. We denote the set of all such row
vectors by F n, and de®ne addition and scalar multiplication on F n by
(x1, : : : , xn)� (x91, : : : , x9n) � (x1 � x91, : : : , xn � x9n),
ë(x1, : : : , xn) � (ëx1, : : : , ëxn):
Then F n is a vector space over F.
Bases of vector spaces
Let v1, . . . , vn be vectors in a vector space V over F. A vector v in V
is a linear combination of v1, . . . , vn if
v � ë1v1 � : : :� ënvn
for some ë1, . . . , ën in F. The vectors v1, . . . , vn are said to span V if
every vector in V is a linear combination of v1, . . . , vn.
We say that v1, . . . , vn are linearly dependent if
ë1v1 � : : :� ënvn � 0
for some ë1, . . . , ën in F, not all of which are zero; otherwise,
v1, . . . , vn are linearly independent.
The vectors v1, . . . , vn form a basis of V if they span V and are
linearly independent.
Throughout this book, we shall consider only vector spaces V which
are ®nite-dimensional ± this means that V has a basis consisting of
®nitely many vectors, as above. It turns out that any two bases of V
have the same number of vectors. The number of vectors in a basis of
V is called the dimension of V and is written as dim V. If V � {0} then
dim V � 0. The vector space V is n-dimensional if dim V � n.
Vector spaces and linear transformations 15
2.3 Example
Let V � F n. Then
(1, 0, 0, : : : , 0), (0, 1, 0, : : : , 0), : : : , (0, 0, 0, : : : , 1)
is a basis of V, so dim V � n. Another basis is
(1, 0, 0, : : : , 0), (1, 1, 0, : : : , 0), : : : , (1, 1, 1, : : : , 1):
Given a basis v1, . . . , vn of a vector space V, each vector v in V can
be written in a unique way as
v � ë1v1 � : : :� ënvn,
with ë1, . . . , ën in F. The vector v therefore determines the scalars
ë1, . . . , ën. Except in the case where V � {0}, there are many bases of
V. Indeed, the next result says that any linearly independent vectors can
be extended to a basis.
(2.4) If v1, . . . , vk are linearly independent vectors in V,
then there exist vk�1, . . . , vn in V such that
v1, . . . , vn form a basis of V.
Subspaces
A subspace of a vector space V over F is a subset of V which is itself
a vector space under the addition and scalar multiplication inherited
from V. For a subset U of V to be a subspace, it is necessary and
suf®cient that all the following conditions hold:
(2.5) (1) 0 2 U;
(2) if u, v 2 U then u � v 2 U;
(3) if ë 2 F and u 2 U then ëu 2 U.
2.6 Examples
(1) {0} and V are subspaces of V.
(2) Let u1, . . . , ur be vectors in V. We de®ne sp (u1, . . . , ur) to be the
set of all linear combinations of u1, . . . , ur; that is,
sp (u1, : : : , ur) � fë1u1 � : : :� ërur: ë1, : : : , ër 2 Fg:By (2.5), sp (u1, . . . , ur) is a subspace of V, and it is called the
subspace spanned by u1, . . . , ur.
16 Representations and characters of groups
Notice that the following fact is a consequence of (2.4).
(2.7) Suppose that U is a subspace of the vector space V.
Then dim U < dim V. Also, dim U � dim V if and
only if U � V.
Direct sums of subspaces
If U1, . . . , Ur are subspaces of a vector space V, then the sum
U1 � . . . � Ur is de®ned by
U1 � : : :� Ur � fu1 � : : :� ur: ui 2 Ui for 1 < i < rg:By (2.5), U1 � . . . � Ur is a subspace of V.
We say that the sum U1 � . . . � Ur is a direct sum if every element
of the sum can be written in a unique way as u1 � . . . � ur with
ui 2 Ui for 1 < i < r. If the sum is direct, then we write it as
U1 � : : :� Ur:
2.8 Examples
(1) Suppose that v1, . . . , vn is a basis of V, and for 1 < i < n, let Ui
be the subspace spanned by vi. Then
V � U1 � : : :� Un:
(2) Let U be a subspace of V and let v1, . . . , vk be a basis of U.
Extend v1, . . . , vk to a basis v1, . . . , vn of V (see (2.4)), and let W �sp (vk�1, . . . , vn). Then
V � U � W :
From this construction it follows that there are in®nitely many sub-
spaces W with V � U � W, unless U is {0} or V.
The next result is frequently useful when dealing with the direct sum
of two subspaces. You should consult the solutions to Exercises 2.3 and
2.4 if you have dif®culty with the proof.
(2.9) Suppose that V � U � W, that u1, . . . , ur is a basis of
U and that w1, . . . , ws is a basis of W. Then the
following three conditions are equivalent:
(1) V � U � W,
(2) u1, . . . , ur, w1, . . . , ws is a basis of V,
(3) U \ W � {0}.
Vector spaces and linear transformations 17
Our next result, involving the direct sum of several subspaces, can
be deduced immediately from the de®nition of a direct sum.
(2.10) Suppose that U, W, U1, . . . , Ua, W1, . . . , Wb are
subspaces of a vector space V. If V � U � W and
also
U � U1 � : : :� Ua, and
W � W1 � : : :� Wb,
thenV � U1 � : : :� Ua � W1 � : : :� Wb:
We now introduce a construction for vector spaces which is analo-
gous to the construction of direct products for groups.
Let U1, . . . , Ur be vector spaces over F, and let
V � f(u1, : : : , ur): ui 2 Ui for 1 < i < rg:De®ne addition and scalar multiplication on V as follows: for all ui, u9iin Ui (1 < i < r) and all ë in F, let
(u1, : : : , ur)� (u91, : : : , u9r) � (u1 � u91, : : : , ur � u9r),
ë(u1, : : : , ur) � (ëu1, : : : , ëur):
With these de®nitions, V is a vector space over F. If, for 1 < i < r, we
putU 9i � f(0, : : : , ui, : : : , 0): ui 2 Uig
(where the ui is in the ith position), then it is immediate that
V � U 91 � : : :� U 9r:
We call V the external direct sum of U1, . . . , Ur, and, abusing notation
slightly, we write
V � U1 � : : :� Ur:
Linear transformations
Let V and W be vector spaces over F. A linear transformation from V
to W is a function W: V! W which satis®es
(u� v)W � uW� vW for all u, v 2 V , and
(ëv)W � ë(vW) for all ë 2 F and v 2 V :
18 Representations and characters of groups
Just as a group homomorphism preserves the group multiplication, so a
linear transformation preserves addition and scalar multiplication.
Notice that if W: V! W is a linear transformation and v1, . . . , vn is
a basis of V, then for ë1, . . . , ën in F we have
(ë1v1 � : : :� ënvn)W � ë1(v1W)� : : :� ën(vnW):
Thus, W is determined by its action on a basis. Furthermore, given any
basis v1, . . . , vn of V and any n vectors w1, . . . , wn in W, there is a
unique linear transformation ö: V! W such that viö� wi for all i; the
linear transformation ö is given by
(ë1v1 � : : :� ënvn)ö � ë1w1 � : : :� ënwn:
We sometimes construct a linear transformation ö: V! W in this way,
by specifying the values of ö on a basis of V, and then saying `extend
the action of ö to be linear'.
Kernels and images
Suppose that W: V! W is a linear transformation. The kernel of W(written Ker W) and the image of W (written Im W) are de®ned as
follows:
Ker W � fv 2 V : vW � 0g,(2:11)
Im W � fvW: v 2 Vg:Using (2.5), it is easy to check that Ker W is a subspace of V and Im Wis a subspace of W. Their dimensions are connected by the following
equation, which is known as the Rank±Nullity Theorem:
dim V � dim (Ker W)� dim (Im W):(2:12)
2.13 Examples
(1) If W: V! W is de®ned by vW � 0 for all v 2 V, then W is a linear
transformation, and
Ker W � V , Im W � f0g:(2) If W: V! V is de®ned by vW � 3v for all v 2 V, then W is a linear
transformation, and
Ker W � f0g, Im W � V :
Vector spaces and linear transformations 19
(3) If W: R3 ! R2 is given by
(x, y, z)W � (x� 2y� z, ÿy� 3z)
for all x, y, z 2 R, then W is a linear transformation; we have
Ker W � sp ((7, ÿ3, ÿ1)), Im W � R2,
so dim (Ker W) � 1 and dim (Im W) � 2.
Invertible linear transformations
Again, let V and W be vector spaces over F. A linear transformation Wfrom V to W is injective if and only if Ker W � {0}, and hence W is
invertible precisely when W is surjective and Ker W � {0}. It turns out
that the inverse of an invertible linear transformation is also a linear
transformation (see Exercise 2.1).
If there exists an invertible linear transformation from V to W, then
V and W are said to be isomorphic vector spaces. By applying (2.12),
we see that isomorphic vector spaces have the same dimension. By
also taking (2.7) into account, we obtain the next result (see
Exercise 2.2).
(2.14) Let W be a linear transformation from V to itself. Then the
following three conditions are equivalent:
(1) W is invertible;
(2) Ker W � {0};
(3) Im W � V.
Endomorphisms
A linear transformation from a vector space V to itself is called an
endomorphism of V.
Suppose that W and ö are endomorphisms of V and ë 2 F. We de®ne
the functions W � ö, Wö and ëW from V to V by
v(W� ö) � vW� vö,(2:15)
v(Wö) � (vW)ö,
v(ëW) � ë(vW),
for all v 2 V. Then W � ö, Wö and ëW are endomorphisms of V. We
write W2 for WW.
20 Representations and characters of groups
2.16 Examples
(1) The identity function 1V de®ned by
1V : v! v for all v 2 V
is an endomorphism of V. If W is an endomorphism of V, then so is
W ÿ ë1V , for all ë 2 F. Note that
Ker (Wÿ ë1V ) � fv 2 V : vW � ëvg:(2) Let V � R2, and let W, ö be the functions from V to V de®ned by
(x, y)W � (x� y, xÿ 2y),
(x, y)ö � (xÿ 2y, ÿ2x� 4y):
Then W and ö are endomorphisms of V, and W � ö, Wö, 3W and W2 are
given by
(x, y)(W� ö) � (2xÿ y, ÿx� 2y),
(x, y)(Wö) � (ÿx� 5y, 2xÿ 10y),
(x, y)(3W) � (3x� 3y, 3xÿ 6y),
(x, y)W2 � (2xÿ y, ÿx� 5y):
Matrices
Let V be a vector space over F, and let W be an endomorphism of V.
Suppose that v1, . . . , vn is a basis of V and call it B . Then there are
scalars aij in F (1 < i < n, 1 < j < n) such that for all i,
viW � ai1v1 � : : :� ainvn:
2.17 De®nition
The n 3 n matrix (aij) is called the matrix of W relative to the basis
B , and is denoted by [W]B .
2.18 Examples
(1) If W � 1V (so that vW � v for all v 2 V), then [W]B � In for all
bases B of V, where In denotes the n 3 n identity matrix.
(2) Let V � R2 and let W be the endomorphism (x, y)! (x� y,
xÿ 2y) of V. If B is the basis (1, 0), (0, 1) of V and B 9 is the basis
Vector spaces and linear transformations 21
(1, 0), (1, 1) of V, then
[W]B � 1 1
1 ÿ2
� �, [W]B 9 � 0 1
3 ÿ1
� �:
If we wish to indicate that the entries in a matrix A come from F,
then we describe A as a matrix over F.
Given two m 3 n matrices A � (aij) and B � (bij) over F, their sum
A � B is the m 3 n matrix over F whose ij-entry is aij � bij for all i,
j; and for ë 2 F, the matrix ëA is the m 3 n matrix over F obtained
from A by multiplying all the entries by ë.
As you know, the product of two matrices is de®ned in a less
transparent way. Given an m 3 n matrix A � (aij) and an n 3 p matrix
B � (bij), their product AB is the m 3 p matrix whose ij-entry isXn
k�1
aikbkj:
2.19 Example
Let
A � ÿ1 2
3 1
� �, B � 0 ÿ4
2 ÿ1
� �:
Then
A� B �ÿ1 ÿ2
5 0
!, 3A �
ÿ3 6
9 3
!, AB �
4 2
2 ÿ13
!,
BA �ÿ12 ÿ4
ÿ5 3
!:
The matrix of the sum or product of two endomorphisms (relative to
some basis) is related to the matrices of the individual endomorphisms
in the way you would expect:
(2.20) Suppose that B is a basis of the vector space V, and
that W and ö are endomorphisms of V. Then
[W� ö]B � [W]B � [ö]B , and
[Wö]B � [W]B [ö]B :
22 Representations and characters of groups
Also, for all scalars ë,
[ëW]B � ë[W]B :
We showed you in (2.17) how to get a matrix from an endomorph-
ism of a vector space V, given a basis of V. It is easy enough to
reverse this process and use a matrix to de®ne an endomorphism. We
concentrate on a particular way of doing this. Suppose that A is an
n 3 n matrix over F, and let V � Fn, the vector space of row vectors
(x1, . . . , xn) with each xi in F. Then for all v in V, the matrix product
vA also lies in V. The following remark is easily justi®ed.
(2.21) If A is an n 3 n matrix over F, then the function
v! vA (v 2 F n)
is an endomorphism of Fn.
2.22 Example
Let
A � 1 ÿ1
3 2
� �:
Then A gives us an endomorphism W of F2, where
(x, y)W � (x, y)1 ÿ1
3 2
� �� (x� 3y, ÿx� 2y):
Invertible matrices
An n 3 n matrix A is said to be invertible if there exists an n 3 n
matrix B with AB � BA � In. Such a matrix B, if it exists, is unique;
it is called the inverse of A and is written as Aÿ1. Write det A for the
determinant of A. Then a necessary and suf®cient condition for A to
be invertible is that det A 6� 0.
The connection between invertible endomorphisms and invertible
matrices is straightforward, and follows from (2.20): given a basis Bof V, an endomorphism W of V is invertible if and only if the matrix
[W]B is invertible.
Invertible matrices turn up when we relate two bases of a vector
space. An invertible matrix converts one basis into another, and this
same matrix is used to describe the way in which the matrix of an
Vector spaces and linear transformations 23
endomorphism depends upon the basis. The precise meaning of these
remarks is revealed in the de®nition (2.23) and the result (2.24) below.
2.23 De®nition
Let v1, . . . , vn be a basis B of the vector space V, and let v91, . . . , v9nbe a basis B 9 of V. Then for 1 < i < n,
v9i � ti1v1 � : : :� tinvn
for certain scalars tij. The n 3 n matrix T � (tij) is invertible, and is
called the change of basis matrix from B to B 9.
The inverse of T is the change of basis matrix from B 9 to B .
(2.24) If B and B 9 are bases of V and W is an endomorphism of V,
then
[W]B � Tÿ1[W]B 9T ,
where T is the change of basis matrix from B to B 9.
2.25 Example
Suppose that V � R2. Let B be the basis (1, 0), (0, 1) and B 9 the
basis (1, 0), (1, 1) of V. Then
T � 1 0
1 1
� �, Tÿ1 � 1 0
ÿ1 1
� �:
If W is the endomorphism
W: (x, y)! (x� y, xÿ 2y)
of V, as in Example 2.18(2), then
[W]B � 1 1
1 ÿ2
� �� 1 0
ÿ1 1
� �0 1
3 ÿ1
� �1 0
1 1
� �� Tÿ1[W]B 9T :
Eigenvalues
Let V be an n-dimensional vector space over F, and suppose that W is
an endomorphism of V. The scalar ë is said to be an eigenvalue of Wif vW � ëv for some non-zero vector v in V. Such a vector v is called
an eigenvector of W.
24 Representations and characters of groups
Now ë is an eigenvalue of W if and only if Ker (W ÿ ë1V ) 6� {0},
which occurs if and only if W ÿ ë1V is not invertible. Therefore, if Bis a basis of V, then the eigenvalues of W are those scalars ë in F
which satisfy the equation
det ([W]B ÿ ëIn) � 0:
Solving this equation involves ®nding the roots of a polynomial of
degree n. Since every non-constant polynomial with coef®cients in C
has a root in C, we deduce the following result.
(2.26) Let V be a non-zero vector space over C, and let W be an
endomorphism of V. Then W has an eigenvalue.
2.27 Examples
(1) Let V � C2 and let W be the endomorphism of V which is given by
(x, y)W � (ÿy, x):
If B is the basis (1, 0), (0, 1) of V, then
[W]B � 0 1
ÿ1 0
� �:
We have det ([W]B ÿ ëI2) � ë2 � 1, so i and ÿi are the eigenvalues of
W. Corresponding eigenvectors are (1, ÿi) and (1, i). Note that if B 9 is
the basis (1, ÿi), (1, i) of V, then
[W]B 9 � i 0
0 ÿi
� �:
(2) Let V � R2 and let W again be the endomorphism which is given
by
(x, y)W � (ÿy, x):
This time, V is a vector space over R, and W has no eigenvalues in R.
Thus we depend upon F being C in result (2.26).
For an n 3 n matrix A over F, the element ë of F is said to be an
eigenvalue of A if vA � ëv for some non-zero row vector v in Fn. The
eigenvalues of A are those elements ë of F which satisfy
det (Aÿ ëIn) � 0:
Vector spaces and linear transformations 25
2.28 Example
We say that an n 3 n matrix A � (aij) is diagonal if aij � 0 for all i
and j with i 6� j. We often display such a matrix in the form
A �ë1
. ..
ën
0B@1CA
which indicates, in addition, that aii � ëi for 1 < i < n. For this
diagonal matrix A, the eigenvalues are ë1, . . . , ën.
Projections
If a vector space V is a direct sum of two subspaces U and W, then
we can construct a special endomorphism of V which depends upon
the expression V � U � W:
2.29 Proposition
Suppose that V � U � W. De®ne ð: V! V by
(u� w)ð � u for all u 2 U , w 2 W :
Then ð is an endomorphism of V. Further,
Imð � U , Kerð � W and ð2 � ð:
Proof Since every vector in V has a unique expression in the form
u � w with u 2 U, w 2 W, it follows that ð is a function on V.
Let v and v9 belong to V. Then v � u � w and v9 � u9 � w9 for
some u, u9 in U and w, w9 in W. We have
(v� v9)ð � (u� u9� w� w9)ð � u� u9
� (u� w)ð� (u9� w9)ð
� vð� v9ð:
Also, for ë in F,
(ëv)ð � (ëu� ëw)ð � ëu � ë(vð):
Therefore, ð is an endomorphism of V.
Clearly Imð # U; and since uð � u for all u in U, we have Im � U.
Also,
0
0
26 Representations and characters of groups
(u� w)ð � 0, u � 0, u� w 2 W ,
and so Kerð � W.
Finally,
(u� w)ð2 � uð � u � (u� w)ð,
and so ð2 � ð. j
2.30 De®nition
An endomorphism ð of a vector space V which satis®es ð2 � ð is
called a projection of V.
2.31 Example
The endomorphism
(x, y)! (2x� 2y, ÿxÿ y)
of R2 is a projection.
We now show that every projection can be constructed using a direct
sum, as in Proposition 2.29.
2.32 Proposition
Suppose that ð is a projection of a vector space V. Then
V � Imð� Kerð:
Proof If v 2 V then v � vð � (v ÿ vð). Clearly the ®rst term vðbelongs to Imð; and the second term v ÿ vð lies in Kerð, since
(vÿ vð)ð � vðÿ vð2 � vðÿ vð � 0:
This establishes that V � Imð � Kerð.
Now suppose that v lies in Imð \ Kerð. As v 2 Imð, we have
v � uð for some u 2 V. Therefore
vð � uð2 � uð � v:
Since v 2 Kerð, it follows that v � vð � 0. Thus
Imð \ Kerð � f0g,and (2.9) now shows that V � Imð � Kerð. j
Vector spaces and linear transformations 27
2.33 Example
If ð: (x, y)! (2x � 2y, ÿx ÿ y) is the projection of R2 which appears
in Example 2.31, then
Imð � f(2x, ÿx): x 2 Rg, Kerð � f(x, ÿx): x 2 Rg:
Summary of Chapter 2
1. All our vector spaces are ®nite-dimensional over F, where F � C or
R. For example, F n is the set of row vectors (x1, . . . , xn) with each
xi in F, and dimF n � n.
2. V � U1 � . . . � Ur if each Ui is a subspace of V, and every element
v of V has a unique expression of the form v � u1 � . . . � ur
(ui 2 Ui).
Also, V � U � W if and only if V � U � W and U \ W � {0}.
3. A linear transformation W: V! W satis®es
(u� v)W � uW� vW and (ëv)W � ë(vW)
for all u, v in V and all ë in F. Ker W is a subspace of V and Im Wis a subspace of W, and
dim V � dim (Ker W)� dim (Im W):
4. A linear transformation W: V! W is invertible if and only if
Ker W � {0} and Im W � W.
5. Given a basis B of the n-dimensional vector space V, there is a
correspondence between the endomorphisms W of V and the n 3 n
matrices [W]B over F.
Given two bases B and B 9 of V, and an endomorphism W of V,
there exists an invertible matrix T such that
[W]B � Tÿ1[W]B 9T :
6. Eigenvalues ë of an endomorphism W satisfy vW � ëv for some non-
zero v in V.
7. A projection is an endomorphism ð of V which satis®es ð2 � ð.
Exercises for Chapter 2
1. Show that if V and W are vector spaces and W: V! W is an
invertible linear transformation then Wÿ1 is a linear transformation.
28 Representations and characters of groups
2. Suppose that W is an endomorphism of the vector space V. Show
that the following are equivalent:
(1) W is invertible;
(2) Ker W � {0};
(3) Im W � V.
3. Let U and W be subspaces of the vector space V. Prove that
V � U � W if and only if V � U � W and U \ W = {0}.
4. Let U and W be subspaces of the vector space V. Suppose that
u1, . . . , ur is a basis of U and w1, . . . , ws is a basis of W. Show
that V � U � W if and only if u1, . . . , ur, w1, . . . , ws is a basis
of V.
5. (a) Let U1, U2 and U3 be subspaces of a vector space V, with
V � U1 � U2 � U3. Show that
V � U1 � U2 � U3 ,U1 \ (U2 � U3) � U2 \ (U1 � U3) � U3 \ (U1 � U2) � f0g:
(b) Give an example of a vector space V with three subspaces U1,
U2 and U3 such that V � U1 � U2 � U3 and
U1 \ U2 � U1 \ U3 � U2 \ U3 � f0g,but V 6� U1 � U2 � U3.
6. Suppose that U1, . . . , Ur are subspaces of the vector space V, and
that V � U1 � . . . � Ur. Prove that
dim V � dim U1 � : : :� dim Ur:
7. Give an example of a vector space V with endomorphisms W and ösuch that V � Im W � Ker W, but V 6� Imö � Kerö.
8. Let V be a vector space and let W be an endomorphism of V. Show
that W is a projection if and only if there is a basis B of V such
that [W]B is diagonal, with all diagonal entries equal to 1 or 0.
9. Suppose that W is an endomorphism of the vector space V and
W2 � 1V . Show that V � U � W, where
U � fv 2 V : vW � vg, W � fv 2 V : vW � ÿvg:Deduce that V has a basis B such that [W]B is diagonal, with all
diagonal entries equal to �1 or ÿ1.
Vector spaces and linear transformations 29
30
3
Group representations
A representation of a group G gives us a way of visualizing G as a
group of matrices. To be precise, a representation is a homomorphism
from G into a group of invertible matrices. We set out this idea in
more detail, and give some examples of representations. We also
introduce the concept of equivalence of representations, and consider
the kernel of a representation.
Representations
Let G be a group and let F be R or C. Recall from the ®rst chapter
that GL (n, F) denotes the group of invertible n 3 n matrices with
entries in F.
3.1 De®nition
A representation of G over F is a homomorphism r from G to
GL (n, F), for some n. The degree of r is the integer n.
Thus if r is a function from G to GL (n, F), then r is a representa-
tion if and only if
(gh)r � (gr)(hr) for all g, h 2 G:
Since a representation is a homomorphism, it follows that for every
representation r: G! GL (n, F), we have
1r � In, and
gÿ1r � (gr)ÿ1 for all g 2 G,
where In denotes the n 3 n identity matrix.
3.2 Examples
(1) Let G be the dihedral group D8 � ka, b: a4 � b2 � 1, bÿ1ab �aÿ1l. De®ne the matrices A and B by
A � 0 1
ÿ1 0
� �, B � 1 0
0 ÿ1
� �and check that
A4 � B2 � I , Bÿ1 AB � Aÿ1:
It follows (see Example 1.4) that the function r: G! GL (2, F) which
is given by
r: aib j ! Ai Bj (0 < i < 3, 0 < j < 1)
is a representation of D8 over F. The degree of r is 2.
The matrices gr for g in D8 are given in the following table:
g 1 a a2 a3
gr1 0
0 1
� �0 1
ÿ1 0
� � ÿ1 0
0 ÿ1
� �0 ÿ1
1 0
� �
g b ab a2b a3b
gr1 0
0 ÿ1
� �0 ÿ1
ÿ1 0
� � ÿ1 0
0 1
� �0 1
1 0
� �
(2) Let G be any group. De®ne r: G! GL (n, F) by
gr � In for all g 2 G,
where In is the n 3 n identity matrix, as usual. Then
(gh)r � In � InIn � (gr)(hr)
for all g, h 2 G, so r is a representation of G. This shows that every
group has representations of arbitrarily large degree.
Equivalent representations
We now look at a way of converting a given representation into another
one.
Group representations 31
Let r: G! GL (n, F) be a representation, and let T be an invertible
n 3 n matrix over F. Note that for all n 3 n matrices A and B, we
have
(Tÿ1 AT )(Tÿ1 BT ) � Tÿ1(AB)T :
We can use this observation to produce a new representation ó from r;
we simply de®ne
gó � Tÿ1(gr)T for all g 2 G:
Then for all g, h 2 G,
(gh)ó � Tÿ1((gh)r)T
� Tÿ1((gr)(hr))T
� Tÿ1(gr)T . Tÿ1(hr)T
� (gó )(hó ),
and so ó is, indeed, a representation.
3.3 De®nition
Let r: G! GL (m, F) and ó : G! GL (n, F) be representations of G
over F. We say that r is equivalent to ó if n � m and there exists an
invertible n 3 n matrix T such that for all g 2 G,
gó � Tÿ1(gr)T :
Note that for all representations r, ó and ô of G over F, we have
(see Exercise 3.4):
(1) r is equivalent to r;
(2) if r is equivalent to ó then ó is equivalent to r;
(3) if r is equivalent to ó and ó is equivalent to ô, then r is
equivalent to ô.
In other words, equivalence of representations is an equivalence rela-
tion.
3.4 Examples
(1) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and consider the
representation r of G which appears in Example 3.2(1). Thus ar � A
32 Representations and characters of groups
and br � B, where
A � 0 1
ÿ1 0
� �, B � 1 0
0 ÿ1
� �:
Assume that F � C, and de®ne
T � 1p2
1 1
i ÿi
� �:
Then
Tÿ1 � 1p2
1 ÿi
1 i
� �:
In fact, T has been constructed so that Tÿ1AT is diagonal; we have
Tÿ1 AT � i 0
0 ÿi
� �, Tÿ1 BT � 0 1
1 0
� �,
and so we obtain a representation ó of D8 for which
aó � i 0
0 ÿi
� �, bó � 0 1
1 0
� �:
The representations r and ó are equivalent.
(2) Let G � C2 � ka: a2 � 1l and let
A � ÿ5 12
ÿ2 5
� �:
Check that A2 � I. Hence r: 1! I, a! A is a representation of G. If
T � 2 ÿ3
1 ÿ1
� �,
then
Tÿ1 AT � 1 0
0 ÿ1
� �,
and so we obtain a representation ó of G for which
1ó � 1 0
0 1
� �, aó � 1 0
0 ÿ1
� �,
and ó is equivalent to r.
Group representations 33
There are two easily recognized situations where the only represent-
ation which is equivalent to r is r itself; these are when the degree of
r is 1, and when gr � In for all g in G. However, there are usually
lots of representations which are equivalent to r.
Kernels of representations
We conclude the chapter with a discussion of the kernel of a
representation r: G! GL (n, F). In agreement with De®nition 1.8, this
consists of the group elements g in G for which gr is the identity
matrix. Thus
Ker r � fg 2 G: gr � Ing:Note that Ker r is a normal subgroup of G.
It can happen that the kernel of a representation is the whole of G,
as is shown by the following de®nition.
3.5 De®nition
The representation r: G! GL (1, F) which is de®ned by
gr � (1) for all g 2 G,
is called the trivial representation of G.
To put the de®nition another way, the trivial representation of G is
the representation where every group element is sent to the 1 3 1
identity matrix.
Of particular interest are those representations whose kernel is just
the identity subgroup.
3.6 De®nition
A representation r: G ! GL (n, F) is said to be faithful if
Ker r � {1}; that is, if the identity element of G is the only element g
for which gr � In.
3.7 Proposition
A representation r of a ®nite group G is faithful if and only if Im r is
isomorphic to G.
34 Representations and characters of groups
Proof We know that Ker r v G and by Theorem 1.10, the factor group
G/ Ker r is isomorphic to Im r. Therefore, if Ker r � {1} then
G � Im r. Conversely, if G � Im r, then these two groups have the
same (®nite) order, and so |Ker r| � 1; that is, r is faithful. j
3.8 Examples
(1) The representation r of D8 given by
(aib j)r � 0 1
ÿ1 0
� �i1 0
0 ÿ1
� � j
as in Example 3.2(1) is faithful, since the identity is the only element
g which satis®es gr � I. The group generated by the matrices
0 1
ÿ1 0
� �and
1 0
0 ÿ1
� �is therefore isomorphic to D8.
(2) Since Tÿ1AT � In if and only if A � In, it follows that all
representations which are equivalent to a faithful representation are
faithful.
(3) The trivial representation of a group G if faithful if and only if
G � {1}.
In Chapter 6 we shall show that every ®nite group has a faithful
representation.
The basic problem of representation theory is to discover and under-
stand representations of ®nite groups.
Summary of Chapter 3
1. A representation of a group G is a homomorphism from G into
GL(n, F), for some n.
2. Representations r and ó of G are equivalent if and only if there
exists an invertible matrix T such that for all g 2 G,
gó � Tÿ1(gr)T :
3. A representation is faithful if it is injective.
Group representations 35
Exercises for Chapter 3
1. Let G be the cyclic group of order m, say G � ka: am � 1l. Suppose
that A 2 GL (n, C), and de®ne r: G! GL (n, C) by
r: ar ! Ar (0 < r < mÿ 1):
Show that r is a representation of G over C if and only if Am � I.
2. Let
A � 1 0
0 1
� �, B � 1 0
0 e2ði=3
� �, C � 0 1
ÿ1 ÿ1
� �and let G � ka: a3 � 1l � C3. Show that each of the functions
r j: G! GL (2, C) (1 < j < 3), de®ned by
r1: ar ! Ar,
r2: ar ! Br,
r3: ar ! Cr (0 < r < 2),
is a representation of G over C. Which of these representations are
faithful?
3. Suppose that G � D2n � ka, b: an � b2 � 1, bÿ1ab � aÿ1l, and
F � R or C. Show that there is a representation r: G! GL (1, F)
such that ar � (1) and br � (ÿ1).
4. Suppose that r, ó and ô are representations of G over F. Prove:
(1) r is equivalent to r;
(2) if r is equivalent to ó, then ó is equivalent to r;
(3) if r is equivalent to ó, and ó is equivalent to ô, then r is
equivalent to ô.
5. Let G � D12 � ka, b: a6 � b2 � 1, bÿ1ab � aÿ1l. De®ne the matrices
A, B, C, D over C by
A � eið=3 0
0 eÿið=3
� �, B � 0 1
1 0
� �,
C � 1=2p
3=2
ÿp3=2 1=2
� �, D � 1 0
0 ÿ1
� �:
Prove that each of the functions rk : G! GL (2, C) (k � 1, 2, 3, 4),
given by
36 Representations and characters of groups
r1: arbs ! ArBs,
r2: arbs ! A3r(ÿB)s,
r3: arbs ! (ÿA)rBs,
r4: arbs ! C r Ds (0 < r < 5, 0 < s < 1),
is a representation of G. Which of these representations are faithful?
Which are equivalent?
6. Give an example of a faithful representation of D8 of degree 3.
7. Suppose that r is a representation of G of degree 1. Prove that
G=Ker r is abelian.
8. Let r be a representation of the group G. Suppose that g and h are
elements of G such that (gr)(hr) � (hr)(gr). Does it follow that
gh � hg?
Group representations 37
38
4
FG-modules
We now introduce the concept of an FG-module, and show that there
is a close connection between FG-modules and representations of G
over F. Much of the material in the remainder of the book will be
presented in terms of FG-modules, as there are several advantages to
this approach to representation theory.
FG-modules
Let G be a group and let F be R or C.
Suppose that r: G! GL (n, F) is a representation of G. Write
V � F n, the vector space of all row vectors (ë1, . . . , ën) with ëi 2 F.
For all v 2 V and g 2 G, the matrix product
v(gr),
of the row vector v with the n 3 n matrix gr, is a row vector in V
(since the product of a 1 3 n matrix with an n 3 n matrix is again a
1 3 n matrix).
We now list some basic properties of the multiplication v(gr). First,
the fact that r is a homomorphism shows that
v((gh)r) � v(gr)(hr)
for all v 2 V and all g, h 2 G. Next, since 1r is the identity matrix, we
have
v(1r) � v
for all v 2 V. Finally, the properties of matrix multiplication give
(ëv)(gr) � ë(v(gr)),
(u� v)(gr) � u(gr)� v(gr)
for all u, v 2 V, ë 2 F and g 2 G.
4.1 Example
Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let r: G!GL (2, F) be the representation of G over F given in Example 3.2(1).
Thus
ar � 0 1
ÿ1 0
� �, br � 1 0
0 ÿ1
� �:
If v � (ë1, ë2) 2 F2 then, for example,
v(ar) � (ÿë2, ë1),
v(br) � (ë1, ÿë2),
v(a3r) � (ë2, ÿë1):
Motivated by the above observations on the product v(gr), we now
de®ne an FG-module.
4.2 De®nition
Let V be a vector space over F and let G be a group. Then V is an
FG-module if a multiplication vg (v 2 V, g 2 G) is de®ned, satisfying
the following conditions for all u, v 2 V, ë 2 F and g, h 2 G:
(1) vg 2 V;
(2) v(gh) � (vg)h;
(3) v1 � v;
(4) (ëv)g � ë(vg);
(5) (u � v)g � ug � vg.
We use the letters F and G in the name `FG-module' to indicate
that V is a vector space over F and that G is the group from which we
are taking the elements g to form the products vg (v 2 V).
Note that conditions (1), (4) and (5) in the de®nition ensure that for
all g 2 G, the function
v! vg (v 2 V )
is an endomorphism of V.
FG-modules 39
4.3 De®nition
Let V be an FG-module, and let B be a basis of V. For each g 2 G,
let
[g]B
denote the matrix of the endomorphism v! vg of V, relative to the
basis B .
The connection between FG-modules and representations of G over
F is revealed in the following basic result.
4.4 Theorem
(1) If r: G! GL(n, F) is a representation of G over F, and V � F n,
then V becomes an FG-module if we de®ne the multiplication vg by
vg � v(gr) (v 2 V , g 2 G):
Moreover, there is a basis B of V such that
gr � [g]B for all g 2 G:
(2) Assume that V is an FG-module and let B be a basis of V.
Then the function
g ! [g]B (g 2 G)
is a representation of G over F.
Proof (1) We have already observed that for all u, v 2 F n, ë 2 F and
g, h 2 G, we have
v(gr) 2 F n,
v((gh)r) � (v(gr))(hr),
v(1r) � v,
(ëv)(gr) � ë(v(gr)),
(u� v)(gr) � u(gr)� v(gr):
Therefore, F n becomes an FG-module if we de®ne
vg � v(gr) for all v 2 F n, g 2 G:
Moreover, if we let B be the basis
(1, 0, 0, : : : , 0), (0, 1, 0, : : : , 0), : : : , (0, 0, 0, : : : , 1)
of F n, then gr � [g]B for all g 2 G.
40 Representations and characters of groups
(2) Let V be an FG-module with basis B . Since v(gh) � (vg)h for
all g, h 2 G and all v in the basis B of V, it follows that
[gh]B � [g]B [h]B :
In particular,
[1]B � [g]B [gÿ1]B
for all g 2 G. Now v1 � v for all v 2 V, so [1]B is the identity matrix.
Therefore each matrix [g]B is invertible (with inverse [gÿ1]B ).
We have proved that the function g! [g]B is a homomorphism
from G to GL (n, F) (where n � dim V ), and hence is a representation
of G over F. j
Our next example illustrates part (1) of Theorem 4.4.
4.5 Examples
(1) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l and let r be the
representation of G over F given in Example 3.2(1), so
ar � 0 1
ÿ1 0
� �, br � 1 0
0 ÿ1
� �:
Write V � F2. By Theorem 4.4(1), V becomes an FG-module if we
de®ne
vg � v(gr) (v 2 V , g 2 G):
For instance,
(1, 0)a � (1, 0)0 1
ÿ1 0
� �� (0, 1):
If v1, v2 is the basis (1, 0), (0, 1) of V, then we have
v1a � v2, v1b � v1,
v2a � ÿv1, v2b � ÿv2:
If B denotes the basis v1, v2, then the representation
g ! [g]B (g 2 G)
is just the representation r (see Theorem 4.4(1) again).
(2) Let G � Q8 � ka, b: a4� 1, a2 � b2, bÿ1ab � aÿ1l. In Example
FG-modules 41
1.2(4) we de®ned Q8 to be the subgroup of GL (2, C) generated by
A � i 0
0 ÿi
� �and B � 0 1
ÿ1 0
� �,
so we already have a representation of G over C. To illustrate Theorem
4.4(1) we must this time take F � C. We then obtain a CG-module
with basis v1, v2 such that
v1a � iv1, v1b � v2,
v2a � ÿiv2, v2b � ÿv1:
Notice that in the above examples, the vectors v1a, v2a, v1b and v2b
determine vg for all v 2 V and g 2 G. For instance, in Example 4.5(1),
(v1 � 2v2)ab � v1ab� 2v2ab
� v2bÿ 2v1b
� ÿv2 ÿ 2v1:
A similar remark holds for all FG-modules V: if v1, . . . , vn is a basis
of V and g1, . . . , gr generate G, then the vectors vi g j (1 < i < n,
1 < j < r) determine vg for all v 2 V and g 2 G.
Shortly, we shall show you various ways of constructing FG-modules
directly, without using a representation. To do this, we turn a vector
space V over F into an FG-module by specifying the action of group
elements on a basis v1, . . . , vn of V and then extending the action to
be linear on the whole of V; that is, we ®rst de®ne vig for each i and
each g in G, and then de®ne
(ë1v1 � : : :� ënvn)g (ëi 2 F)
to be
ë1(v1 g)� : : :� ën(vng):
As you might expect, there are restrictions on how we may de®ne the
vectors vig. The next result will often be used to show that our chosen
multiplication turns V into an FG-module.
4.6 Proposition
Assume that v1, . . . , vn is a basis of a vector space V over F. Suppose
that we have a multiplication vg for all v in V and g in G which
42 Representations and characters of groups
satis®es the following conditions for all i with 1 < i < n, for all
g, h 2 G, and for all ë1, . . . , ën 2 F:
(1) vig 2 V;
(2) vi(gh) � (vi g)h;
(3) vi1 � vi;
(4) (ë1v1 � . . . � ënvn)g � ë1(v1 g) � . . . � ën(vn g).
Then V is an FG-module.
Proof It is clear from (3) and (4) that v1 � v for all v 2 V.
Conditions (1) and (4) ensure that for all g in G, the function
v! vg (v 2 V) is an endomorphism of V. That is,
vg 2 V ,
(ëv)g � ë(vg),
(u� v)g � ug � vg,
for all u, v 2 V, ë 2 F and g 2 G. Hence
(ë1u1 � : : :� ënun)h � ë1(u1 h)� : : :� ën(unh)(4:7)
for all ë1, . . . , ën 2 F, all u1, . . . , un 2 V and all h 2 G.
Now let v 2 V and g, h 2 G. Then v � ë1v1 � . . . � ënvn for some
ë1, . . . , ën 2 F, and
v(gh) � ë1(v1(gh))� : : :� ën(vn(gh)) by condition (4)
� ë1((v1 g)h)� : : :� ën((vng)h) by condition (2)
� (ë1(v1 g)� : : :� ën(vng))h by (4:7)
� (vg)h by condition (4):
We have now checked all the axioms which are required for V to be
an FG-module. j
Our next de®nitions translate the concepts of the trivial represent-
ation and a faithful representation into module terms.
4.8 De®nitions
(1) The trivial FG-module is the 1-dimensional vector space V over F
with
vg � v for all v 2 V , g 2 G:
FG-modules 43
(2) An FG-module V is faithful if the identity element of G is the
only element g for which
vg � v for all v 2 V :
For instance, the FD8-module which appears in Example 4.5(1) is
faithful.
Our next aim is to use Proposition 4.6 to construct faithful FG-
modules for all subgroups of symmetric groups.
Permutation modules
Let G be a subgroup of Sn, so that G is a group of permutations of
{1, . . . , n}. Let V be an n-dimensional vector space over F, with basis
v1, . . . , vn. For each i with 1 < i < n and each permutation g in G,
de®ne
vig � vig:
Then vig 2 V and vi1 � vi. Also, for g, h in G,
vi(gh) � vi( gh) � v(ig)h � (vig)h:
We now extend the action of each g linearly to the whole of V; that is,
for all ë1, . . . , ën in F and g in G, we de®ne
(ë1v1 � : : :� ënvn)g � ë1(v1 g)� : : :� ën(vng):
Then V is an FG-module, by Proposition 4.6.
4.9 Example
Let G � S4 and let B denote the basis v1, v2, v3, v4 of V. If g � (1 2),
then
v1 g � v2, v2 g � v1, v3 g � v3, v4 g � v4:
And if h � (1 3 4), then
v1 h � v3, v2 h � v2, v3 h � v4, v4 h � v1:
We have
[g]B �0 1 0 0
1 0 0 0
0 0 1 0
0 0 0 1
0BB@1CCA, [h]B �
0 0 1 0
0 1 0 0
0 0 0 1
1 0 0 0
0BB@1CCA:
44 Representations and characters of groups
4.10 De®nition
Let G be a subgroup of Sn. The FG-module V with basis v1, . . . , vn
such that
vig � vig for all i, and all g 2 G,
is called the permutation module for G over F. We call v1, . . . , vn the
natural basis of V.
Note that if we write B for the basis v1, . . . , vn of the permutation
module, then for all g in G, the matrix [g]B has precisely one non-
zero entry in each row and column, and this entry is 1. Such a matrix
is called a permutation matrix.
Since the only element of G which ®xes every vi is the identity, we see
that the permutation module is a faithful FG-module. If you are aware of
the fact that every group G of order n is isomorphic to a subgroup of Sn,
then you should be able to see that G has a faithful FG-module of
dimension n. We shall go into this in more detail in Chapter 6.
4.11 Example
Let G � C3 � ka: a3 � 1l. Then G is isomorphic to the cyclic subgroup
of S3 which is generated by the permutation (1 2 3). This alerts us to
the fact that if V is a 3-dimensional vector space over F, with basis v1,
v2, v3, then we may make V into an FG-module in which
v11 � v1, v21 � v2, v31 � v3,
v1a � v2, v2a � v3, v3a � v1,
v1a2 � v3, v2a2 � v1, v3a2 � v2:
Of course, we de®ne vg, for v an arbitrary vector in V and g � 1, a
or a2, by
(ë1v1 � ë2v2 � ë3v3)g � ë1(v1 g)� ë2(v2 g)� ë3(v3 g)
for all ë1, ë2, ë3 2 F. Proposition 4.6 can be used to verify that V is an
FG-module, but we have been motivated by the de®nition of permuta-
tion modules in our construction.
FG-modules and equivalent representations
We conclude the chapter with a discussion of the relationship between
FG-modules and equivalent representations of G over F. An FG-
FG-modules 45
module gives us many representations, all of the form
g ! [g]B (g 2 G)
for some basis B of V. The next result shows that all these representa-
tions are equivalent to each other (see De®nition 3.3); and moreover,
any two equivalent representations of G arise from some FG-module in
this way.
4.12 Theorem
Suppose that V is an FG-module with basis B , and let r be the
representation of G over F de®ned by
r: g ! [g]B (g 2 G):
(1) If B 9 is a basis of V, then the representation
ö: g ! [g]B 9 (g 2 G)
of G is equivalent to r.
(2) If ó is a representation of G which is equivalent to r, then there
is a basis B 0 of V such that
ó : g ! [g]B 0 (g 2 G):
Proof (1) Let T be the change of basis matrix from B to B 9 (see
De®nition 2.23). Then by (2.24), for all g 2 G, we have
[g]B � Tÿ1[g]B 9T :
Therefore ö is equivalent to r.
(2) Suppose that r and ó are equivalent representations of G. Then
for some invertible matrix T, we have
gr � Tÿ1(gó )T for all g 2 G:
Let B 0 be the basis of V such that the change of basis matrix from Bto B 0 is T. Then for all g 2 G,
[g]B � Tÿ1[g]B 0T ,
and so gó � [g]B 0. j
4.13 Example
Again let G � C3 � ka: a3 � 1l. There is a representation r of G which
is given by
46 Representations and characters of groups
1r � 1 0
0 1
� �, ar � 0 1
ÿ1 ÿ1
� �, a2r � ÿ1 ÿ1
1 0
� �:
(To see this, simply note that (ar)2 � a2r and (ar)3 � I; see Exercise
3.2.)
If V is a 2-dimensional vector space over C, with basis v1, v2 (which
we call B ), then we can turn V into a CG-module as in Theorem
4.4(1) by de®ning
v11 � v1, v1a � v2, v1a2 � ÿv1 ÿ v2,
v21 � v2, v2a � ÿv1 ÿ v2, v2a2 � v1:
We then have
[1]B � 1 0
0 1
� �, [a]B � 0 1
ÿ1 ÿ1
� �, [a2]B � ÿ1 ÿ1
1 0
� �:
Now let u1 � v1 and u2 � v1 � v2. Then u1, u2 is another basis of V,
which we call B 9. Since
u11 � u1, u1a � ÿu1 � u2, u1a2 � ÿu2,
u21 � u2, u2a � ÿu1, u2a2 � u1 ÿ u2,
we obtain the representation ö: g! [g]B 9 where
[1]B 9 � 1 0
0 1
� �, [a]B 9 � ÿ1 1
ÿ1 0
� �, [a2]B 9 � 0 ÿ1
1 ÿ1
� �:
Note that if
T � 1 0
1 1
� �then for all g in G, we have
[g]B � Tÿ1[g]B 9T ,
and so r and ö are equivalent, in agreement with Theorem 4.12(1).
Summary of Chapter 4
1. An FG-module is a vector space over F, together with a multi-
plication by elements of G on the right. The multiplication satis®es
properties (1)±(5) of De®nition 4.2.
FG-modules 47
2. There is a correspondence between representations of G over F
and FG-modules, as follows.
(a) Suppose that r: G! GL (n, F) is a representation of G.
Then F n is an FG-module, if we de®ne
vg � v(gr) (v 2 F n, g 2 G):
(b) If V is an FG-module, with basis B , then r: g! [g]B is a
representation of G over F.
3. If G is a subgroup of Sn, then the permutation FG-module has
basis v1, . . . , vn, and vi g � vig for all i with 1 < i < n, and all g
in G.
Exercises for Chapter 4
1. Suppose that G � S3, and that V � sp (v1, v2, v3) is the permutation
module for G over C, as in De®nition 4.10. Let B 1 be the basis
v1, v2, v3 of V and let B 2 be the basis v1 � v2 � v3, v1 ÿ v2,
v1 ÿ v3. Calculate the 3 3 3 matrices [g]B 1and [g]B 2
for all g in
S3. What do you notice about the matices [g]B 2?
2. Let G � Sn and let V be a vector space over F. Show that V
becomes an FG-module if we de®ne, for all v in V,
vg � v, if g is an even permutation,
ÿv, if g is an odd permutation.
�3. Let Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l, the quaternion
group of order 8. Show that there is an RQ8-module V of
dimension 4 with basis v1, v2, v3, v4 such that
v1a � v2, v2a � ÿv1, v3a � ÿv4, v4a � v3, and
v1b � v3, v2b � v4, v3b � ÿv1, v4b � ÿv2:
4. Let A be an n 3 n matrix and let B be a matrix obtained from A
by permuting the rows. Show that there is an n 3 n permutation
matrix P such that B � PA. Find a similar result for a matrix
obtained from A by permuting the columns.
48 Representations and characters of groups
49
5
FG-submodules and reducibility
We begin the study of FG-modules by introducing the basic building
blocks of the theory ± the irreducible FG-modules. First we require
the notion of an FG-submodule of an FG-module.
Throughout, G is a group and F is R or C.
FG-submodules
5.1 De®nition
Let V be an FG-module. A subset W of V is said to be an FG-
submodule of V if W is a subspace and wg 2 W for all w 2 W and all
g 2 G.
Thus an FG-submodule of V is a subspace which is also an FG-
module.
5.2 Examples
(1) For every FG-module V, the zero subspace {0}, and V itself, are
FG-submodules of V.
(2) Let G � C3 � ka: a3 � 1l, and let V be the 3-dimensional FG-
module de®ned in Example 4.11. Thus, V has basis v1, v2, v3, and
v11 � v1, v21 � v2, v31 � v3,
v1a � v2, v2a � v3, v3a � v1,
v1a2 � v3, v2a2 � v1, v3a2 � v2:
Put w � v1 � v2 � v3, and let W � sp (w), the 1-dimensional subspace
spanned by w. Since
w1 � wa � wa2 � w,
W is an FG-submodule of V. However, sp (v1 � v2) is not an FG-
submodule, since
(v1 � v2)a � v2 � v3 =2 sp (v1 � v2):
Irreducible FG-modules
5.3 De®nition
An FG-module V is said to be irreducible if it is non-zero and it has
no FG-submodules apart from {0} and V.
If V has an FG-submodule W with W not equal to {0} or V, then V
is reducible.
Similarly, a representation r: G! GL (n, F) is irreducible if the
corresponding FG-module F n given by
vg � v(gr) (v 2 F n, g 2 G)
(see Theorem 4.4(1)) is irreducible; and r is reducible if F n is
reducible.
Suppose that V is a reducible FG-module, so that there is an FG-
submodule W with 0 , dim W , dim V. Take a basis B 1 of W and
extend it to a basis B of V. Then for all g in G, the matrix [g]B has
the form
X g 0
Yg Zg
0@ 1A(5:4)
for some matrices Xg, Yg and Zg, where Xg is k 3 k (k � dim W).
A representation of degree n is reducible if and only if it is
equivalent to a representation of the form (5.4), where Xg is k 3 k and
0 , k , n.
Notice that in (5.4), the functions g! Xg and g! Zg are represen-
tations of G: to see this, let g, h 2 G and multiply the matrices [g]B
and [h]B given by (5.4). Notice also that if V is reducible then
dim V > 2.
5.5 Examples
(1) Let G � C3 � ka: a3 � 1l and let V be the 3-dimensional FG-
module with basis v1, v2, v3 such that
50 Representations and characters of groups
v1a � v2, v2a � v3, v3a � v1,
as in Example 4.11. We saw in Example 5.2(2) that V is a reducible
FG-module, and has an FG-submodule W � sp (v1 � v2 � v3). Let Bbe the basis v1 � v2 � v3, v1, v2 of V. Then
[1]B �1 0 0
0 1 0
0 0 1
0BB@1CCA, [a]B �
1 0 0
0 0 1
1 ÿ1 ÿ1
0BB@1CCA,
[a2]B �1 0 0
1 ÿ1 ÿ1
0 1 0
0BB@1CCA:
This reducible representation gives us two other representations: at the
`top left' we have the trivial representation and at the `bottom right'
we have the representation which is given by
1! 1 0
0 1
� �, a! 0 1
ÿ1 ÿ1
� �, a2 ! ÿ1 ÿ1
1 0
� �:
(2) Let G � D8 and let V � F2 be the 2-dimensional FG-module
described in Example 4.5(1). Thus G � ka, bl, and for all (ë, ì) 2 V we
have
(ë, ì)a � (ÿì, ë), (ë, ì)b � (ë, ÿì):
We claim that V is an irreducible FG-module. To see this, suppose
that there is an FG-submodule U which is not equal to V. Then
dim U < 1, so U � sp ((á, â)) for some á, â 2 F. As U is an FG-
module, (á, â)b is a scalar multiple of (á, â), and hence either á � 0
or â � 0. Since (á, â)a is also a scalar multiple of (á, â), this forces
á � â � 0, so U � {0}. Consequently V is irreducible, as claimed.
Summary of Chapter 5
1. If V is an FG-module, and W is a subspace of V which is itself an
FG-module, then W is an FG-submodule of V.
2. The FG-module V is irreducible if it is non-zero and the only FG-
submodules are {0} and V.
FG-submodules and reducibility 51
Exercises for Chapter 5
1. Let G � C2 � ka: a2 � 1l, and let V � F2. For (á, â) 2 V, de®ne
(á, â)1 � (á, â) and (á, â)a � (â, á). Verify that V is an FG-module
and ®nd all the FG-submodules of V.
2. Let r and ó be equivalent representations of the group G over F.
Prove that if r is reducible then ó is reducible.
3. Which of the four representations of D12 de®ned in Exercise 3.5 are
irreducible?
4. De®ne the permutations a, b, c 2 S6 by
a � (1 2 3), b � (4 5 6), c � (2 3)(4 5),
and let G � ka, b, cl.(a) Check that
a3 � b3 � c2 � 1, ab � ba,
cÿ1ac � aÿ1 and cÿ1bc � bÿ1:
Deduce that G has order 18.
(b) Suppose that å and ç are complex cube roots of unity. Prove
that there is a representation r of G over C such that
ar � å 0
0 åÿ1
� �, br � ç 0
0 çÿ1
� �, cr � 0 1
1 0
� �:
(c) For which values of å, ç is r faithful?
(d) For which values of å, ç is r irreducible?
5. Let G � C13. Find a CG-module which is neither reducible nor
irreducible.
52 Representations and characters of groups
53
6
Group algebras
The group algebra of a ®nite group G is a vector space of dimension
|G| which also carries extra structure involving the product operation
on G. In a sense, group algebras are the source of all you need to
know about representation theory. In particular, the ultimate goal of
representation theory ± that of understanding all the representations of
®nite groups ± would be achieved if group algebras could be fully
analysed. Group algebras are therefore of great interest.
After de®ning the group algebra of G, we shall use it to construct
an important faithful representation, known as the regular representation
of G, which will be explored in greater detail later on.
The group algebra of G
Let G be a ®nite group whose elements are g1, . . . , gn, and let F be
R or C.
We de®ne a vector space over F with g1, . . . , gn as a basis, and
we call this vector space FG. Take as the elements of FG all
expressions of the form
ë1 g1 � : : :� ën gn (all ëi 2 F):
The rules for addition and scalar multiplication in FG are the natural
ones: namely, if
u �Xn
i�1
ëi gi and v �Xn
i�1
ìi gi
are elements of FG, and ë 2 F, then
u� v �Xn
i�1
(ëi � ìi)gi and ëu �Xn
i�1
(ëëi)gi:
With these rules, FG is a vector space over F of dimension n, with
basis g1, . . . , gn. The basis g1, . . . , gn is called the natural basis of
FG.
6.1 Example
Let G � C3 � ka: a3 � el. (To avoid confusion with the element 1 of
F, we write e for the identity element of G, in this example.) The
vector space CG contains
u � eÿ a� 2a2 and v � 12e� 5a:
We have
u� v � 32e� 4a� 2a2, 1
3u � 1
3eÿ 1
3a� 2
3a2:
Sometimes we write elements of FG in the formXg2G
ë g g (ë g 2 F):
Now, FG carries more structure than that of a vector space ± we
can use the product operation on G to de®ne multiplication in FG as
follows: Xg2G
ë g g
! Xh2G
ìh h
!�
Xg,h2G
ë gìh(gh)
�Xg2G
Xh2G
(ëhìhÿ1 g)g
where all ë g, ìh 2 F.
6.2 Example
If G � C3 and u, v are the elements of CG which appear in Example
6.1, then
uv � (eÿ a� 2a2)(12e� 5a)
� 12e� 5aÿ 1
2aÿ 5a2 � a2 � 10a3
� 212
e� 92aÿ 4a2:
54 Representations and characters of groups
6.3 De®nition
The vector space FG, with multiplication de®ned byXg2G
ë g g
! Xh2G
ìh h
!�X
g,h2G
ë gìh(gh)
(ë g, ìh 2 F), is called the group algebra of G over F.
The group algebra FG contains an identity for multiplication, namely
the element 1e (where 1 is the identity of F and e is the identity of
G). We write this element simply as 1.
6.4 Proposition
Multiplication in FG satis®es the following properties, for all
r, s, t 2 FG and ë 2 F:
(1) rs 2 FG;
(2) r(st) � (rs)t;
(3) r1 � 1r � r;
(4) (ër)s � ë(rs) � r(ës);
(5) (r � s)t � rt � st;
(6) r(s � t) � rs � rt;
(7) r0 � 0r � 0.
Proof (1) It follows immediately from the de®nition of rs that
rs 2 FG.
(2) Let
r �Xg2G
ë g g, s �Xg2G
ì g g, t �Xg2G
í g g,
(ë g, ì g, í g 2 F). Then
(rs)t �X
g,h,k2G
ë gìhík(gh)k
�X
g,h,k2G
ë gìhík g(hk)
� r(st):
We leave the proofs of the other equations as easy exercises. j
Group algebras 55
In fact, any vector space equipped with a multiplication satisfying
properties (1)±(7) of Proposition 6.4 is called an algebra. We shall be
concerned only with group algebras, but it is worth pointing out that
the axioms for an algebra mean that it is both a vector space and a
ring.
The regular FG-module
We now use the group algebra to de®ne an important FG-module.
Let V � FG, so that V is a vector space of dimension n over F,
where n � |G|. For all u, v 2 V, ë 2 F and g, h 2 G, we have
vg 2 V ,
v(gh) � (vg)h,
v1 � v,
(ëv)g � ë(vg),
(u� v)g � ug � vg,
by parts (1), (2), (3), (4) and (5) of Proposition 6.4, respectively.
Therefore V is an FG-module.
6.5 De®nition
Let G be a ®nite group and F be R or C. The vector space FG, with
the natural multiplication vg (v 2 FG, g 2 G), is called the regular
FG-module.
The representation g! [g]B obtained by taking B to be the natural
basis of FG is called the regular representation of G over F.
Note that the regular FG-module has dimension equal to |G|.
6.6 Proposition
The regular FG-module is faithful.
Proof Suppose that g 2 G and vg � v for all v 2 FG. Then 1g � 1, so
g � 1, and the result follows. j
6.7 Example
Let G � C3 � ka: a3 � el. The elements of FG have the form
56 Representations and characters of groups
ë1e� ë2a� ë3a2 (ëi 2 F):
We have
(ë1e� ë2a� ë3a2)e � ë1e� ë2a� ë3a2,
(ë1e� ë2a� ë3a2)a � ë3e� ë1a� ë2a2,
(ë1e� ë2a� ë3a2)a2 � ë2e� ë3a� ë1a2:
By taking matrices relative to the basis e, a, a2 of FG, we obtain the
regular representation of G:
e!1 0 0
0 1 0
0 0 1
0@ 1A, a!0 1 0
0 0 1
1 0 0
0@ 1A, a2 !0 0 1
1 0 0
0 1 0
0@ 1A:FG acts on an FG-module
You will remember that an FG-module is a vector space over F,
together with a multiplication vg for v 2 V and g 2 G (and the multi-
plication satis®es various axioms). Now, it is sometimes helpful to
extend the de®nition of the multiplication so that we have an element
vr of V for all elements r in the group algebra FG. This is done in
the following natural way.
6.8 De®nition
Suppose that V is an FG-module, and that v 2 V and r 2 FG; say
r �P g2G ì g g (ì g 2 F). De®ne vr by
vr �Xg2G
ì g(vg):
6.9 Examples
(1) Let V be the permutation module for S4, as described in Example
4.9. If
r � ë(1 2)� ì(1 3 4) (ë, ì 2 F)
then
v1 r � ëv1(1 2)� ìv1(1 3 4) � ëv2 � ìv3,
v2 r � ëv1 � ìv2,
(2v1 � v2)r � ëv1 � (2ë� ì)v2 � 2ìv3:
Group algebras 57
(2) If V is the regular FG-module, then for all v 2 V and r 2 FG, the
element vr is simply the product of v and r as elements of the group
algebra, given by De®nition 6.3.
Compare the next result with Proposition 6.4.
6.10 Proposition
Suppose that V is an FG-module. Then the following properties hold
for all u, v 2 V, all ë 2 F and all r, s 2 FG:
(1) vr 2 V;
(2) v(rs) � (vr)s;
(3) v1 � v;
(4) (ëv)r � ë(vr) � v(ër);
(5) (u � v)r � ur � vr;
(6) v(r � s) � vr � vs;
(7) v0 � 0r � 0.
Proof All parts except (2) are straightforward, and we leave them to
you. We shall give a proof of part (2), assuming the other parts.
Let v 2 V, and let r, s 2 FG with
r �Xg2G
ë g g, s �Xh2G
ìh h:
Then
v(rs) � vXg,h
ë gìh(gh)
!
�Xg,h
ë gìh(v(gh)) by (4) and (6)
�Xg,h
ë gìh((vg)h)
�X
g
ë g(vg)
! Xh
ìh h
!by (4), (5), (6)
� (vr)s: j
58 Representations and characters of groups
Summary of Chapter 6
1. The group algebra FG of G over F consists of all linear combina-
tions of elements of G, and has a natural multiplication de®ned
on it.
2. The vector space FG, with the natural multiplication vg (v 2 FG,
g 2 G) is the regular FG-module.
3. The regular FG-module is faithful.
Exercises for Chapter 6
1. Suppose that G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l.(a) Let x and y be the following elements of CG:
x � a� 2a2, y � b� abÿ a2:
Calculate xy, yx and x2.
(b) Let z � b � a2b. Show that zg � gz for all g in G. Deduce that
zr � rz for all r in CG.
2. Work out matrices for the regular representation of C2 3 C2 over F.
3. Let G � C2. For r and s in CG, does rs � 0 imply that r � 0 or
s � 0?
4. Assume that G is a ®nite group, say G � {g1, . . . , gn}, and write c
for the elementP
ni�1 gi of CG.
(a) Prove that ch � hc � c for all h in G.
(b) Deduce that c2 � |G|c.
(c) Let W: CG! CG be the linear transformation sending v to vc
for all v in CG. What is the matrix [W]B , where B is the basis
g1, . . . , gn of CG?
5. If V is an FG-module, prove from the de®nition that
0r � 0 for all r 2 FG, and
v0 � 0 for all v 2 V ,
where the symbol 0 is used for the zero elements of V and FG.
Show that for every ®nite group G, with |G| . 1, there exists an
FG-module V and elements v 2 V, r 2 FG such that vr � 0, but
neither v nor r is 0.
Group algebras 59
6. Suppose that G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l, and let
ù � e2ði=3. Prove that the 2-dimensional subspace W of CG, de®ned
by
W � sp (1� ù2a� ùa2, b� ù2ab� ùa2b),
is an irreducible CG-submodule of the regular CG-module.
60 Representations and characters of groups
61
7
FG-homomorphisms
For groups and vector spaces, the `structure-preserving' functions are,
respectively, group homomorphisms and linear transformations. The
analogous functions for FG-modules are called FG-homomorphisms,
and we introduce these in this chapter.
FG-homomorphisms
7.1 De®nition
Let V and W be FG-modules. A function W: V! W is said to be an
FG-homomorphism if W is a linear transformation and
(vg)W � (vW)g for all v 2 V , g 2 G:
In other words, if W sends v to w then it sends vg to wg.
Note that if G is a ®nite group and W: V! W is an FG-homomorph-
ism, then for all v 2 V and r �P g2G ë g g 2 FG, we have
(vr)W � (vW)r
since
(vr)W �Xg2G
ë g(vg)W �Xg2G
ë g(vW)g � (vW)r:
The next result shows that FG-homomorphisms give rise to FG-
submodules in a natural way.
7.2 Proposition
Let V and W be FG-modules and let W: V! W be an FG-homomorph-
ism. Then Ker W is an FG-submodule of V, and Im W is an FG-
submodule of W.
Proof First note that Ker W is a subspace of V and Im W is a subspace
of W, since W is a linear transformation.
Let v 2 Ker W and g 2 G. Then
(vg)W � (vW)g � 0g � 0,
so vg 2 Ker W. Therefore Ker W is an FG-submodule of V.
Now let w 2 Im W, so that w � vW for some v 2 V. For all g 2 G,
wg � (vW)g � (vg)W 2 Im W,
and so Im W is an FG-submodule of W. j
7.3 Examples
(1) If W: V! W is de®ned by vW � 0 for all v 2 V, then W is an FG-
homomorphism, and Ker W � V, Im W � {0}.
(2) Let ë 2 F, and de®ne W: V! V by vW � ëv for all v 2 V. Then W is
an FG-homomorphism. Provided ë 6� 0, we have Ker W � {0}, Im W � V.
(3) Suppose that G is a subgroup of Sn. Let V � sp (v1, . . . , vn) be the
permutation module for G over F (see De®nition 4.10), and let
W � sp (w) be the trivial FG-module (see De®nition 4.8). We construct
an FG-homomorphism W from V to W. De®ne
W:Xn
i�1
ëivi !Xn
i�1
ëi
!w (ëi 2 F):
Thus viW � w for all i. Then W is a linear transformation, and for all
v �Pëivi 2 V and all g 2 G, we have
(vg)W �X
ëivig
!W �
Xëi
!w,
and
(vW)g �X
ëi
!wg �
Xëi
!w:
Therefore W is an FG-homomorphism. Here,
Ker W �Xn
i�1
ëivi:Xn
i�1
ëi � 0
( ),
Im W � W :
62 Representations and characters of groups
By Proposition 7.2, Ker W is an FG-submodule of the permutation
module V.
Isomorphic FG-modules
7.4 De®nition
Let V and W be FG-modules. We call a function W: V! W an FG-
isomorphism if W is an FG-homomorphism and W is invertible. If there
is such an FG-isomorphism, then we say that V and W are isomorphic
FG-modules and write V � W.
In the next result, we check that if V � W then W � V.
7.5 Proposition
If W: V! W is an FG-isomorphism, then the inverse Wÿ1: W! V is
also an FG-isomorphism.
Proof Certainly Wÿ1 is an invertible linear transformation, so we need
only show that Wÿ1 is an FG-homomorphism. For w 2 W and g 2 G,
((wWÿ1)g)W � ((wWÿ1)W)g as W is an FG-homomorphism
� wg
� ((wg)Wÿ1)W:
Hence (wWÿ1)g � (wg)Wÿ1, as required. j
Suppose that W: V! W is an FG-isomorphism. Then we may use Wand Wÿ1 to translate back and forth between the isomorphic FG-
modules V and W, and prove that V and W share the same structural
properties. We list some examples below:
(1) dim V � dim W (since v1, . . . , vn is a basis of V if and only if
v1W, . . . , vnW is a basis of W);
(2) V is irreducible if and only if W is irreducible (since X is an FG-
submodule of V if and only if XW is an FG-submodule of W);
(3) V contains a trivial FG-submodule if and only if W contains a
trivial FG-submodule (since X is a trivial FG-submodule of V if
and only if XW is a trivial FG-submodule of W).
FG-homomorphisms 63
Just as we often regard isomorphic groups as being identical, we
frequently disdain to distinguish between isomorphic FG-modules. For
the moment, though, we continue simply to emphasize the similarity
between isomorphic FG-modules. In the next result, we show that iso-
morphic FG-modules correspond to equivalent representations.
7.6 Theorem
Suppose that V is an FG-module with basis B , and W is an FG-
module with basis B 9. Then V and W are isomorphic if and only if the
representations
r: g ! [g]B and ó : g ! [g]B 9
are equivalent.
Proof We ®rst establish the following fact:
(7.7) The FG-modules V and W are isomorphic if and only
if there are a basis B 1 of V and a basis B 2 of W
such that
[g]B 1� [g]B 2
for all g 2 G:
To see this, suppose ®rst that W is an FG-isomorphism from V to W,
and let v1, . . . , vn be a basis B 1 of V; then v1W, . . . , vnW is a basis
B 2 of W. Let g 2 G. Since (vig)W � (viW)g for each i, it follows that
[g]B 1� [g]B 2
.
Conversely, suppose that v1, . . . , vn is a basis B 1 of V and w1,
. . . , wn is a basis B 2 of W such that [g]B 1� [g]B 2
for all g 2 G.
Let W be the invertible linear transformation from V to W for which
viW � wi for all i. Let g 2 G. Since [g]B 1� [g]B 2
, we deduce that
(vig)W � (viW)g for all i, and hence W is an FG-isomorphism. This
completes the proof of (7.7).
Now assume that V and W are isomorphic FG-modules. By (7.7),
there are a basis B 1 of V and a basis B 2 of W such that
[g]B 1� [g]B 2
for all g 2 G. De®ne a representation ö of G by
ö: g! [g]B 1. Then by Theorem 4.12(1), ö is equivalent to both r and
ó. Hence r and ó are equivalent.
Conversely, suppose that r and ó are equivalent. Then by Theorem
4.12(2), there is a basis B 0 of V such that gó � [g]B 0 for all g 2 G;
64 Representations and characters of groups
that is, [g]B 9 � [g]B 0 for all g 2 G. Therefore V and W are isomorphic
FG-modules, by (7.7). j
7.8 Example
Let G � ka: a3 � 1l, a cyclic group of order 3, and let W denote the
regular FG-module. Then 1, a, a2 is a basis of W; call it B 9. We have
[1]B 9 �1 0 0
0 1 0
0 0 1
0BB@1CCA, [a]B 9 �
0 1 0
0 0 1
1 0 0
0BB@1CCA,
[a2]B 9 �0 0 1
1 0 0
0 1 0
0BB@1CCA:
Compare the FG-module V de®ned in Example 4.11, with basis
v1, v2, v3 such that
v1a � v2, v2a � v3, v3a � v1:
Writing B for the basis v1, v2, v3 of V, we have
[g]B � [g]B 9 for all g 2 G:
According to (7.7), the FG-modules V and W are therefore isomorphic.
Indeed, the function
W: ë1v1 � ë2v2 � ë3v3 ! ë11� ë2a� ë3a2 (ëi 2 F)
is an FG-isomorphism from V to W.
7.9 Example
Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. In Example 3.4(1) we
encountered two equivalent representations r and ó of G, where
ar � 0 1
ÿ1 0
� �, br � 1 0
0 ÿ1
� �and
aó � i 0
0 ÿi
� �, bó � 0 1
1 0
� �:
FG-homomorphisms 65
Let V be the CG-module with basis v1, v2 for which
v1a � v2, v1b � v1,
v2a � ÿv1, v2b � ÿv2
(see Example 4.5(1)), and, in a similar way, let W be the CG-module
with basis w1, w2 for which
w1a � iw1, w1b � w2,
w2a � ÿiw2, w2b � w1
Thus, if we write B for the basis v1, v2 of V and B 9 for the basis w1,
w2 of W, then for all g 2 G we have
r: g ! [g]B and ó : g ! [g]B 9:
According to Theorem 7.6, the CG-modules V and W are isomorphic,
since r and ó are equivalent. To verify this directly, let W: V! W be
the invertible linear transformation such that
W: v1 ! w1 � w2,
v2 ! iw1 ÿ iw2:
Then (v ja)W � (v jW)a and (v jb)W � (v jW)b for j � 1, 2, and hence W is
a CG-isomorphism from V to W. (Compare Example 3.4(1).)
Direct sums
We conclude the chapter with a discussion of direct sums of FG-
modules, and we show that these give rise to FG-homomorphisms.
Let V be an FG-module, and suppose that
V � U � W ,
where U and W are FG-submodules of V. Let u1, . . . , um be a basis
B 1 of U, and w1, . . . , wn be a basis B 2 of W. Then by (2.9),
u1, . . . , um, w1, . . . , wn is a basis B of V, and for g 2 G,
[g]B �[g]B 1
0
0 [g]B 2
0@ 1A:More generally, if V � U1 � . . . � Ur, a direct sum of FG-submodules
Ui, and B i is a basis of Ui, then we can amalgamate B 1, . . . , B r to
66 Representations and characters of groups
obtain a basis B of V, and for g 2 G,
[g]B �[g]B 1
. ..
[g]B r
0B@1CA:(7:10)
The next result shows that direct sums give rise naturally to FG-
homomorphisms.
7.11 Proposition
Let V be an FG-module, and suppose that
V � U1 � : : :� Ur
where each Ui is an FG-submodule of V. For v 2 V, we have
v � u1 � . . . � ur for unique vectors ui 2 Ui, and we de®ne ði: V! V
(1 < i < r) by setting
vði � ui:
Then each ði is an FG-homomorphism, and is also a projection of V.
Proof Clearly ði is a linear transformation; and ði is an FG-homo-
morphism, since for v 2 V with v � u1 � . . . � ur (u j 2 Uj for all j),
and g 2 G, we have
(vg)ði � (u1 g � : : :� urg)ði � uig � (vði)g:
Also,
vð2i � uiði � ui � vði,
so ð2i � ði. Thus ði is a projection (see De®nition 2.30). j
We now present a technical result concerning sums of irreducible
FG-modules which will be used at a later stage.
7.12 Proposition
Let V be an FG-module, and suppose that
V � U1 � : : :� Ur,
where each Ui is an irreducible FG-submodule of V. Then V is a direct
sum of some of the FG-submodules Ui.
0
0
FG-homomorphisms 67
Proof The idea is to choose as many as we can of the FG-submodules
U1, . . . , Ur so that the sum of our chosen FG-submodules is direct.
To this end, choose a subset Y � {W1, . . . , Ws} of {U1, . . . , Ur} which
has the properties that
W1 � : : :� Ws is direct (i:e: equal to W1 � : : :� Ws), but
W1 � : : :� Ws � Ui is not direct, if Ui =2 Y :
Let
W � W1 � : : :� Ws:
We claim that Ui # W for all i. If Ui 2 Y this is clear, so assume that
Ui =2 Y. Then W � Ui is not a direct sum, so W \ Ui 6� {0}. But W \ Ui
is an FG-submodule of Ui, and Ui is irreducible; therefore
W \ Ui � Ui, and so Ui # W, as claimed.
Since Ui � W for all i with 1 < i < r, we have V � W �W1� . . . �Ws, as required. j
Finally, we remark that if V1, . . . , Vr are FG-modules, then we can
make the external direct sum V1� . . . �Vr (see Chapter 2) into an FG-
module by de®ning
(v1, : : : , vr)g � (v1 g, : : : , vrg)
for all vi 2 Vi (1 < i < r) and all g 2 G.
Summary of Chapter 7
1. If V and W are FG-modules and W: V! W is a linear transformation
which satis®es
(vg)W � (vW)g
for all v 2 V, g 2 G, then W is an FG-homomorphism.
2. Kernels and images of FG-homomorphisms are FG-modules.
3. Isomorphic FG-modules correspond to equivalent representations.
Exercises for Chapter 7
1. Let U, V and W be FG-modules, and let W: U! V and ö: V! W
be FG-homomorphisms. Prove that Wö: U! W is an FG-homo-
morphism.
68 Representations and characters of groups
2. Let G be the subgroup of S5 which is generated by (1 2 3 4 5).
Prove that the permutation module for G over F is isomorphic to
the regular FG-module.
3. Assume that V is an FG-module. Prove that the subset
V0 � fv 2 V : vg � v for all g 2 Ggis an FG-submodule of V. Show that the function
W: v!Xg2G
vg (v 2 V )
is an FG-homomorphism from V to V0. Is it necessarily surjective?
4. Suppose that V and W are isomorphic FG-modules. De®ne the FG-
submodules V0 and W0 of V and W as in Exercise 3. Prove that V0
and W0 are isomorphic FG-modules.
5. Let G be the subgroup of S4 which is generated by (1 2) and (3 4).
Is the permutation module for G over F isomorphic to the regular
FG-module?
6. Let G � C2 � kx: x2 � 1l.(a) Show that the function
W: á1� âx! (áÿ â)(1ÿ x) (á, â 2 F)
is an FG-homomorphism from the regular FG-module to itself.
(b) Prove that W2 � 2W.
(c) Find a basis B of FG such that
[W]B � 2 0
0 0
� �:
FG-homomorphisms 69
70
8
Maschke's Theorem
We now come to our ®rst major result in representation theory, namely
Maschke's Theorem. A consequence of this theorem is that every FG-
module is a direct sum of irreducible FG-submodules, where as usual
F � R or C. (The assumption on F is important ± see Example 8.2(2)
below.) This essentially reduces representation theory to the study of
irreducible FG-modules.
Maschke's Theorem
8.1 Maschke's Theorem
Let G be a ®nite group, let F be R or C, and let V be an FG-module.
If U is an FG-submodule of V, then there is an FG-submodule W of V
such that
V � U � W :
Before proving Maschke's Theorem, we illustrate it with some
examples.
8.2 Examples
(1) Let G � S3 and let V � sp (v1, v2, v3) be the permutation module
for G over F (see De®nition 4.10). Put
u � v1 � v2 � v3 and U � sp (u):
Then U is an FG-submodule of V, since ug � u for all g 2 G.
There are many subspaces W of V such that V � U � W, for instance
sp (v2, v3) and sp (v1, v2 ÿ 2v3). But there is, in fact, only one FG-
submodule W of V with V � U � W. We shall ®nd this W in an
example after proving Maschke's Theorem (but you may like to look
for it yourself now).
(2) The conclusion of Maschke's Theorem can fail if F is not R or C.
For example, let p be a prime number, let G � C p � ka: a p � 1l, and
take F to be the ®eld of integers modulo p. Check that the function
a j ! 1 0
j 1
� �( j � 0, 1, : : : , pÿ 1)
is a representation from G to GL (2, F). The corresponding FG-module
is V � sp (v1, v2), where, for 0 < j < p ÿ 1,
v1a j � v1,
v2a j � jv1 � v2:
Clearly, U � sp (v1) is an FG-submodule of V. But there is no FG-
submodule W such that V � U � W, since U is the only 1-dimensional
FG-submodule of V, as can easily be seen.
Proof of Maschke's Theorem 8.1 We are given U, an FG-submodule of
the FG-module V. Choose any subspace W0 of V such that
V � U � W0:
(There are many choices for W0 ± simply take a basis v1, . . . , vm
of U, extend it to a basis v1, . . . , vn of V, and let W0 �sp (vm�1, . . . , vn).)
For v 2 V, we have v � u � w for unique vectors u 2 U and w 2 W0,
and we de®ne ö: V! V by setting vö � u. By Proposition 2.29, ö is
a projection of V with kernel W0 and image U.
We aim to modify the projection ö to create an FG-homomorphism
from V to V with image U. To this end, de®ne W: V! V by
vW � 1
jGjXg2G
vgögÿ1 (v 2 V ):(8:3)
It is clear that W is an endomorphism of V and Im W # U.
We show ®rst that W is an FG-homomorphism. For v 2 V and x 2 G,
(vx)W � 1
jGjXg2G
(vx)gögÿ1:
Maschke's Theorem 71
As g runs over the elements of G, so does h � xg. Hence
(vx)W � 1
jGjXh2G
vhöhÿ1x
� 1
jGjXh2G
vhöhÿ1
!x
� (vW)x:
Thus W is an FG-homomorphism.
Next, we prove that W2 � W. First note that for u 2 U, g 2 G, we
have ug 2 U, and so (ug)ö � ug. Using this,
uW � 1
jGjXg2G
ugögÿ1 � 1
jGjXg2G
(ug)gÿ1 � 1
jGjXg2G
u � u:(8:4)
Now let v 2 V. Then vW 2 U, so by (8.4) we have (vW)W � vW.
Consequently W2 � W, as claimed.
We have now established that W: V! V is a projection and an FG-
homomorphism. Moreover, (8.4) shows that Im W � U. Let W � Ker W.
Then W is an FG-submodule of V by Proposition 7.2, and V � U � W
by Proposition 2.32.
This completes the proof of Maschke's Theorem. j
8.5 Example
Let G � S3 and let V � sp (v1, v2, v3) be the permutation module, with
submodule U � sp (v1 � v2 � v3), as in Example 8.2(1). We use the
proof of Maschke's Theorem to ®nd an FG-submodule W of V such
that V � U � W.
First, let W0 � sp (v1, v2). Then V � U � W0 (but of course W0 is not
an FG-submodule). The projection ö onto U is given by
ö: v1 ! 0, v2 ! 0, v3 ! v1 � v2 � v3:
Check now that the FG-homomorphism W given by (8.3) is
W: vi ! 13(v1 � v2 � v3) (i � 1, 2, 3):
The required FG-submodule W is then Ker W, so
W � sp (v1 ÿ v2, v2 ÿ v3):
(In fact, W � Pëivi:
Pëi � 0
� , the FG-submodule constructed in
Example 7.3(3).)
72 Representations and characters of groups
Note that if B is the basis v1 � v2 � v3, v1, v2 of V, then for all
g 2 G, the matrix [g]B has the form
[g]B �j 0 0
j j j
j j j
0@ 1A:The zeros re¯ect the fact that U is an FG-submodule of V (see (5.4)).
If instead we use v1 � v2 � v3, v1 ÿ v2, v2 ÿ v3 as a basis B 9, then
we get
[g]B 9 �j 0 0
0 j j
0 j j
0@ 1A,
because sp (v1 ÿ v2, v2 ÿ v3) is also an FG-submodule of V.
This example illustrates the matrix version of Maschke's Theorem:
for an arbitrary ®nite group G, if we can choose a basis B of an FG-
module V such that [g]B has the form
� 0
� �
0@ 1Afor all g 2 G (see (5.4)), then we can ®nd a basis B 9 such that [g]B 9
has the form
� 0
0 �
0@ 1Afor all g 2 G.
To put this another way, suppose that r is a reducible representation
of a ®nite group G over F of degree n. Then we know that r is
equivalent to a representation of the form
g !Xg 0
Y g Z g
0@ 1A (g 2 G),
for some matrices Xg, Yg, Zg, where Xg is k 3 k with 0 , k , n.
Maschke's Theorem 73
Maschke's Theorem asserts further that r is equivalent to a represent-
ation of the form
g !Ag 0
0 Bg
0@ 1A,
where A g is also a k 3 k matrix.
Consequences of Maschke's Theorem
We now use Maschke's Theorem to show that every non-zero FG-
module is a direct sum of irreducible FG-submodules. (By an irreduci-
ble FG-submodule, we simply mean an FG-submodule which is an
irreducible FG-module.)
8.6 De®nition
An FG-module V is said to be completely reducible if V �U1� . . . �Ur, where each Ui is an irreducible FG-submodule of V.
8.7 Theorem
If G is a ®nite group and F � R or C, then every non-zero FG-module
is completely reducible.
Proof Let V be a non-zero FG-module. The proof goes by induction
on dim V. The result is true if dim V � 1, since V is irreducible in this
case.
If V is irreducible then the result holds, so suppose that V is
reducible. Then V has an FG-submodule U not equal to {0} or V.
By Maschke's Theorem, there is an FG-submodule W such that
V � U � W. Since dim U , dim V and dim W , dim V, we have, by
induction,
U � U1 � : : :� Ur, W � W1 � : : :� Ws,
where each Ui and Wj is an irreducible FG-module. Then by (2.10),
V � U1 � : : :� Ur � W1 � : : :� Ws,
a direct sum of irreducible FG-modules. j
Another useful consequence of Maschke's Theorem is the next
proposition.
74 Representations and characters of groups
8.8 Proposition
Let V be an FG-module, where F � R or C and G is a ®nite group.
Suppose that U is an FG-submodule of V. Then there exists a surjective
FG-homomorphism from V onto U.
Proof By Maschke's Theorem, there is an FG-submodule W of V such
that V � U � W. Then the function ð: V! U which is de®ned by
ð: u� w! u (u 2 U , w 2 W )
is an FG-homomorphism onto U, by Proposition 7.11. j
Theorem 8.7 tells us that every non-zero FG-module is a direct sum
of irreducible FG-modules. Thus, in order to understand FG-modules,
we may concentrate upon the irreducible FG-modules. We begin our
study of these in the next chapter.
Summary of Chapter 8
Assume that G is a ®nite group and F � R or C.
1. Maschke's Theorem says that for every FG-submodule U of an FG-
module V, there is an FG-submodule W with
V � U � W :
2. Every non-zero FG-module V is a direct sum of irreducible FG-
modules:
V � U1 � : : :� Ur:
Exercises for Chapter 8
1. Let G � kx: x3 � 1l � C3, and let V be the 2-dimensional CG-
module with basis v1, v2, where
v1x � v2, v2x � ÿv1 ÿ v2:
(This is a CG-module, by Exercise 3.2.)
Express V as a direct sum of irreducible CG-submodules.
2. If G � C2 3 C2, express the group algebra RG as a direct sum of
1-dimensional RG-submodules.
Maschke's Theorem 75
3. Find a group G, a CG-module V and a CG-homomorphism
W: V! V such that V 6� Ker W � Im W.
4. Let G be a ®nite group and let r: G! GL (2, C) be a representation
of G. Suppose that there are elements g, h in G such that the
matrices gr and hr do not commute. Prove that r is irreducible.
(You may care to revisit Example 5.5(2) and Exercises 5.1, 5.3, 5.4,
6.6 in the light of this result.)
5. Suppose that G is the in®nite group
1 0
n 1
� �: n 2 Z
� �and let V be the CG-module C2, with the natural multiplication by
elements of G (so that for v 2 V, g 2 G, the vector vg is just the
product of the row vector v with the matrix g).
Show that V is not completely reducible.
(This shows that Maschke's Theorem fails for in®nite groups ±
compare Example 8.2(2).)
6. An alternative proof of Maschke's Theorem for CG-modules.
Let V be a CG-module with basis v1, . . . , vn and suppose that U
is a CG-submodule of V. De®ne a complex inner product ( , ) on
V as follows (see (14.2) for the de®nition of a complex inner
product): for ëi, ì j 2 C,Xn
i�1
ëivi,Xn
j�1
ì jv j
!�Xn
i�1
ëiìi:
De®ne another complex inner product [ , ] on V by
[u, v] �Xx2G
(ux, vx) (u, v 2 V ):
(1) Verify that [ , ] is a complex inner product, which satis®es
[ug, vg] � [u, v] for all u, v 2 V and g 2 G:
(2) Suppose that U is a CG-submodule of V, and de®ne
U? � fv 2 V : [u, v] � 0 for all u 2 Ug:Show that U? is a CG-submodule of V.
76 Representations and characters of groups
(3) Deduce Maschke's Theorem. (Hint: it is a standard property
of complex inner products that V � U � U? for all subspaces U
of V.)
7. Prove that for every ®nite simple group G, there exists a faithful
irreducible CG-module.
Maschke's Theorem 77
78
9
Schur's Lemma
Schur's Lemma is a basic result concerning irreducible modules.
Though simple in both statement and proof, Schur's Lemma is funda-
mental to representation theory, and we give an immediate application
by determining all the irreducible representations of ®nite abelian
groups.
Schur's Lemma concerns CG-modules rather than RG-modules, and
since much of the ensuing theory depends on it, we shall deal with
CG-modules for the remainder of the book (except in Chapter 23).
Throughout, G denotes a ®nite group.
Schur's Lemma
9.1 Schur's Lemma
Let V and W be irreducible CG-modules.
(1) If W: V! W is a CG-homomorphism, then either W is a CG-
isomorphism, or vW � 0 for all v 2 V.
(2) If W: V! V is a CG-isomorphism, then W is a scalar multiple of
the identity endomorphism 1V .
Proof (1) Suppose that vW 6� 0 for some v 2 V. Then Im W 6� {0}. As
Im W is a CG-submodule of W by Proposition 7.2, and W is irreducible,
we have Im W � W. Also by Proposition 7.2, Ker W is a CG-submodule
of V; as Ker W 6� V and V is irreducible, Ker W � {0}. Thus W is
invertible, and hence is a CG-isomorphism.
(2) By (2.26), the endomorphism W has an eigenvalue ë 2 C, and so
Ker (W ÿ ë1V ) 6� {0}. Thus Ker (W ÿ ë1V ) is a non-zero CG-submodule
of V. Since V is irreducible, Ker (W ÿ ë1V ) � V. Therefore
v(Wÿ ë1V ) � 0 for all v 2 V :
That is, W � ë1V , as required. j
Part (2) of Schur's Lemma has the following converse.
9.2 Proposition
Let V be a non-zero CG-module, and suppose that every CG-homo-
morphism from V to V is a scalar multiple of 1V . Then V is
irreducible.
Proof Suppose that V is reducible, so that V has a CG-submodule U
not equal to {0} or V. By Maschke's Theorem, there is a CG-
submodule W of V such that
V � U � W :
Then the projection ð: V! V de®ned by (u � w)ð � u for all u 2 U,
w 2 W is a CG-homomorphism (see Proposition 7.11), and is not a
scalar multiple of 1V , which is a contradiction. Hence V is
irreducible. j
We next interpret Schur's Lemma and its converse in terms of
representations.
9.3 Corollary
Let r: G! GL (n, C) be a representation of G. Then r is irreducible if
and only if every n 3 n matrix A which satis®es
(gr)A � A(gr) for all g 2 G
has the form A � ëIn with ë 2 C.
Proof As in Theorem 4.4(1), regard Cn as a CG-module by de®ning
vg � v(gr) for all v 2 Cn, g 2 G.
Let A be an n 3 n matrix over C. The endomorphism v! vA of
Cn is a CG-homomorphism if and only if
(vg)A � (vA)g for all v 2 Cn, g 2 G;
Schur's Lemma 79
that is, if and only if
(gr)A � A(gr) for all g 2 G:
The result now follows from Schur's Lemma 9.1 and Proposition 9.2.
j
9.4 Examples
(1) Let G � C3 � ka: a3 � 1l, and let r: G! GL (2, C) be the repre-
sentation for which
ar � 0 1
ÿ1 ÿ1
� �(see Exercise 3.2). Since the matrix
0 1
ÿ1 ÿ1
� �commutes with all gr (g 2 G), Corollary 9.3 implies that r is
reducible.
(2) Let G � D10 � ka, b: a5 � b2 � 1, bÿ1ab � aÿ1l, and let ù � e2ði=5.
Check that there is a representation r: G! GL (2, C) for which
ar � ù 0
0 ùÿ1
� �, br � 0 1
1 0
� �:
Assume that the matrix
A � á âã ä
� �commutes with both ar and br. The fact that (ar)A � A(ar) forces
â � ã � 0; and then (br)A � A(br) gives á � ä. Hence
A � á 0
0 á
� �� áI :
Consequently r is irreducible, by Corollary 9.3.
80 Representations and characters of groups
Representation theory of ®nite abelian groups
Let G be a ®nite abelian group, and let V be an irreducible CG-
module. Pick x 2 G. Since G is abelian,
vgx � vxg for all g 2 G,
and hence the endomorphism v! vx of V is a CG-homomorphism. By
Schur's Lemma 9.1(2), this endomorphism is a scalar multiple of the
identity 1V , say ëx1V . Thus
vx � ëxv for all v 2 V :
This implies that every subspace of V is a CG-submodule. As V is
irreducible, we deduce that dim V � 1. Thus we have proved
9.5 Proposition
If G is a ®nite abelian group, then every irreducible CG-module has
dimension 1.
The next result is a major structure theorem for ®nite abelian groups.
We shall not prove it here, but refer you to Chapter 9 of the book of
J. B. Fraleigh listed in the Bibliography.
9.6 Theorem
Every ®nite abelian group is isomorphic to a direct product of cyclic
groups.
We shall determine the irreducible representations of all direct
products
Cn13 Cn2
3 : : :3 Cnr
where n1, . . . , nr are positive integers. By Theorem 9.6, this covers
the irreducible representations of all ®nite abelian groups.
Let G � Cn13 . . . 3 Cn r
, and for 1 < i < r, let ci be a generator
for Cni. Write
gi � (1, : : : , ci, : : : , 1) (ci in ith position):
Then
G � hg1, : : : , g ri, with g ni
i � 1 and gigj � gjgi for all i, j:
Now let r: G! GL (n, C) be an irreducible representation of G
Schur's Lemma 81
over C. Then n � 1 by Proposition 9.5, so for 1 < i < r, there exists
ëi 2 C such that
gir � (ëi)
(where of course (ëi) is a 1 3 1 matrix). As gi has order ni, we have
ëni
i � 1; that is, ëi is an nith root of unity. Also, the values ë1, . . . , ër
determine r, since for g 2 G, we have g � gi11 . . . gir
r for some
integers i1, . . . , ir, and then
gr � (gi11 : : :g
irr )r � (ëi1
1 : : :ëi r
r ):(9:7)
For a representation r of G satisfying (9.7) for all i1, . . . , ir, write
r � rë1,:::,ë r:
Conversely, given any nith roots of unity ëi (1 < i < r), the function
gi11 : : :g
irr ! (ëi1
1 : : :ëi r
r )
is a representation of G. There are n1 n2 . . . nr such representations,
and no two of them are equivalent.
We have proved the following theorem.
9.8 Theorem
Let G be the abelian group Cn13 . . . 3 Cn r
. The representations
rë1,:::,ë rof G constructed above are irreducible and have degree 1.
There are |G| of these representations, and every irreducible represent-
ation of G over C is equivalent to precisely one of them.
9.9 Examples
(1) Let G � Cn � ka: an � 1l, and put ù � e2ði=n. The n irreducible
representations of G over C are rù j (0 < j < n ÿ 1), where
akrù j � (ù jk) (0 < k < nÿ 1):
(2) The four irreducible CG-modules for G � C2 3 C2 � kg1, g2l are
V1, V2, V3, V4, where Vi is a 1-dimensional space with basis vi
(i � 1, 2, 3, 4) and
v1 g1 � v1, v1 g2 � v2;
v2 g1 � v2, v2 g2 � ÿv2;
v3 g1 � ÿv3, v3 g2 � v3;
v4 g1 � ÿv4, v4 g2 � ÿv4:
82 Representations and characters of groups
Diagonalization
Let H � kgl be a cyclic group of order n, and let V be a non-zero
CH-module. By Theorem 8.7,
V � U1 � : : :� Ur,
a direct sum of irreducible CH-submodules Ui of V. Each Ui has
dimension 1, by Proposition 9.5; let ui be a vector spanning Ui. Put
ù � e2ði=n. Then for each i, there exists an integer mi such that
uig � ùmi ui:
Thus if B is the basis u1, . . . , ur of V, then
[g]B �ùm1
. ..
ùmr
0B@1CA:(9:10)
The following useful result is an immediate consequence of this.
9.11 Proposition
Let G be a ®nite group and V a CG-module. If g 2 G, then there is a
basis B of V such that the matrix [g]B is diagonal. If g has order n,
then the entries on the diagonal of [g]B are nth roots of unity.
Proof Let H � kgl. As V is also a CH-module, the result follows from
(9.10). j
Some further applications of Schur's Lemma
Our next application concerns an important subspace of the group
algebra CG.
9.12 De®nition
Let G be a ®nite group. The centre of the group algebra CG, written
Z(CG), is de®ned by
Z(CG) � fz 2 CG: zr � rz for all r 2 CGg:Using (2.5), it is easy to check that Z(CG) is a subspace of CG.
For abelian groups G, the centre Z(CG) is the whole group algebra.
For arbitrary groups G, we shall see that Z(CG) plays a crucial role in
0
0
Schur's Lemma 83
the study of representations of G (for example, its dimension is equal
to the number of irreducible representations of G ± see Chapter 15).
9.13 Example
The elements 1 andP
g2G g lie in Z(CG). Indeed, if H is any normal
subgroup of G, then Xh2H
h 2 Z(CG):
To see this, write z �Ph2H h. Then for all g 2 G,
gÿ1zg �Xh2H
gÿ1 hg �Xh2H
h � z,
and so zg � gz. Consequently zr � rz for all r 2 CG.
For example, if G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l, then
{1}, kal and G are normal subgroups of G, so the elements
1, 1� a� a2 and 1� a� a2 � b� ab� a2b
lie in Z(CG). We shall see later that these elements in fact form a
basis of Z(CG).
We use Schur's Lemma to prove the following important property of
the elements of Z(CG).
9.14 Proposition
Let V be an irreducible CG-module, and let z 2 Z(CG). Then there
exists ë 2 C such that
vz � ëv for all v 2 V :
Proof For all r 2 CG and v 2 V, we have
vrz � vzr,
and hence the function v! vz is a CG-homomorphism from V to V.
By Schur's Lemma 9.1(2), this CG-homomorphism is equal to ë1V for
some ë 2 C, and the result follows. j
Some elements of the centre of CG are provided by the centre of G,
which we now de®ne.
84 Representations and characters of groups
9.15 De®nition
The centre of G, written Z(G), is de®ned by
Z(G) � fz 2 G: zg � gz for all g 2 Gg:Clearly Z(G) is a normal subgroup of G, and is a subset of Z(CG).
Although we have seen in Proposition 6.6 that for every ®nite group
G there is a faithful CG-module, it is not necessarily the case that
there is a faithful irreducible CG-module. Indeed, the following result
shows that the existence of a faithful irreducible CG-module imposes a
strong restriction on the structure of G.
9.16 Proposition
If there exists a faithful irreducible CG-module, then Z(G) is cyclic.
Proof Let V be a faithful irreducible CG-module. If z 2 Z(G) then z
lies in Z(CG), and hence by Proposition 9.14, there exists ëz 2 C such
that
vz � ëzv for all v 2 V :
Since V is faithful, the function
z! ëz (z 2 Z(G))
is an injective homomorphism from Z(G) into the multiplicative group
C� of non-zero complex numbers. Therefore Z(G) � {ëz: z 2 Z(G)},
which, being a ®nite subgroup of C�, is cyclic (see Exercise 1.7). j
We remark that the converse of Proposition 9.16 is false, since in
Exercise 25.6, we give an example of a group G such that Z(G) is
cyclic but there exists no faithful irreducible CG-module.
9.17 Example
If G is an abelian group, then G � Z(G), and so by Proposition 9.16,
there is no faithful irreducible CG-module unless G is cyclic. For
example, C2 3 C2 has no faithful irreducible representation (compare
Example 9.9(2)).
The irreducible representations of non-abelian groups are more
dif®cult to construct than those of abelian groups. In particular, they
Schur's Lemma 85
do not all have degree 1, as is shown by the following converse to
Proposition 9.5.
9.18 Proposition
Suppose that G is a ®nite group such that every irreducible CG-module
has dimension 1. Then G is abelian.
Proof By Theorem 8.7, we can write
CG � V1 � : : :� Vn,
where each Vi is an irreducible CG-submodule of the regular CG-
module CG. Then dim Vi � 1 for all i, since we are assuming that all
irreducible CG-modules have dimension 1. For 1 < i < n, let vi be a
vector spanning Vi. Then v1, . . . , vn is a basis of CG; call it B . For
all x, y 2 G, the matrices [x]B and [y]B are diagonal, and hence they
commute. Since the representation
g ! [g]B (g 2 G)
of G is faithful (see Proposition 6.6), we deduce that x and y
commute. Hence G is abelian, as required. j
Summary of Chapter 9
1. Schur's Lemma states that every CG-homomorphism between irredu-
cible CG-modules is either zero or a CG-isomorphism. Also, the
only CG-homomorphisms from an irreducible CG-module to itself
are scalar multiples of the identity.
2. The centre Z(CG) of the group algebra CG consists of those
elements which commute with all elements of CG. The elements of
Z(CG) act as scalar multiples of the identity on all irreducible CG-
modules.
3. All irreducible CG-modules for a ®nite abelian group G have
dimension 1, and there are precisely |G| of them.
Exercises for Chapter 9
1. Write down the irreducible representations over C of the groups C2,
C3 and C2 3 C2.
86 Representations and characters of groups
2. Let G � C4 3 C4.
(a) Find a non-trivial irreducible representation r of G such that
g2r � (1) for all g 2 G.
(b) Prove that there is no irreducible representation ó of G such
that gó � (ÿ1) for all elements g of order 2 in G.
3. Let G be the ®nite abelian group Cn13 . . . 3 Cn r
. Prove that G has
a faithful representation of degree r. Can G have a faithful
representation of degree less than r?
4. Suppose that G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. Check that
there is a representation r of G over C such that
ar � ÿ7 10
ÿ5 7
� �, br � ÿ5 6
ÿ4 5
� �:
Find all 2 3 2 matrices M such that M(gr) � (gr)M for all g 2 G.
Hence determine whether or not r is irreducible.
Do the same for the representation ó of G, where
aó � 5 ÿ6
4 ÿ5
� �, bó � ÿ5 6
ÿ4 5
� �:
5. Show that if V is an irreducible CG-module, then there exists ë 2 C
such that
vXg2G
g
!� ëv for all v 2 V :
6. Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l. Write ù � e2ði=3,
and let W be the irreducible CG-submodule of the regular CG-
module de®ned by
W � sp (1� ù2a� ùa2, b� ù2ab� ùa2b)
(see Exercise 6.6).
(a) Show that a � aÿ1 2 Z(CG).
(b) Find ë 2 C such that
w(a� aÿ1) � ëw
for all w 2 W. (Compare Proposition 9.14.)
Schur's Lemma 87
7. Which of the following groups have a faithful irreducible represent-
ation?
(a) Cn (n a positive integer);
(b) D8;
(c) C2 3 D8;
(d) C3 3 D8.
88 Representations and characters of groups
89
10
Irreducible modules and the group algebra
Let G be a ®nite group and CG be the group algebra of G over C.
Consider CG as the regular CG-module. By Theorem 8.7, we can
write
CG � U1 � : : :� Ur
where each Ui is an irreducible CG-module. We shall show in this
chapter that every irreducible CG-module is isomorphic to one of the
CG-modules U1, . . . , Ur. As a consequence, there are only ®nitely
many non-isomorphic irreducible CG-modules (a result which has
already been established for abelian groups in Theorem 9.8). Also, in
theory, to ®nd all irreducible CG-modules, it is suf®cient to decompose
CG as a direct sum of irreducible CG-submodules. However, this is
not really a practical way of ®nding the irreducible CG-modules, unless
G is a small group.
Irreducible submodules of CG
We begin with another consequence of Maschke's Theorem.
10.1 Proposition
Let V and W be CG-modules and let W: V! W be a CG-homomorph-
ism. Then there is a CG-submodule U of V such that V � Ker W � U
and U � Im W.
Proof Since Ker W is a CG-submodule of V by Proposition 7.2, there is
by Maschke's Theorem a CG-submodule U of V such that V �Ker W� U . De®ne a function W: U! Im W by
uW � uW (u 2 U ):
We show that W is a CG-isomorphism from U to Im W. Clearly W is a
CG-homomorphism, since W is a CG-homomorphism. If u 2 Ker W then
u 2 Ker W \ U � {0}; hence Ker W � {0}. Now let w 2 Im W; so w � vWfor some v 2 V. Write v � k � u with k 2 Ker W, u 2 U. Then
w � vW � kW� uW � uW � uW:
Therefore Im W � Im W. We have now established that W: U! Im W is
an invertible CG-homomorphism. Thus U � Im W, as required.
10.2 Proposition
Let V be a CG-module, and write
V � U1 � : : :� Us,
a direct sum of irreducible CG-submodules Ui. If U is any irreducible
CG-submodule of V, then U � Ui for some i.
Proof For u 2 U, we have u � u1 � . . . � us for unique vectors ui 2 Ui
(1 < i < s). De®ne ði: U! Ui by setting uði � ui. Choosing i such
that ui 6� 0 for some u 2 U, we have ði 6� 0.
Now ði is a CG-homomorphism (see Proposition 7.11). As U and
Ui are irreducible, and ði 6� 0, Schur's Lemma 9.1(1) implies that ði is
a CG-isomorphism. Therefore U � Ui. j
Of course it can happen that U is an irreducible CG-submodule of
U1 � . . . � Us (each Ui irreducible) without U being equal to any Ui,
as the following example shows.
10.3 Example
Let G be any group and let V be a 2-dimensional CG-module, with
basis v1, v2, such that vg � v for all v 2 V and g 2 G. Then
V � U1 � U2,
where U1 � sp (v1) and U2 � sp (v2) are irreducible CG-submodules.
However, U � sp (v1 � v2) is an irreducible CG-submodule which is
not equal to U1 or U2.
10.4 De®nitions
(1) If V is a CG-module and U is an irreducible CG-module, then we
say that U is a composition factor of V if V has a CG-submodule
which is isomorphic to U.
90 Representations and characters of groups
(2) Two CG-modules V and W are said to have a common composition
factor if there is an irreducible CG-module which is a composition
factor of both V and W.
We now come to the main result of the chapter, which shows that
every irreducible CG-module is a composition factor of the regular
CG-module.
10.5 Theorem
Regard CG as the regular CG-module, and write
CG � U1 � : : :� Ur,
a direct sum of irreducible CG-submodules. Then every irreducible
CG-module is isomorphic to one of the CG-modules Ui.
Proof Let W be an irreducible CG-module, and choose a non-zero
vector w 2 W. Observe that {wr: r 2 CG} is a CG-submodule of W;
since W is irreducible, it follows that
W � fwr: r 2 CGg:(10:6)
Now de®ne W: CG! W by
rW � wr (r 2 CG):
Clearly W is a linear transformation, and Im W � W by (10.6). Moreover,
W is a CG-homomorphism, since for r, s 2 CG,
(rs)W � w(rs) � (wr)s � (rW)s:
By Proposition 10.1, there is a CG-submodule U of CG such that
CG � U � Ker W and U � Im W � W :
As W is irreducible, so is U. By Proposition 10.2 we have U � Ui for
some i; then W � Ui, and the result is proved. j
Theorem 10.5 shows that there is a ®nite set of irreducible CG-
modules such that every irreducible CG-module is isomorphic to one
of them. We record this fact in the following corollary.
10.7 Corollary
If G is a ®nite group, then there are only ®nitely many non-isomorphic
irreducible CG-modules.
Irreducible modules and the group algebra 91
According to Theorem 10.5, to ®nd all the irreducible CG-modules
we need only decompose the regular CG-module as a direct sum of
irreducible CG-submodules. We now do this for a couple of examples;
however, this is not a practical method for studying CG-modules in
general.
10.8 Examples
(1) Let G � C3 � ka: a3 � 1l, and write ù � e2ði=3. De®ne v0, v1,
v2 2 CG by
v0 � 1� a� a2,
v1 � 1� ù2a� ùa2,
v2 � 1� ùa� ù2a2,
and let Ui � sp (vi) for i � 0, 1, 2. Then v1a � a � ù2a2 � ù1 � ùv1,
and similarly
via � ùivi for i � 0, 1, 2:
Hence Ui is a CG-submodule of CG for i � 0, 1, 2.
It is easy to check that v0, v1, v2 is a basis of CG, and hence
CG � U0 � U1 � U2,
a direct sum of irreducible CG-submodules Ui. By Theorem 10.5,
every irreducible CG-module is isomorphic to U0, U1 or U2. The
irreducible representation of G corresponding to Ui is the representation
rùi of Example 9.9(1).
(2) Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l. We decompose CG
as a direct sum of irreducible CG-submodules. Let ù � e2ði=3 and
de®ne
v0 � 1� a� a2, w0 � bv0 (� b� ba� ba2),
v1 � 1� ù2a� ùa2, w1 � bv1,
v2 � 1� ùa� ù2a2, w2 � bv2:
As in (1) above, via � ùivi for i � 0, 1, 2, and so sp (vi) and sp (wi)
are Ckal-modules. Next, note that
v0b � w0, w0b � v0,
v1b � w2, w1b � v2,
v2b � w1, w2b � v1:
92 Representations and characters of groups
Therefore, sp (v0, w0), sp (v1, w2) and sp (v2, w1) are Ckbl-modules, and
hence are CG-submodules of CG. By the argument in Example 5.5(2),
the CG-submodules U3 � sp (v1, w2) and U4 � sp (v2, w1) are irreduci-
ble. However, sp(v0, w0) is reducible, as U1 � sp(v0 � w0) and U2 �sp(v0 ÿ w0) are CG-submodules.
Now v0, v1, v2, w0, w1, w2 is a basis of CG, and hence
CG � U1 � U2 � U3 � U4,
a direct sum of irreducible CG-submodules. Note that U1 is the trivial
CG-module, and U1 is not isomorphic to U2, the other 1-dimensional
Ui. But U3 � U4 (there is a CG-isomorphism sending v1 ! w1,
w2 ! v2).
We conclude from Theorem 10.5 that there are exactly three non-
isomorphic irreducible CG-modules, namely U1, U2 and U3. Corre-
spondingly, every irreducible representation of D6 over C is equivalent
to precisely one of the following:
r1: a! (1), b! (1);
r2: a! (1), b! (ÿ1);
r3: a! ù 0
0 ùÿ1
� �, b! 0 1
1 0
� �:
Summary of Chapter 10
1. Every irreducible CG-module occurs as a composition factor of the
regular CG-module.
2. There are only ®nitely many non-isomorphic irreducible CG-
modules.
Exercises for Chapter 10
1. Let G be a ®nite group. Find a CG-submodule of CG which is
isomorphic to the trivial CG-module. Is there only one such CG-
submodule?
2. Let G � C4. Express CG as a direct sum of irreducible CG-
submodules. (Hint: copy the method of Example 10.8(1).)
Irreducible modules and the group algebra 93
3. Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. Find a 1-dimen-
sional CG-submodule, sp (u1) say, of CG such that
u1a � u1, u1b � ÿu1:
Find also 1-dimensional CG-submodules, sp (u2) and sp (u3), such
that
u2a � ÿu2, u2b � u2, and
u3a � ÿu3, u3b � ÿu3:
4. Use the method of Example 10.8(2) to ®nd all the irreducible
representations of D8 over C.
5. Suppose that V is a non-zero CG-module such that V � U1 � U2,
where U1 and U2 are isomorphic CG-modules. Show that there is a
CG-submodule U of V which is not equal to U1 or U2, but is
isomorphic to both of them.
6. Let G � Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l, and let V be the
CG-module given in Example 4.5(2). Thus V has basis v1, v2 and
v1a � iv1, v1b � v2,
v2a � ÿiv2, v2b � ÿv1:
Show that V is irreducible, and ®nd a CG-submodule of CG which
is isomorphic to V.
94 Representations and characters of groups
95
11
More on the group algebra
We now go further into the structure of the group algebra CG of a
®nite group G. As in Chapter 10, we write
CG � U1 � : : :� Ur,
a direct sum of irreducible CG-modules Ui. In Theorem 10.5 we
proved that every irreducible CG-module U is isomorphic to one of
the Ui. The question arises: how many of the Ui are isomorphic to U?
There is an elegant and signi®cant answer to this question: the number
is precisely dim U (see Theorem 11.9).
Our proof of Theorem 11.9 is based on a study of the vector space
of CG-homomorphisms from one CG-module to another.
The space of CG-homomorphisms
11.1 De®nition
Let V and W be CG-modules. We write HomCG (V , W ) for the set of
all CG-homomorphisms from V to W.
De®ne addition and scalar multiplication on HomCG (V , W ) as
follows: for W, ö 2 HomCG (V , W ) and ë 2 C, de®ne W � ö and ëW by
v(W� ö) � vW� vö,
v(ëW) � ë(vW)
for all v 2 V . Then W� ö, ëW 2 HomCG (V , W ). With these de®nitions,
it is easily checked that HomCG (V, W ) is a vector space over C.
We begin our study of the vector space HomCG (V, W) with an easy
consequence of Schur's Lemma.
11.2 Proposition
Suppose that V and W are irreducible CG-modules. Then
dim (HomCG (V , W )) � 1, if V � W ,
0, if V 6� W :
�
Proof If V 6� W then this is immediate from Schur's Lemma 9.1(1).
Now suppose that V � W, and let W: V! W be a CG-isomorphism.
If ö 2 HomCG (V , W ), then öWÿ1 is a CG-isomorphism from V to V,
so by Schur's Lemma 9.1(2), there exists ë 2 C such that
öWÿ1 � ë1V :
Then ö � ëW, and so HomCG (V , W ) � fëW: ë 2 Cg, a 1-dimensional
space. j
For the next result, recall the de®nition of a composition factor of a
CG-module from 10.4.
11.3 Proposition
Let V and W be CG-modules, and suppose that HomCG(V , W ) 6� f0g.Then V and W have a common composition factor.
Proof Let W be a non-zero element of HomCG (V , W ). Then
V � Ker W� U for some non-zero CG-module U, by Maschke's Theo-
rem. Let X be an irreducible CG-submodule of U. Since XW 6� {0},
Schur's Lemma 9.1(1) implies that XW � X. Therefore X is a common
composition factor of V and W. j
The next few results show how to calculate the dimension of
HomCG (V , W ) in general. The key step is the following proposition.
11.4 Proposition
Let V , V1, V2 and W , W1, W2 be CG-modules. Then
(1) dim (HomCG (V, W1 � W2)) �dim (HomCG (V, W1)) � dim (HomCG(V, W2)),
(2) dim (HomCG (V1 � V2, W )) �dim (HomCG (V1, W)) � dim (HomCG(V2, W )).
96 Representations and characters of groups
Proof (1) De®ne the functions ð1: W1 � W2 ! W1 and ð2:
W1 � W2 ! W2 by
(w1 � w2)ð1 � w1, (w1 � w2)ð2 � w2
for all w1 2 W1, w2 2 W2. By Proposition 7.11, ð1 and ð2 are
CG-homomorphisms. If W 2 HomCG (V , W1 � W2), then Wð1 2HomCG(V , W1) and Wð2 2 HomCG (V, W2) (see Exercise 7.1).
We now de®ne a function f from HomCG (V, W1 � W2) to the
(external) direct sum of HomCG (V, W1) and HomCG (V, W2) by
f : W! (Wð1, Wð2) (W 2 HomCG (V , W1 � W2)):
Clearly f is a linear transformation. We show that f is invertible.
Given öi 2 HomCG (V, Wi) (i � 1, 2), the function
ö: v! vö1 � vö2 (v 2 V )
lies in HomCG (V, W1 � W2), and the image of ö under f is (ö1, ö2).
Hence f is surjective.
If W 2 Ker f, then vWð1 � 0 and vWð2 � 0 for all v 2 V, so vW �vW(ð1 � ð2) � 0. Therefore W � 0, so Ker f � {0} and f is injective.
We have established that f is an invertible linear transformation from
HomCG (V, W1 � W2) to HomCG (V, W1) � HomCG (V, W2). Consequently
these two vector spaces have equal dimensions, and (1) follows.
(2) For W 2 HomCG (V1 � V2, W ), de®ne WVi: Vi ! W (i � 1, 2) to
be the restriction of W to Vi; that is, WViis the function
viWVi� viW (vi 2 Vi):
Then WVi2 HomCG (Vi, W ) for i � 1, 2.
Now let h be the function from HomCG (V1 � V2, W ) to
HomCG (V1, W )� HomCG (V2, W ) which is given by
h: W! (WV1, WV2
) (W 2 HomCG (V1 � V2, W )):
Clearly h is an injective linear transformation. Given öi 2HomCG(Vi, W ) (i � 1, 2), the function
ö: v1 � v2 ! v1ö1 � v2ö2 (vi 2 Vi for i � 1, 2)
lies in HomCG (V1 � V2, W ) and has image (ö1, ö2) under h. Hence h
is surjective. We have shown that h is an invertible linear transforma-
tion, and (2) follows. j
More on the group algebra 97
Now suppose that we have CG-modules V, W, Vi, Wj (1 < i < r,
1 < j < s). By an obvious induction using Proposition 11.4, we have
(11.5) (1) dim (HomCG (V, W1 � . . . � Ws))
�Xs
j�1
dim (HomCG (V, Wj)),
(2) dim(HomCG(V1� . . . �Vr, W ))
�Xr
i�1
dim (HomCG (Vi, W)).
These in turn imply
(3) dim (HomCG (V1 � . . . � Vr, W1 � . . . � Ws))
�Xr
i�1
Xs
j�1
dim (HomCG (Vi, Wj)).
By applying (3) when all Vi and Wj are irreducible, and using
Proposition 11.2, we can ®nd dim (HomCG (V, W )) in general. In the
following corollary we single out the case where one of the CG-
modules is irreducible.
11.6 Corollary
Let V be a CG-module with
V � U1 � : : :� Us,
where each Ui is an irreducible CG-module. Let W be any irreducible
CG-module. Then the dimensions of HomCG (V , W ) and
HomCG (W , V ) are both equal to the number of CG-modules Ui such
that Ui � W.
Proof By (11.5),
dim (HomCG (V , W )) �Xs
i�1
dim (HomCG (Ui, W )), and
dim (HomCG (W , V )) �Xs
i�1
dim (HomCG (W , Ui)):
98 Representations and characters of groups
And by Proposition 11.2,
dim (HomCG (Ui, W )) � dim (HomCG (W , Ui)) � 1, if Ui � W ,
0, if Ui 6� W :
�The result follows. j
11.7 Example
For G � D6, we saw in Example 10.8(2) that
CG � U1 � U2 � U3 � U4,
a direct sum of irreducible CG-submodules, with U3 � U4 but U3 not
isomorphic to U1 or U2. Thus by Corollary 11.6, we have
dim (HomCG (CG, U3)) � dim (HomCG (U3, CG)) � 2:
You are asked in Exercise 11.5 to ®nd bases for these two vector
spaces of CG-homomorphisms.
The next proposition investigates the space of CG-homomorphisms
from the regular CG-module to any other CG-module. When combined
with Corollary 11.6, it will give the main result of this chapter.
11.8 Proposition
If U is a CG-module, then
dim (HomCG (CG, U )) � dim U :
Proof Let d � dim U. Choose a basis u1, . . . , ud of U. For 1 < i < d,
de®ne öi: CG! U by
röi � uir (r 2 CG):
Then öi 2 HomCG (CG, U) since for all r, s 2 CG,
(rs)öi � ui(rs) � (uir)s � (röi)s:
We shall prove that ö1, . . . , öd is a basis of HomCG (CG, U ).
Suppose that ö 2 HomCG(CG, U ). Then
1ö � ë1u1 � : : :� ëd ud
for some ëi 2 C. Since ö is a CG-homomorphism, for all r 2 CG we
have
More on the group algebra 99
rö � (1r)ö � (1ö)r
� ë1u1 r � : : :� ëd ud r
� r(ë1ö1 � : : :� ëdöd):
Hence ö � ë1ö1 � . . . � ëdöd . Therefore ö1, . . . , öd span
HomCG (CG, U ).
Now assume that
ë1ö1 � : : :� ëdöd � 0 (ëi 2 C):
Evaluating both sides at the identity 1, we have
0 � 1(ë1ö1 � : : :� ëdöd)
� ë1u1 � : : :� ëd ud ,
which forces ëi � 0 for all i. Hence ö1, . . . , öd is a basis of
HomCG (CG, U ), which therefore has dimension d. j
We now come to the main theorem of the chapter, which tells us
how often each irreducible CG-module occurs in the regular CG-
module.
11.9 Theorem
Suppose that
CG � U1 � : : :� Ur,
a direct sum of irreducible CG-submodules. If U is any irreducible
CG-module, then the number of CG-modules Ui with Ui � U is equal
to dim U.
Proof By Proposition 11.8,
dim U � dim (HomCG (CG, U )),
and by Corollary 11.6, this is equal to the number of Ui with Ui � U.
j
11.10 Example
Recall again from Example 10.8(2) that if G � D6 then
CG � U1 � U2 � U3 � U4,
where U1, U2 are non-isomorphic 1-dimensional CG-modules, and
100 Representations and characters of groups
U3, U4 are isomorphic irreducible 2-dimensional CG-modules. This
illustrates Theorem 11.9:
U1 occurs once, dim U1 � 1;
U2 occurs once, dim U2 � 1;
U3 occurs twice, dim U3 � 2:
We conclude the chapter with a signi®cant consequence of
Theorem 11.9 concerning the dimensions of all irreducible CG-modules.
11.11 De®nition
We say that the irreducible CG-modules V1, : : : , Vk form a com-
plete set of non-isomorphic irreducible CG-modules if every irreducible
CG-module is isomorphic to some Vi, and no two of V1, : : : , Vk
are isomorphic. (By Corollary 10.7, for any ®nite group G there exists
a complete set of non-isomorphic irreducible CG-modules.)
11.12 Theorem
Let V1, : : : , Vk form a complete set of non-isomorphic irreducible
CG-modules. Then Xk
i�1
(dim Vi)2 � jGj:
Proof Let CG � U1 � . . . � Ur, a direct sum of irreducible CG-sub-
modules. For 1 < i < k, write di � dim Vi. By Theorem 11.9, for each
i, the number of CG-modules Uj with Uj � Vi is equal to di. Therefore
dim CG � dim U1 � : : :� dim Ur
�Xk
i�1
di(dim Vi) �Xk
i�1
d2i :
As dim CG � |G|, the result follows. j
11.13 Example
Let G be a group of order 8, and let d1, . . . , dk be the dimensions of
all the irreducible CG-modules. By Theorem 11.12,Xk
i�1
d2i � 8:
More on the group algebra 101
Observe that the trivial CG-module is irreducible of dimension 1, and
so di � 1 for some i. Hence the possibilities for d1, . . . , dk are
1, 1, 1, 1, 1, 1, 1, 1 and
1, 1, 1, 1, 2:
Both these possibilities do occur: the ®rst holds when G is an abelian
group (see Proposition 9.5), and the second when G � D8 (see
Exercise 10.4).
We shall see later that dim Vi divides |G| for all i, and this fact,
combined with Theorem 11.12, is quite a powerful tool in ®nding the
dimensions of irreducible CG-modules.
Summary of Chapter 11
1. dim (HomCG (V1 � : : :� Vr, W1 � : : :� Ws))
�Xr
i�1
Xs
j�1
dim (HomCG (Vi, Wj)):
2. dim (HomCG (CG, U )) � dim U .
3. Let CG � U1 � . . . � Ur, a direct sum of irreducible CG-modules,
and let U be any irreducible CG-module. Then the number of Ui
with Ui � U is equal to dim U.
4. If V1, : : : , Vk is a complete set of non-isomorphic irreducible CG-
modules, then Xk
i�1
(dim Vi)2 � jGj:
Exercises for Chapter 11
1. If G is a non-abelian group of order 6, ®nd the dimensions of all
the irreducible CG-modules.
2. If G is a group of order 12, what are the possible degrees of all the
irreducible representations of G?
Find the degrees of the irreducible representations of D12.
(Hint: use Exercise 5.3.)
3. Let G be a ®nite group. Find a basis for HomCG (CG, CG).
102 Representations and characters of groups
4. Suppose that G � Sn and V is the n-dimensional permutation
module for G over C, as de®ned in 4.10. If U is the trivial CG-
module, show that HomCG (V, U) has dimension 1.
5. Let G � D6 and let CG � U1 � U2 � U3 � U4, a direct sum of
irreducible CG-modules, as in Example 10.8(2). Find a basis for
HomCG (CG, U3) and a basis for HomCG (U3, CG).
6. Let V1, : : : , Vk be a complete set of non-isomorphic irreducible
CG-modules, and let V, W be arbitrary CG-modules. Assume that
for 1 < i < k,
di � dim (HomCG (V , Vi)) and ei � dim (HomCG (W , Vi)):
Show that dim (HomCG (V , W )) �Pki�1 diei.
More on the group algebra 103
104
12
Conjugacy classes
We take a break from representation theory to discuss some topics in
group theory which will be relevant in our further study of represent-
ations. After de®ning conjugacy classes, we develop enough theory to
determine the conjugacy classes of dihedral, symmetric and alternating
groups. At the end of the chapter we prove a result linking the
conjugacy classes of a group to the structure of its group algebra.
Throughout the chapter, G is a ®nite group.
Conjugacy classes
12.1 De®nition
Let x, y 2 G. We say that x is conjugate to y in G if
y � gÿ1xg for some g 2 G:
The set of all elements conjugate to x in G is
xG � fgÿ1xg: g 2 Gg,and is called the conjugacy class of x in G.
Our ®rst result shows that two distinct conjugacy classes have no
elements in common.
12.2 Proposition
If x, y 2 G, then either xG � yG or xG \ yG is empty.
Proof Suppose that xG \ yG is not empty, and pick z 2 xG \ yG. Then
there exist g, h 2 G such that
z � gÿ1xg � hÿ1 yh:
Hence x � ghÿ1 yhgÿ1 � kÿ1 yk, where k � hgÿ1. So
a 2 xG ) a � bÿ1xb for some b 2 G
) a � bÿ1 kÿ1 ykb
) a � cÿ1 yc where c � kb
) a 2 yG:
Therefore xG # yG. Similarly yG # xG (using y � kxkÿ1), and so
xG � yG. j
Since every element x of G lies in the conjugacy class xG (as
x � 1ÿ1x1 with 1 2 G), G is the union of its conjugacy classes and so
we deduce immediately
12.3 Corollary
Every group is a union of conjugacy classes, and distinct conjugacy
classes are disjoint.
Another way of seeing this Corollary 12.3 is to observe that
conjugacy is an equivalence relation, and that the conjugacy classes are
the equivalence classes.
12.4 De®nition
If G � xG1 [ . . . [ xG
l , where the conjugacy classes xG1 , . . . , xG
l are
distinct, then we call x1, . . . , xl representatives of the conjugacy classes
of G.
12.5 Examples
(1) For every group G, 1G � {1} is a conjugacy class of G.
(2) Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l. The elements of G
are 1, a, a2, b, ab, a2b. Since gÿ1ag is a or a2 for every g 2 G, and
bÿ1ab � a2, we have
aG � fa, a2g:Also, aÿibai � aÿ2ib for all integers i, so
bG � fb, ab, a2bg:Thus the conjugacy classes of G are
f1g, fa, a2g, fb, ab, a2bg:
Conjugacy classes 105
(3) If G is abelian then gÿ1xg � x for all x, g 2 G, and so xG � {x}.
Hence every conjugacy class of G consists of just one element.
The next proposition is often useful when calculating conjugacy
classes.
12.6 Proposition
Let x, y 2 G. If x is conjugate to y in G, then xn is conjugate to yn in
G for every integer n, and x and y have the same order.
Proof Observe that for a, b 2 G, we have
gÿ1abg � (gÿ1ag)(gÿ1bg):
Hence gÿ1xng � (gÿ1xg)n. Suppose that x is conjugate to y in G, so
that y � gÿ1xg for some g 2 G. Then yn � gÿ1xng and therefore xn is
conjugate to yn in G. Let x have order m. Then ym � gÿ1xmg � 1, and
for 0 , r , m, yr � gÿ1xrg 6� 1, so y also has order m. j
Conjugacy class sizes
The next theorem determines the sizes of the conjugacy classes in G
in terms of certain subgroups which we now de®ne.
12.7 De®nition
Let x 2 G. The centralizer of x in G, written CG(x), is the set of
elements of G which commute with x; that is,
CG(x) � fg 2 G: xg � gxg:(So also CG(x) � {g 2 G: gÿ1xg � x}.)
It is easy to check that CG(x) is a subgroup of G (Exercise 12.1).
Observe that x 2 CG(x) and indeed, kxl # CG(x) for all x 2 G.
12.8 Theorem
Let x 2 G. Then the size of the conjugacy class xG is given by
jxGj � jG: CG(x)j � jGj=jCG(x)j:In particular, |xG | divides |G|.
106 Representations and characters of groups
Proof Observe ®rst that for g, h 2 G, we have
gÿ1xg � hÿ1xh, hgÿ1x � xhgÿ1
, hgÿ1 2 CG(x)
, CG(x)g � CG(x)h:
By dint of this, we may de®ne an injective function f from xG to the
set of right cosets of CG(x) in G by
f : gÿ1xg ! CG(x)g (g 2 G):
Clearly f is surjective. Hence f is a bijection, proving that |xG | �|G:CG(x)|. j
Before summarizing our results on conjugacy classes, we make the
observation that
jxGj � 1, gÿ1xg � x for all g 2 G(12:9)
, x 2 Z(G),
where Z(G) is the centre of G, as de®ned in 9.15.
We have now proved all parts of the following result.
12.10 The Class Equation
Let x1, . . . , xl be representatives of the conjugacy classes of G. Then
jGj � jZ(G)j �X
xi=2Z(G)
jxGi j,
where |xGi | � |G:CG(xi)|, and both |Z(G)| and |xG
i | divide |G|.
Conjugacy classes of dihedral groups
We illustrate the use of Theorem 12.8 by ®nding the conjugacy classes
of all dihedral groups.
Let G � D2n, the dihedral group of order 2n. Thus
G � ha, b: an � b2 � 1, bÿ1ab � aÿ1i:In ®nding the conjugacy classes of G, it is convenient to consider
separately the cases where n is odd and where n is even.
(1) n odd
First consider ai (1 < i < n ÿ 1). Since CG(ai) contains kal,jG: CG(ai)j < jG: haij � 2:
Conjugacy classes 107
Also bÿ1aib � aÿi, so {ai, aÿi} # (ai)G. As n is odd, ai 6� aÿi, and so
|(ai)G | > 2. Using Theorem 12.8, we have
2 > jG: CG(ai)j � j(ai)Gj > 2:
Hence equality holds here, and
CG(ai) � hai, (ai)G � fai, aÿig:Next, CG(b) contains {1, b}; and as bÿ1aib � aÿi, no element ai or aib
(with 1 < i < n ÿ 1) commutes with b. Thus
CG(b) � f1, bg:
Therefore by Theorem 12.8, |bG | � n. Since all the elements ai have
been accounted for, bG must consist of the remaining n elements of G.
That is,
bG � fb, ab, : : : , anÿ1bg:
We have shown
(12.11) The dihedral group D2n (n odd) has precisely 12(n� 3) con-
jugacy classes:
{1}, {a, aÿ1}, . . . , {a(nÿ1)=2, aÿ(nÿ1)=2}, {b, ab, . . . , anÿ1b}.
(2) n even
Write n � 2m. As bÿ1amb � aÿm � am, the centralizer of am in G
contains both a and b, and hence CG(am) � G. Therefore the con-
jugacy class of am in G is just famg. As in case (1), (ai)G � {ai, aÿi}
for 1 < i < m ÿ 1.
For every integer j,
a jbaÿ j � a2 jb, a j(ab)aÿ j � a2 j�1b:
It follows that
bG � fa2 jb: 0 < j < mÿ 1g, (ab)G � fa2 j�1b: 0 < j < mÿ 1g:Hence
(12.12) The dihedral group D2n (n even, n � 2m) has precisely m � 3
conjugacy classes:
{1}, {am}, {a, aÿ1}, . . . , {amÿ1, aÿm�1},
{a2 jb: 0 < j < m ÿ 1}, {a2 j�1b: 0 < j < m ÿ 1}.
108 Representations and characters of groups
Conjugacy classes of Sn
We shall later need to know the conjugacy classes of the symmetric
group Sn. Our ®rst observation is simple but crucial.
12.13 Proposition
Let x be a k-cycle (i1 i2 . . . ik) in Sn, and let g 2 Sn. Then gÿ1xg is the
k-cycle (i1 g i2 g . . . ikg).
Proof Write A � {i1, . . . , ik}. For ir 2 A,
irg(gÿ1xg) � ir xg � ir�1 g (or i1 g if r � k):
Also, for 1 < i < n and i =2 A,
ig(gÿ1xg) � ixg � ig:
Hence gÿ1(i1 i2 . . . ik)g � (i1 g i2 g . . . ikg), as required. j
Now consider an arbitrary permutation x 2 Sn. Write
x � (a1 : : : ak 1)(b1 : : : bk 2
) : : : (c1 : : : cks),
a product of disjoint cycles, with k1 > k2 > . . . > ks. By Proposition
12.13, for g 2 Sn we have
(12.14)
gÿ1xg � gÿ1(a1 : : : ak1)ggÿ1(b1 : : : bk2
)g : : : gÿ1(c1 : : : cks)g
� (a1 g : : : ak1g)(b1 g : : : bk2
g) : : : (c1 g : : : cksg):
We call (k1, . . . , ks) the cycle-shape of x, and note that x and gÿ1xg
have the same cycle-shape. On the other hand, given any two permuta-
tions x, y of the same cycle-shape, say
x � (a1 : : : ak1) : : : (c1 : : : cks
),
y � (a91 : : : a9k1) : : : (c91 : : : c9ks
),
(products of disjoint cycles), there exists g 2 Sn sending
a1 ! a91, . . . , cks! c9ks
, and so by (12.14),
gÿ1xg � y:
We have proved the following result.
Conjugacy classes 109
12.15 Theorem
For x 2 Sn, the conjugacy class xSn of x in Sn consists of all permuta-
tions in Sn which have the same cycle-shape as x.
12.16 Examples
(1) The conjugacy classes of S3 are
(2) The conjugacy class of (1 2)(3 4) in S4 consists of all the elements
of cycle-shape (2, 2) and is
f(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:(3) There are precisely ®ve conjugacy classes of S4, with represent-
atives (see De®nition 12.4):
1, (1 2), (1 2 3), (1 2)(3 4), (1 2 3 4):
To calculate the sizes of the conjugacy classes, we simply count the
number of 2-cycles, 3-cycles, and so on. The number of 2-cycles is
equal to the number of pairs that can be chosen from {1, 2, 3, 4},
which is 42
ÿ � � 6. (The notation nr� � means the binomial coef®cient
n!=(r!(nÿ r)!).) The number of 3-cycles is 4 3 2 (4 for the choice of
®xed point and 2 because there are two 3-cycles ®xing a given point).
Similarly, there are three elements of cycle-shape (2, 2) and there are
six 4-cycles. Thus for G � S4, the conjugacy class representatives g,
the conjugacy class sizes |gG | and the centralizer orders |CG(g)|
(obtained using Theorem 12.8) are as follows:
Representative g 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)Class size | gG | 1 6 8 3 6
|CG(g)| 24 4 3 8 4
We check our arithmetic by noting that
jS4j � 1� 6� 8� 3� 6:
Class Cycle-shape
{1} (1){(1 2), (1 3), (2 3)} (2)
{(1 2 3), (1 3 2)} (3)
110 Representations and characters of groups
(4) Similarly, the corresponding table for G � S5 is
Rep. g 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)
|gG | 1 10 20 15 30 20 24
|CG(g)| 120 12 6 8 4 6 5
Conjugacy classes of An
Given an even permutation x 2 An, we have seen in Theorem 12.15
that the conjugacy class xSn consists of all permutations in Sn which
have the same cycle-shape as x. The conjugacy class x An of x in An,
given by
x An � fgÿ1xg: g 2 Ang,is of course contained in xS n ; however, x An might not be equal to xSn .
For an easy example where equality does not hold, consider x �(1 2 3) 2 A3; here x A3 � fxg, while xS3 � {x, xÿ1}.
The next result determines precisely when xAn and xSn are equal, and
what happens when equality fails.
12.17 Proposition
Let x 2 An with n . 1.
(1) If x commutes with some odd permutation in Sn, then xSn � xAn .
(2) If x does not commute with any odd permutation in Sn then xSn
splits into two conjugacy classes in An of equal size, with represent-
atives x and (1 2)ÿ1x(1 2).
Proof (1) Assume that x commutes with an odd permutation g. Let
y 2 xSn , so that y � hÿ1xh for some h 2 Sn. If h is even then y 2 xAn ;
and if h is odd then gh 2 An and
y � hÿ1xh � hÿ1 gÿ1xgh � (gh)ÿ1x(gh),
so again y 2 xAn . Thus xSn # xAn , and so xSn � xAn .
(2) Assume that x does not commute with any odd permutation.
Then
CSn(x) � CAn
(x):
Conjugacy classes 111
Hence by Theorem 12.8,
jxAn j � jAn: CAn(x)j � 1
2jSn: CAn
(x)j (as jAnj � 12jSnj)
� 12jSn: CSn
(x)j � 12jxSn j:
Next, we observe that
fhÿ1xh: h is oddg � ((1 2)ÿ1x(1 2))An
since every odd permutation has the form (1 2)a for some a 2 An. Now
xSn � fhÿ1xh: h is eveng [ fhÿ1xh: h is oddg� xAn [ ((1 2)ÿ1x(1 2))An :
Since |xAn | � 12|xSn |, the conjugacy classes xAn and ((1 2)ÿ1x(1 2))An must
be disjoint and of equal size, as we wished to show. j
12.18 Examples
(1) We ®nd the conjugacy classes of A4. The elements of A4 are the
identity, together with the permutations of cycle-shapes (2, 2) and (3).
Since (1 2)(3 4) commutes with the odd permutation (1 2), Proposition
12.17 implies that
(1 2)(3 4)A4 � (1 2)(3 4)S4 � f(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:However, the 3-cycle (1 2 3) commutes with no odd permutation: for if
gÿ1(1 2 3)g � (1 2 3) then (1 2 3) � (1g 2g 3g) by Proposition 12.13,
so g is 1, (1 2 3) or (1 3 2), an even permutation. Hence by Proposition
12.17, (1 2 3)S4 splits into two conjugacy classes in A4 of size 4, with
representatives (1 2 3) and (1 2)ÿ1(1 2 3)(1 2) � (1 3 2).
Thus the conjugacy classes of A4 are
Representative 1 (1 2)(3 4) (1 2 3) (1 3 2)Class size 1 3 4 4Centralizer order 12 4 3 3
(2) We ®nd the conjugacy classes of A5. The non-identity even
permutations in S5 are those of cycle-shapes (3), (2, 2) and (5). The
elements (1 2 3) and (2 3)(4 5) commute with the odd permutation
(4 5); but (1 2 3 4 5) commutes with no odd permutation. (Check this
by using the argument in (1) above.) Hence by Proposition 12.17, the
112 Representations and characters of groups
conjugacy classes of A5 are represented by 1, (1 2 3), (1 2)(3 4),
(1 2 3 4 5) and (1 2)ÿ1(1 2 3 4 5)(1 2) � (1 3 4 5 2). Using Proposition
12.17(2), we see that the class sizes and centralizer orders are as
follows:
Representative 1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)Class size 1 20 15 12 12Centralizer order 60 3 4 5 5
Normal subgroups
Normal subgroups are related to conjugacy classes by the following
elementary result.
12.19 Proposition
Let H be a subgroup of G. Then H v G if and only if H is a union of
conjugacy classes of G.
Proof If H is a union of conjugacy classes, then
h 2 H , g 2 G) gÿ1 hg 2 H ,
so gÿ1 Hg # H. Thus H v G.
Conversely, if H v G then for all h 2 H, g 2 G, we have
gÿ1 hg 2 H, and so hG # H. Therefore
H �[h2H
hG,
and so H is a union of conjugacy classes of G. j
12.20 Example
We ®nd all the normal subgroups of S4. Let H v S4. Then by
Proposition 12.19, H is a union of conjugacy classes of S4. As we saw
in Example 12.16(3), these conjugacy classes have sizes 1, 6, 8, 3, 6.
Since jH j divides 24 by Lagrange's Theorem, and 1 2 H, there are just
four possibilities:
jH j � 1, 1� 3, 1� 8� 3 or 1� 6� 8� 3� 6:
Conjugacy classes 113
In the ®rst case H � {1}, in the last case H � S4, and in the third
case H � A4. In the case where jH j � 1 � 3, we have
H � 1S4 [ (1 2)(3 4)S4 � f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:This is easily checked to be a subgroup of S4; we write it as V4
(V stands for `Viergruppe', meaning `four-group').
We have now shown that S4 has exactly four normal subgroups:
f1g, S4, A4 and V4 � f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:
The centre of a group algebra
In this ®nal section we link the conjugacy classes of the group G to
the centre of the group algebra CG. Recall from De®nition 9.12 that
the centre of CG is
Z(CG) � fz 2 CG: zr � rz for all r 2 CGg:We know that Z(CG) is a subspace of the vector space CG. There is a
convenient basis for this subspace which can be described in terms of
the conjugacy classes of G.
12.21 De®nition
Let C1, . . . , Cl be the distinct conjugacy classes of G. For 1 < i < l,
de®ne
Ci �Xg2Ci
g 2 CG:
The elements C1, . . . , Cl of CG are called class sums.
12.22 Proposition
The class sums C1, . . . , Cl form a basis of Z(CG).
Proof First we show that each Ci belongs to Z(CG). Let Ci consist of
the r distinct conjugates yÿ11 gy1, . . . , yÿ1
r gyr of an element g, so
Ci �Xr
j�1
yÿ1j gyj:
For all h 2 G,
hÿ1Ci h �Xr
j�1
hÿ1 yÿ1j gyjh:
114 Representations and characters of groups
As j runs from 1 to r, the elements hÿ1 yÿ1j gyjh run through Ci, since
hÿ1 yÿ1j gyjh � hÿ1 yÿ1
k gykh, yÿ1j gyj � yÿ1
k gyk :
Hence Xr
j�1
hÿ1 yÿ1j gyjh � Ci,
and so hÿ1Ci h � Ci. That is,
Ci h � hCi:
Therefore each Ci commutes with all h 2 G, hence with allPh2G ëh h 2 CG, and so Ci 2 Z(CG).
Next, observe that C1, . . . , Cl are linearly independent: for ifPli�1 ëiCi � 0 (ëi 2 C), then all ëi � 0 as the classes C1, . . . , Cl are
pairwise disjoint by Corollary 12.3.
It remains to show that C1, . . . , Cl span Z(CG). Let r �Pg2G ë gg 2 Z(CG). For h 2 G, we have rh � hr, so hÿ1 rh � r. That
is, Xg2G
ë g hÿ1 gh �Xg2G
ë g g:
So for every h 2 G, the coef®cient ë g of g is equal to the coef®cient
ëhÿ1 gh of hÿ1 gh. That is to say, the function g! ë g is constant on
conjugacy classes of G. It follows that r �P li�1 ëiCi where ëi is the
coef®cient ë gifor some gi 2 Ci. This completes the proof. j
12.23 Examples
(1) From Example 12.16(1), a basis for Z(CS3) is
1, (1 2)� (1 3)� (2 3), (1 2 3)� (1 3 2):
(2) From (12.12), a basis for Z(CD8) is
1, a2, a� a3, b� a2b, ab� a3b:
Summary of Chapter 12
1. Every group is a union of conjugacy classes, and distinct conjugacy
classes are disjoint.
2. For an element x of a group G, the centralizer CG(x) is the set of
Conjugacy classes 115
elements of G which commute with x. It is a subgroup of G, and
the number of elements in the conjugacy class xG is equal to
|G:CG(x)|.
3. The conjugacy classes of Sn correspond to the cycle-shapes of
permutations in Sn.
4. If x 2 An then xSn � xAn if and only if x commutes with some odd
permutation in Sn.
5. The class sums in CG form a basis for the centre of CG.
Exercises for Chapter 12
1. If G is a group and x 2 G, show that CG(x) is a subgroup of G
which contains Z(G).
2. Let G be a ®nite group and suppose that g 2 G and z 2 Z(G). Prove
that the conjugacy classes gG and (gz)G have the same size.
3. Let G � Sn.
(a) Prove that j(1 2)Gj � n2� � and ®nd CG((1 2)). Verify that your
solution satis®es Theorem 12.8.
(b) Show that j(1 2 3)Gj � 2 n3� � and j(1 2)(3 4)Gj � 3 n
4� �.(c) Now let n � 6. Show that
j(1 2 3)(4 5 6)Gj � 40 and j(1 2)(3 4)(5 6)Gj � 15,
and ®nd the sizes of the other conjugacy classes of S6. (There
are 11 conjugacy classes in all.)
4. What are the cycle-shapes of those permutations x 2 A6 for which
x A6 6� xS6 ?
5. Show that A5 is a simple group. (Hint: use the method of Example
12.20.)
6. Find the conjugacy classes of the quaternion group Q8. Give a basis
of the centre of the group algebra CQ8.
7. Let p be a prime number, and let n be a positive integer. Suppose
that G is a group of order pn.
(a) Use the Class Equation 12.10 to show that Z(G) 6� {1}.
(b) Suppose that n > 3 and that |Z(G)| � p. Prove that G has a
conjugacy class of size p.
116 Representations and characters of groups
117
13
Characters
Suppose that r: G! GL(n, C) is a representation of the ®nite group
G. With each n 3 n matrix gr (g 2 G) we associate the complex
number given by adding all the diagonal entries of the matrix, and call
this number ÷(g). The function ÷: G! C is called the character of the
representation r. Characters of representations have many remarkable
properties, and they are the fundamental tools for performing calcula-
tions in representation theory. For example, we shall show later that
two representations have the same character if and only if they are
equivalent. Moreover, basic problems, such as deciding whether or not
a given representation is irreducible, can be resolved by doing some
easy arithmetic with the character of the representation. These facts are
surprising, since from the de®nition of a representation r: G!GL(n, C), it appears that we must keep track of all the n2 entries in
each matrix gr, whereas the character records just one number for
each matrix.
The theory of characters will occupy a considerable portion of the
rest of the book. In this chapter we present some basic properties and
examples.
The trace of a matrix
13.1 De®nition
If A � (aij) is an n 3 n matrix, then the trace of A, written tr A, is
given by
tr A �Xn
i�1
aii:
That is, the trace of A is the sum of the diagonal entries of A.
13.2 Proposition
Let A � (aij) and B � (bij) be n 3 n matrices. Then
tr (A� B) � tr A� tr B, and
tr (AB) � tr (BA):
Moreover, if T is an invertible n 3 n matrix, then
tr (Tÿ1 AT ) � tr A:
Proof The ii-entry of A � B is aii � bii, and the ii-entry of AB isPnj�1 aijbji. Therefore
tr (A� B) �Xn
i�1
(aii � bii) �Xn
i�1
aii �Xn
i�1
bii � tr A� tr B,
and
tr (AB) �Xn
i�1
Xn
j�1
aijbji �Xn
j�1
Xn
i�1
bjiaij � tr (BA):
For the last part,
tr (Tÿ1 AT ) � tr ((Tÿ1 A)T )
� tr (T (Tÿ1 A)) (by the second part )
� tr A: j
Notice that, unlike the determinant function, the trace function is not
multiplicative; that is, tr (AB) need not equal (tr A)(tr B).
Characters
13.3 De®nition
Suppose that V is a CG-module with a basis B . Then the character of
V is the function ÷: G! C de®ned by
÷(g) � tr [g]B (g 2 G):
The character of V does not depend on the basis B , since if B and
B 9 are bases of V, then
[g]B 9 � Tÿ1[g]B T
118 Representations and characters of groups
for some invertible matrix T (see (2.24)), and so by Proposition 13.2,
tr [g]B 9 � tr [g]B for all g 2 G:
Naturally enough, we de®ne the character of a representation
r: G! GL(n, C) to be the character ÷ of the corresponding CG-
module Cn, namely
÷(g) � tr (gr) (g 2 G):
13.4 De®nition
We say that ÷ is a character of G if ÷ is the character of some CG-
module. Further, ÷ is an irreducible character of G if ÷ is the character
of an irreducible CG-module; and ÷ is reducible if it is the character
of a reducible CG-module.
You will have noticed that we are writing characters as functions on
the left. That is, we write ÷(g) and not g÷.
13.5 Proposition
(1) Isomorphic CG-modules have the same character.
(2) If x and y are conjugate elements of the group G, then
÷(x) � ÷(y)
for all characters ÷ of G.
Proof (1) Suppose that V and W are isomorphic CG-modules. Then by
(7.7), there are a basis B 1 of V and a basis B 2 of W such that
[g]B 1� [g]B 2
for all g 2 G:
Consequently tr [g]B 1� tr [g]B 2
for all g 2 G, and so V and W have
the same character.
(2) Assume that x and y are conjugate elements of G, so that
x � gÿ1 yg for some g 2 G. Let V be a CG-module, and let B be a
basis of V. Then
[x]B � [gÿ1 yg]B � [g]ÿ1B [y]B [g]B :
Hence by Proposition 13.2, we have tr [x]B � tr [y]B . Therefore ÷(x) �÷(y), where ÷ is the character of V. j
The result corresponding to Proposition 13.5(1) for representations is
that equivalent representations have the same character.
Characters 119
Later, we shall prove an astonishing converse of Proposition 13.5(1):
if two CG-modules have the same character, then they are isomorphic.
13.6 Examples
(1) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let r: G!GL(2, C) be the representation for which
ar � 0 1
ÿ1 0
� �, br � 1 0
0 ÿ1
� �(see Example 3.2(1)). Let ÷ be the character of this representation. The
following table records g, gr and ÷(g) as g runs through G. (We
obtain ÷(g) by adding the two entries on the diagonal of gr.)
g 1 a a2 a3
gr1 0
0 1
� �0 1
ÿ1 0
� � ÿ1 0
0 ÿ1
� �0 ÿ1
1 0
� �÷(g) 2 0 ÿ2 0
g b ab a2b a3b
gr1 0
0 ÿ1
� �0 ÿ1
ÿ1 0
� � ÿ1 0
0 1
� �0 1
1 0
� �÷(g) 0 0 0 0
(2) Let G �S3, and take V to be the 3-dimensional permutation module
for G over C (see De®nition 4.10). Let B be the natural basis of V;
thus B is the basis v1, v2, v3, where vig � vig for 1 < i < 3 and all
g 2 G. The matrices [g]B (g 2 G) are given by Exercise 4.1. We
record these matrices, together with the character ÷ of V.
g 1 (1 2) (1 3)
[g]B
1 0 0
0 1 0
0 0 1
0@ 1A 0 1 0
1 0 0
0 0 1
0@ 1A 0 0 1
0 1 0
1 0 0
0@ 1A÷(g) 3 1 1
120 Representations and characters of groups
g (2 3) (1 2 3) (1 3 2)
[g]B
1 0 0
0 0 1
0 1 0
0@ 1A 0 1 0
0 0 1
1 0 0
0@ 1A 0 0 1
1 0 0
0 1 0
0@ 1A÷(g) 1 0 0
(3) Let G � C3 � ha: a3 � 1 i. By Theorem 9.8, G has just three
irreducible characters ÷1, ÷2, ÷3, with values
g 1 a a2
÷1(g) 1 1 1÷2(g) 1 ù ù2
÷3(g) 1 ù2 ù
where ù � e2ði=3.
(4) Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l (so G � S3). In
Example 10.8(2), we found a complete set of non-isomorphic irreduci-
ble CG-modules U1, U2, U3. Thus if ÷i is the character of Ui for
1 < i < 3, then the irreducible characters of G are ÷1, ÷2 and ÷3. The
values of these characters on the elements of G can be calculated from
the corresponding representations r1, r2, r3 given in Example 10.8(2),
and they are as follows:
g 1 a a2 b ab a2b
÷1(g) 1 1 1 1 1 1÷2(g) 1 1 1 ÿ1 ÿ1 ÿ1÷3(g) 2 ÿ1 ÿ1 0 0 0
Notice that in all the above examples, the characters given take few
distinct values. This re¯ects the fact that by Proposition 13.5(2), every
character is constant on conjugacy classes of G. Moreover, it is much
quicker to write down the single complex number ÷(g) for the group
element g than to record the matrix which corresponds to g. Never-
theless, the character encapsulates a great deal of information about
the representation. This will become clear as the theory of characters
develops.
Characters 121
13.7 De®nition
If ÷ is the character of the CG-module V, then the dimension of V is
called the degree of ÷.
13.8 Examples
(1) In Example 13.6(1) we gave a character of D8 of degree 2; in
13.6(2) we gave a character of S3 of degree 3; and in 13.6(4) we saw
that the irreducible characters of D6 have degrees 1, 1 and 2.
(2) If V is any 1-dimensional CG-module, then for each g 2 G there is
a complex number ë g such that
vg � ë gv for all v 2 V :
The character ÷ of V is given by
÷(g) � ë g (g 2 G)
and ÷ has degree 1. Characters of degree 1 are called linear characters;
they are, of course, irreducible characters.
Observe that Theorem 9.8 gives all the irreducible characters of
®nite abelian groups; in particular, they are all linear characters.
Every linear character of G is a homomorphism from G to the
multiplicative group of non-zero complex numbers. In fact, these are
the only non-zero characters of G which are homomorphisms (see
Exercise 13.4).
(3) The character of the trivial CG-module (see De®nition 4.8(1)) is a
linear character, called the trivial character of G. We denote it by 1G.
Thus1G: g ! 1 for all g 2 G:
Given any group G, we therefore know at least one of the irreducible
characters of G, namely the trivial character. Finding all the irreducible
characters is usually dif®cult.
The values of a character
The next result gives information about the complex numbers ÷(g),
where ÷ is a character of G and g 2 G.
13.9 Proposition
Let ÷ be the character of a CG-module V. Suppose that g 2 G and g
has order m. Then
122 Representations and characters of groups
(1) ÷(1) � dim V;
(2) ÷(g) is a sum of mth roots of unity;
(3) ÷(gÿ1) � ÷(g);
(4) ÷(g) is a real number if g is conjugate to gÿ1.
Proof (1) Let n � dim V, and let B be a basis of V. Then the matrix
[1]B of the identity element 1 relative to B is equal to In, the n 3 n
identity matrix. Consequently
÷(1) � tr [1]B � tr In � n,
and so ÷(1) � dim V.
(2) By Proposition 9.11 there is a basis B of V such that
[g]B �ù1
. ..
ùn
0B@1CA
where each ùi is an mth root of unity. Therefore
÷(g) � ù1 � : : :� ùn,
a sum of mth roots of unity.
(3) We have
[gÿ1]B �ùÿ1
1
. ..
ùÿ1n
0B@1CA
and so ÷(gÿ1) � ùÿ11 � . . . � ùÿ1
n . Every complex mth root of unity ùsatis®es ùÿ1 � ù, since for all real W,
(eiW)ÿ1 � eÿiW,
which is the complex conjugate of eiW. Therefore
÷(gÿ1) � ù1 � : : :� ùn � ÷(g):
(4) If g is conjugate to gÿ1 then ÷(g) � ÷(gÿ1) by Proposition
13.5(2). Also ÷(gÿ1) � ÷(g) by (3), and so ÷(g) � ÷(g); that is, ÷(g) is
real. j
When the element g of G has order 2, we can be much more
speci®c about the possibilities for ÷(g):
0
0
0
0
Characters 123
13.10 Corollary
Let ÷ be a character of G, and let g be an element of order 2 in G.
Then ÷(g) is an integer, and
÷(g) � ÷(1) mod 2:
Proof By Proposition 13.9, we have
÷(g) � ù1 � : : :� ùn,
where n � ÷(1) and each ùi is a square root of unity. Then each ùi is
�1 or ÿ1. Suppose r of them are �1, and s are ÿ1, so that
÷(g) � r ÿ s, and ÷(1) � r � s:
Certainly then, ÷(g) 2 Z, and since r ÿ s � r � s ÿ 2s � r � s mod 2,
we have ÷(g) � ÷(1) mod 2. j
Our next result gives the ®rst inkling of the importance of characters,
showing that we can determine the kernel of a representation just from
knowledge of its character.
13.11 Theorem
Let r: G! GL(n, C) be a representation of G, and let ÷ be the
character of r.
(1) For g 2 G,
j÷(g)j � ÷(1), gr � ëIn for some ë 2 C:
(2) Ker r � {g 2 G: ÷(g) � ÷(1)}.
Proof (1) Let g 2 G, and suppose that g has order m. If gr � ëIn with
ë 2 C, then ë is an mth root of unity, and ÷(g) � në, so
|÷(g)| � n � ÷(1).
Conversely, suppose that |÷(g)| � ÷(1). By Proposition 9.11, there is
a basis B of Cn such that
[g]B �ù1
. ..
ùn
0B@1CA
where each ùi is an mth root of unity. Then
j÷(g)j � jù1 � : : :� ùnj � ÷(1) � n:(13:12)
0
0
124 Representations and characters of groups
Note now that for any complex numbers z1, . . . , zn, we have
jz1 � : : :� znj < jz1j � : : :� jznj,with equality if and only if the arguments of z1, . . . , zn are all equal.
(To see this, consider the picture
in the Argand diagram.) Since |ùi| � 1 for all i, we deduce from
(13.12) that ùi � ù j for all i, j. Thus
[g]B �ù1
. ..
ù1
0B@1CA � ù1 In:
Hence for all bases B 9 of Cn we have [g]B 9 � ù1 In, and so
gr � ù1 In. This completes the proof of (1).
(2) If g 2 Ker r then gr � In, and so ÷(g) � n � ÷(1).
Conversely, suppose that ÷(g) � ÷(1). Then by (1), we have gr � ëIn
for some ë 2 C. This implies that ÷(g) � ë÷(1), whence ë � 1. There-
fore gr � In, and so g 2 Ker r. Part (2) follows. j
Motivated by Theorem 13.11(2), we de®ne the kernel of a character
as follows.
13.13 De®nition
If ÷ is a character of G, then the kernel of ÷, written Ker ÷, is de®ned
by
Ker ÷ � fg 2 G: ÷(g) � ÷(1)g:By Theorem 13.11(2), if r is a representation of G with character ÷,
then Ker r � Ker ÷. In particular, Ker ÷ v G. We call ÷ a faithful
character if Ker ÷ � {1}.
13.14 Examples
(1) According to Example 13.6(4), the irreducible characters of the
group G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l are ÷1, ÷2, ÷3, with
the following values:
0
0
Characters 125
g 1 a a2 b ab a2b
÷1(g) 1 1 1 1 1 1÷2(g) 1 1 1 ÿ1 ÿ1 ÿ1÷3(g) 2 ÿ1 ÿ1 0 0 0
Then Ker ÷1 � G, Ker ÷2 � kal and Ker ÷3 � {1}. In particular, ÷3 is a
faithful irreducible character of D6.
(2) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let ÷ be the
character of G given in Example 13.6(1):
g 1 a a2 a3 b ab a2b a3b
÷(g) 2 0 ÿ2 0 0 0 0 0
Then Ker ÷ � {1}, so ÷ is a faithful character. And since |÷(a2)| �|ÿ2| � ÷(1), Theorem 13.11(1) implies that if r: G! GL(2, C) is a
representation with character ÷, then a2r � ÿI.
We next prove a result which is sometimes useful for constructing a
new character from a given one. For a character ÷ of G, de®ne
÷: G! C by
÷(g) � ÷(g) (g 2 G):
Thus the values of ÷ are the complex conjugates of the values of ÷.
13.15 Proposition
Let ÷ be a character of G. Then ÷ is a character of G. If ÷ is
irreducible, then so is ÷.
Proof Suppose that ÷ is the character of a representation r: G!GL(n, C). Thus
÷(g) � tr (gr) (g 2 G):
If A � (aij) is an n 3 n matrix over C, then we de®ne A to be the
n 3 n matrix (aij). Observe that if A � (aij) and B � (bij) are n 3 n
matrices over C, then
(AB) � A B,(13:16)
126 Representations and characters of groups
since the ij-entry of AB is Xn
k�1
aik bkj,
which is equal to the complex conjugate ofP
nk�1 aikbkj, the ij-entry of
AB.
It follows from (13.16) that the function r: G! GL(n, C) de®ned by
gr � (gr) (g 2 G)
is a representation of G. Since
tr (gr) � tr (gr) � tr (gr) � ÷(g) (g 2 G),
the character of the representation r is ÷.
It is clear that if r is reducible then r is reducible. Hence ÷ is
irreducible if and only if ÷ is irreducible. j
The regular character
13.17 De®nition
The regular character of G is the character of the regular CG-module.
We write the regular character as ÷reg.
In Theorem 13.19, we shall express the regular character in terms of
the irreducible characters of G. First we need a preliminary result.
13.18 Proposition
Let V be a CG-module, and suppose that
V � U1 � : : :� Ur,
a direct sum of irreducible CG-modules Ui. Then the character of V is
equal to the sum of the characters of the CG-modules U1, . . . , Ur.
Proof This is immediate from (7.10). j
13.19 Theorem
Let V1, : : : , Vk be a complete set of non-isomorphic irreducible
CG-modules (see De®nition 11.11), and for i � 1, . . . , k let ÷i be the
character of Vi and di � ÷i(1). Then
÷reg � d1÷1 � : : :� dk÷k :
Characters 127
Proof By Theorem 11.9,
CG � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk),
where for each i there are di factors Vi. Now the result follows from
Proposition 13.18. j
The values of ÷reg on the elements of G are easily described, and are
given in the next result.
13.20 Proposition
If ÷reg is the regular character of G, then
÷reg(1) � jGj, and
÷reg(g) � 0 if g 6� 1:
Proof Let g1, . . . , gn be the elements of G, and let B be the basis
g1, . . . , gn of CG. By Proposition 13.9(1), ÷reg(1) � dim CG � |G|.
Now let g 2 G with g 6� 1. Then for 1 < i < n, we have gig � gj
for some j with j 6� i. Therefore the ith row of the matrix [g]B has
zeros in every place except column j; in particular, the ii-entry is zero
for all i. It follows that
÷reg(g) � tr [g]B � 0: j
13.21 Example
We illustrate Theorem 13.19 and Proposition 13.20 for the group
G � D6. By Example 13.6(4), the irreducible characters of G are ÷1,
÷2, ÷3:
g 1 a a2 b ab a2b
÷1(g) 1 1 1 1 1 1÷2(g) 1 1 1 ÿ1 ÿ1 ÿ1÷3(g) 2 ÿ1 ÿ1 0 0 0
We calculate ÷1 � ÷2 � 2÷3:
(÷1 � ÷2 � 2÷3)(g) 6 0 0 0 0 0
128 Representations and characters of groups
This is the regular character of G, by Theorem 13.19; and it takes the
value |G| on 1, and the value 0 on all non-identity elements of G,
illustrating Proposition 13.20.
Permutation characters
In the case where G is a subgroup of the symmetric group Sn, there is
an easy construction using the permutation module which produces a
character of degree n, and we now describe this.
Suppose that G is a subgroup of Sn, so that G is a group of
permutations of {1, . . . , n}. The permutation module V for G over C
has basis v1, . . . , vn, where for all g 2 G,
vi g � vig (1 < i < n)
(see De®nition 4.10). Let B denote the basis v1, . . . , vn. Then the ii-
entry in the matrix [g]B is 0 if ig 6� i, and is 1 if ig � i. Therefore
the character ð of the permutation module V is given by
ð(g) � (the number of i such that ig � i):
For g 2 G, let
fix (g) � fi: 1 < i < n and ig � ig:Then
ð(g) � jfix(g)j (g 2 G):(13:22)
We call ð the permutation character of G.
13.23 Example
Let G � S4. Then by Example 12.16(3), G has ®ve conjugacy classes,
with representatives
1, (1 2), (1 2 3), (1 2)(3 4), (1 2 3 4):
The permutation character ð takes the values
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
ð(gi) 4 2 1 0 0
Characters 129
13.24 Proposition
Let G be a subgroup of Sn. Then the function í: G! C de®ned by
í(g) � jfix (g)j ÿ 1 (g 2 G)
is a character of G.
Proof Let v1, . . . , vn be the basis of the permutation module V as
above, and let
u � v1 � : : :� vn, and U � sp (u):
Observe that ug � u for all g 2 G, so U is a CG-submodule of V.
Indeed, U is isomorphic to the trivial CG-module, so the character of
U is the trivial character 1G (see Example 13.8(3)). By Maschke's
Theorem 8.1, there is a CG-submodule W of V such that
V � U � W :
Let í be the character of W. Then
ð � 1G � í,
so |®x( g)| � 1 � í(g) for all g 2 G, and therefore
í(g) � jfix(g)j ÿ 1 (g 2 G): j
13.25 Example
Let G � A4, a subgroup of S4. By Example 12.18(1), the conjugacy
classes of G are represented by
1, (1 2)(3 4), (1 2 3), (1 3 2):
The values of the character í of G are
gi 1 (1 2)(3 4) (1 2 3) (1 3 2)
í(gi) 3 ÿ1 0 0
Summary of Chapter 13
1. A character is obtained from a representation by taking the trace of
each matrix.
2. Characters are constant on conjugacy classes.
130 Representations and characters of groups
3. Isomorphic CG-modules have the same character.
4. For all characters ÷ of G, and all g 2 G, the complex number ÷(g)
is a sum of roots of unity, and ÷(gÿ1) � ÷(g).
5. The character of a representation determines the kernel of the
representation.
6. The regular character ÷reg of G takes the value |G| on the identity
and the value 0 on all other elements of G.
7. If G is a subgroup of Sn, then the function í which is given by
í(g) � jfix(g)j ÿ 1 (g 2 G)
is a character of G.
Exercises for Chapter 13
1. Let G � D12 � ka, b: a6 � b2 � 1, bÿ1ab � aÿ1l, and let r1, r2 be
the representations of G for which
ar1 � ù 0
0 ùÿ1
� �, br1 � 0 1
1 0
� �(where ù � e2ði=3)
and
ar2 � ÿ1 0
0 1
� �, br2 � 1 0
0 ÿ1
� �:
Find the characters of r1 and r2. Find also Ker r1 and Ker r2;
check that your answers are consistent with Theorem 13.11.
2. Find all the irreducible characters of C4. Write the regular char-
acter of C4 as a linear combination of these.
3. Let ÷ be the character of the 7-dimensional permutation module
for S7. Find ÷(x) for x � (1 2) and for x � (1 6)(2 3 5).
4. Prove that the only non-zero characters of G which are homo-
morphisms are the linear characters.
5. Assume that ÷ is an irreducible character of G. Suppose that
z 2 Z(G) and that z has order m. Prove that there exists an mth
root of unity ë 2 C such that for all g 2 G,
÷(zg) � ë÷(g):
6. Prove that if ÷ is a faithful irreducible character of the group G,
then Z(G) � {g 2 G: |÷(g)| � ÷(1)}.
Characters 131
7. Let r be a representation of the group G over C.
(a) Show that ä: g! det (gr) (g 2 G) is a linear character of G.
(b) Prove that G/Ker ä is abelian.
(c) Assume that ä(g) � ÿ1 for some g 2 G. Show that G has a
normal subgroup of index 2.
8. Let g be a group of order 2k, where k is an odd integer. By
considering the regular representation of G, show that G has a
normal subgroup of index 2.
9. Let ÷ be a character of a group G, and let g be an element of
order 2 in G. Show that either
(1) ÷(g) � ÷(1) mod 4, or
(2) G has a normal subgroup of index 2.
(Compare Corollary 13.10. Hint: use Exercise 7.)
10. Prove that if x is a non-identity element of the group G, then
÷(x) 6� ÷(1) for some irreducible character ÷ of G.
132 Representations and characters of groups
133
14
Inner products of characters
We establish some signi®cant properties of characters in this chapter,
and in particular we prove the striking result (Theorem 14.21) that if
two CG-modules have the same character then they are isomorphic.
Also, we describe a method for decomposing a given CG-module as a
direct sum of CG-submodules, using characters.
The proofs rely on an inner product involving the characters of a
group, and we describe this ®rst.
Inner products
The characters of a ®nite group G are certain functions from G to C.
The set of all functions from G to C forms a vector space over C, if
we adopt the natural rules for adding functions and multiplying func-
tions by complex numbers. That is, if W, ö are functions from G to C,
and ë 2 C, then we de®ne W � ö: G! C by
(W� ö)(g) � W(g)� ö(g) (g 2 G)
and we de®ne ëW: G! C by
ëW(g) � ë(W(g)) (g 2 G):
(We write these functions on the left to agree with our notation for
characters.)
14.1 Example
Let G � C3 � ka: a3 � 1l, and suppose that W: G! C and ö: G! C
are given by
This means that W(1) � 2, W(a) � i, W(a2) � ÿ1 and ö(1) � ö(a) �ö(a2) � 1. Then W � ö and 3W are given by
We shall often think of functions from G to C as row vectors, as in
this example.
The vector space of all functions from G to C can be equipped with
an inner product in a way which we shall describe shortly. The
de®nition of an inner product on a vector space over C runs as
follows. With every ordered pair of vectors W, ö in the vector space,
there is associated a complex number kW, öl which satis®es the follow-
ing conditions:
(14.2) (a) kW, öl � hö, Wi for all W, ö;
(b) kë1W1 � ë2W2, öl � ë1kW1, öl � ë2kW2, öl for all ë1, ë2 2 C
and all vectors W1, W2, ö;
(c) kW, Wl . 0 if W 6� 0.
Notice that condition (a) implies that kW, Wl is always real, and that
conditions (a) and (b) give
hö, ë1è1 � ë2W2i � ë1hö, W1i � ë2hö, W2ifor all ë1, ë2 2 C and all vectors W1, W2, ö.
We now introduce an inner product on the vector space of all
functions from G to C. This will be of basic importance in our study
of characters.
14.3 De®nition
Suppose that W and ö are functions from G to C. De®ne
hW, öi � 1
jGjXg2G
W(g)ö(g):
1 a a2
W � ö 3 1 � i 03W 6 3i ÿ3
1 a a2
W 2 i ÿ1ö 1 1 1
134 Representations and characters of groups
It is transparent that the conditions of (14.2) hold, so k , l is an inner
product on the vector space of functions from G to C.
14.4 Example
As in Example 14.1, suppose that G � C3 � ka: a3 � 1l and that W and
ö are given by
Then
hè, öi � 13(2 . 1� i . 1ÿ 1 . 1) � 1
3(1� i),
hè, èi � 13(2 . 2� i . i� (ÿ1) . (ÿ1)) � 2,
hö, öi � 13(1 . 1� 1 . 1� 1 . 1) � 1:
Inner products of characters
We can exploit the fact that characters are constant on conjugacy
classes to simplify slightly the calculation of the inner product of two
characters.
14.5 Proposition
Assume that G has exactly l conjugacy classes, with representatives
g1, . . . , gl. Let ÷ and ø be characters of G.
(1) h÷, øi � hø, ÷i � 1
jGjXg2G
÷(g)ø(gÿ1), and this is a real number:
(2) h÷, øi �Xl
i�1
÷(gi)ø(gi)
jCG(gi)j :
Proof (1) We have ø(g) � ø(gÿ1) for all g 2 G, by Proposition
13.9(3). Therefore
h÷, øi � 1
jGjXg2G
÷(g)ø(gÿ1):
1 a a2
W 2 i ÿ1ö 1 1 1
Inner products of characters 135
Since {gÿ1: g 2 G} � G, we also have
h÷, øi � 1
jGjXg2G
÷(gÿ1)ø(g) � hø, ÷i:
Since kø, ÷l � h÷, øi, it follows that h÷, øi is real. (We shall prove
later that h÷, øi is, in fact, an integer.)
(2) Recall that gGi denotes the conjugacy class of G which contains
gi. Since characters are constant on conjugacy classes,Xg2 gG
i
÷(g)ø(g) � jgGi j÷(gi)ø(gi):
Now
G �[l
i�1
gGi and jgG
i j � jGj=jCG(gi)j,
by Corollary 12.3 and Theorem 12.8. Hence
h÷, øi � 1
jGjXg2G
÷(g)ø(g) � 1
jGjXl
i�1
Xg2 gG
i
÷(g)ø(g)
�Xl
i�1
jgGi jjGj ÷(gi)ø(gi)
�Xl
i�1
1
jCG(gi)j ÷(gi)ø(gi): j
14.6 Example
The alternating group A4 has four conjugacy classes, with represent-
atives
g1 � 1, g2 � (1 2)(3 4), g3 � (1 2 3), g4 � (1 3 2)
(see Example 12.18(1)). We shall see in Chapter 18 that there are
characters ÷ and ø of A4 which take the following values on the
representatives gi:
gi g1 g2 g3 g4
|CG(gi)| 12 4 3 3
÷ 1 1 ù ù2
ø 4 0 ù2 ù
136 Representations and characters of groups
(where ù � e2ði=3). Using part (2) of Proposition 14.5, we have
h÷, øi � 1 . 4
12� 1 . 0
4� ù . ù2
3� ù2 . ù
3� 0,
hø, øi � 4 . 4
12� 0 . 0
4� ù2 . ù2
3� ù . ù
3� 2:
We advise you to check also that k÷, ÷l � 1, and to ®nd the inner
products of ÷ and ø with the trivial character (which takes the value 1
on all elements of A4).
We are now going to pave the way to proving the key fact (Theorem
14.12) that the irreducible characters of G form an orthonormal set of
vectors in the vector space of functions from G to C; that is, for
distinct irreducible characters ÷ and ø of G, we have h÷, ÷i � 1 and
h÷, øi � 0.
Recall from Chapter 10 that the regular CG-module is a direct sum
of irreducible CG-submodules, say
CG � U1 � : : :� Ur,
and that every irreducible CG-module is isomorphic to one of the CG-
modules U1, . . . , Ur. There are several ways of choosing CG-sub-
modules W1 and W2 of CG such that CG � W1 � W2 and W1 and W2
have no common composition factor (see De®nition 10.4). For example,
we may take W1 to be the sum of those irreducible CG-submodules Ui
which are isomorphic to a given irreducible CG-module, and then let
W2 be the sum of the remaining CG-modules Ui. We shall investigate
some consequences of writing CG like this; therefore, we temporarily
adopt the following Hypothesis:
14.7 Hypothesis
Let CG � W1 � W2, where W1 and W2 are CG-submodules which have
no common composition factor. Write 1 � e1 � e2 where e1 2 W1 and
e2 2 W2.
Among other results, we shall derive a formula for e1 in terms of
the character of W1.
We ®rst look at the effect of applying the elements e1 and e2 of CG
to W1 and W2.
Inner products of characters 137
14.8 Proposition
For all w1 2 W1 and w2 2 W2, we have
w1e1 � w1, w2e1 � 0,
w1e2 � 0, w2e2 � w2:
Proof If w1 2 W1 then the function w2 ! w1w2 (w2 2 W2) is clearly a
CG-homomorphism from W2 to W1. But W2 and W1 have no common
composition factor, so every CG-homomorphism from W2 to W1 is
zero, by Proposition 11.3. Therefore w1w2 � 0 for all w1 2 W1,
w2 2 W2. Similarly w2w1 � 0. In particular, w1e2 � w2e1 � 0. Now
w1 � w11 � w1(e1 � e2) � w1e1, and
w2 � w21 � w2(e1 � e2) � w2e2,
and this completes the proof. j
14.9 Corollary
For the elements e1 and e2 of CG which appear in Hypothesis 14.7,
we have
e21 � e1, e2
2 � e2 and e1e2 � e2e1 � 0:
Proof In Proposition 14.8, take w1 � e1 and w2 � e2. j
Next, we evaluate e1.
14.10. Proposition
Let ÷ be the character of the CG-module W1 which appears in
Hypothesis 14.7. Then
e1 � 1
jGjXg2G
÷(gÿ1)g:
Proof Let x 2 G. The function
W: w! we1xÿ1 (w 2 CG)
is an endomorphism of CG. We shall calculate the trace of W in two
ways.
138 Representations and characters of groups
First, for w1 2 W1 and w2 2 W2 we have, in view of Proposition 14.8,
w1W � w1e1xÿ1 � w1xÿ1,
w2W � w2e1xÿ1 � 0:
Thus W acts on W1 by w1 ! w1xÿ1 and on W2 by w2 ! 0. By the
de®nition of the character ÷ of W1, the endomorphism w1 ! w1xÿ1 of
W1 has trace equal to ÷(xÿ1), and of course the endomorphism w2 ! 0
of W2 has trace 0. Therefore
tr W � ÷(xÿ1):
Secondly, e1 2 CG, so
e1 �Xg2G
ë g g
for some ë g 2 C. By Proposition 13.20, the endomorphism w! wgxÿ1
(w 2 CG) of CG has trace 0 if g 6� x and has trace |G| if g � x.
Hence, as W: w! wP
g2G ë g gxÿ1, we have
tr W � ëxjGj:Comparing our two expressions for tr W, we see that for all x 2 G,
ëx � ÷(xÿ1)=jGj:Therefore
e1 � 1
jGjXg2G
÷(gÿ1)g:j
14.11 Corollary
Let ÷ be the character of the CG-module W1 which appears in
Hypothesis 14.7. Then
h÷, ÷i � ÷(1):
Proof Using the de®nition 6.3 of the multiplication in CG, we deduce
from Proposition 14.10 that the coef®cient of 1 in e21 is
1
jGj2Xg2G
÷(gÿ1)÷(g) � 1
jGj h÷, ÷i:
Inner products of characters 139
On the other hand, we know from Corollary 14.9 that e21 � e1, and the
coef®cient of 1 in e1 is ÷(1)/|G|. Hence k÷, ÷l � ÷(1), as required.
j
We can now prove the main theorem concerning the inner product
k , l.
14.12 Theorem
Let U and V be non-isomorphic irreducible CG-modules, with char-
acters ÷ and ø, respectively. Then
h÷, ÷i � 1, and
h÷, øi � 0:
Proof Recall from Theorem 11.9 that CG is a direct sum of irreducible
CG-submodules, say
CG � U1 � : : :� Ur,
where the number of CG-submodules Ui which are isomorphic to U is
dim U. Let m � dim U, and de®ne W to be the sum of the m
irreducible CG-submodules Ui which are isomorphic to U; let X be the
sum of the remaining CG-submodules Ui. Then
CG � W � X :
Moreover, every composition factor of W is isomorphic to U, and no
composition factor of X is isomorphic to U. In particular, W and X
have no common composition factor. The character of W is m÷, since
W is the direct sum of m CG-submodules, each of which has
character ÷.
We now apply Corollary 14.11 to the character of W, and obtain
hm÷, m÷i � m÷(1):
As ÷(1) � dim U � m, this yields h÷, ÷i � 1.
Next, let Y be the sum of those CG-submodules Ui of CG which
are isomorphic to either U or V, and let Z be the sum of the remaining
CG-submodules Ui. Then
CG � Y � Z,
140 Representations and characters of groups
and Y and Z have no common composition factor. The character of Y
is m÷ � nø, where n � dim V. By Corollary 14.11,
m÷(1)� nø(1) � hm÷� nø, m÷� nøi� m2h÷, ÷i � n2hø, øi � mn(h÷, øi � hø, ÷i):
Now h÷, ÷i � hø, øi � 1, by the part of the theorem which we have
already proved; and ÷(1) � m, ø(1) � n. Therefore
h÷, øi � hø, ÷i � 0:
By Proposition 14.5(1), k÷, øl � kø, ÷l, and hence k÷, øl � 0. j
Applications of Theorem 14.12
Let G be a ®nite group, and let V1, : : : , Vk be a complete set of non-
isomorphic irreducible CG-modules (see De®nition 11.11). If ÷i is the
character of Vi (1 < i < k), then by Theorem 14.12, we have
h÷i, ÷ ji � äij for all i, j,(14:13)
where äij is the Kronecker delta function (that is, äij is 1 if i � j and
is 0 if i 6� j). In particular, this implies that the irreducible characters
÷1, . . . , ÷k are all distinct.
Now let V be a CG-module. By Theorem 8.7, V is equal to a direct
sum of irreducible CG-submodules. Each of these is isomorphic to
some Vi, so there are non-negative integers d1, . . . , dk such that
V � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :(14:14)
� (Vk � : : :� Vk),
where for each i, there are di factors Vi:
Therefore the character ø of V is given by
ø � d1÷1 � : : :� dk÷k :(14:15)
Using (14.13), we obtain from this
hø, ÷ii � h÷i, øi � di for 1 < i < k, and(14:16)
hø, øi � d21 � : : :� d2
k :
Summarizing, we have
Inner products of characters 141
14.17 Theorem
Let ÷1, . . . , ÷k be the irreducible characters of G. If ø is any
character of G, thenø � d1÷1 � : : :� dk÷k
for some non-negative integers d1, . . . , dk . Moreover,
di � hø, ÷ii for 1 < i < k, and
hø, øi �Xk
i�1
d2i :
14.18 Example
Recall from Example 13.6(4) that the irreducible characters of S3 � D6
are ÷1, ÷2, ÷3, taking the following values on the conjugacy class
representatives 1, (1 2), (1 2 3):
Now let ø be the character of the 3-dimensional permutation module
for S3. By Example 13.6(2), we know that
ø(1) � 3, ø(1 2) � 1, ø(1 2 3) � 0:
Therefore, by Proposition 14.5(2),
hø, ÷1i � 3 . 1
6� 1 . 1
2� 0 � 1:
Similarly, kø, ÷2l � 0 and kø, ÷3l � 1. Thus by Theorem 14.17,
ø � ÷1 � ÷3:
(This can of course be checked immediately by comparing the values
of ø and ÷1 � ÷3 on each conjugacy class representative.)
A more substantial calculation along these lines is given in Example
15.7.
gi 1 (1 2) (1 2 3)|CS3
(gi)| 6 2 3
÷1 1 1 1÷2 1 ÿ1 1÷3 2 0 ÿ1
142 Representations and characters of groups
We shall see many more applications of the important Theorem
14.17.
14.19 De®nition
Suppose that ø is a character of G, and that ÷ is an irreducible
character of G. We say that ÷ is a constituent of ø if kø, ÷l 6� 0. Thus,
the constituents of ø are the irreducible characters ÷i of G for which
the integer di in the expression ø � d1÷1 � . . . � dk÷k is non-zero.
The next result is another signi®cant consequence of Theorem 14.12.
It gives us a quick and effective method of determining whether or not
a given CG-module is irreducible.
14.20 Theorem
Let V be a CG-module with character ø. Then V is irreducible if and
only if kø, øl � 1.
Proof If V is irreducible then kø, øl � 1 by Theorem 14.12.
Conversely, assume that kø, øl � 1. We have
ø � d1÷1 � : : :� dk÷k
for some non-negative integers di, and by (14.16),
1 � hø, øi � d21 � : : :� d2
k :
It follows that one of the integers di is 1 and the rest are zero. Then
by (14.14), V � Vi for some i, and so V is irreducible. j
We are now in a position to prove the remarkable result that `a CG-
module is determined by its character'. It is this fact which motivates
our study of characters in much of the rest of the book, for it means
that many questions about CG-modules can be answered using char-
acter theory.
14.21 Theorem
Suppose that V and W are CG-modules, with characters ÷ and ø,
respectively. Then V and W are isomorphic if and only if ÷ � ø.
Proof In Proposition 13.5 we proved the elementary fact that if V � W
then ÷ � ø. It is the converse which is the substantial part of this
theorem.
Inner products of characters 143
Thus, suppose that ÷ � ø. Again let V1, : : : , Vk be a complete set
of non-isomorphic irreducible CG-modules with characters ÷1, . . . , ÷k .
We know by (14.14) that there are non-negative integers ci, di
(1 < i < k) such that
V � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk)
with ci factors Vi for each i, and
W � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk)
with di factors Vi for each i. By (14.16),
ci � h÷, ÷ii, di � hø, ÷ii (1 < i < k):
Since ÷ � ø, it follows that ci � di for all i, and hence V � W. j
14.22 Example
Let G � C3 � ka: a3 � 1l, and let r1, r2, r3, r4 be the representations
of G over C for which
ar1 �ù 0
0 ù
!, ar2 �
ù 0
0 ùÿ1
!,
ar3 �0 1
ÿ1 ÿ1
!, ar4 �
1 ùÿ1
0 ù
!(ù � e2ði=3). The characters øi of the representations ri (i � 1, 2, 3, 4)
are
Hence by Theorem 14.21, the representations r2 and r3 are equivalent,
but there are no other equivalences among r1, r2, r3 and r4.
The next theorem is another consequence of Theorem 14.12.
1 a a2
ø1 2 2ù 2ù2
ø2 2 ÿ1 ÿ1ø3 2 ÿ1 ÿ1ø4 2 1 � ù 1 � ù2
144 Representations and characters of groups
14.23 Theorem
Let ÷1, . . . , ÷k be the irreducible characters of G. Then ÷1, . . . , ÷k
are linearly independent vectors in the vector space of all functions
from G to C.
Proof Assume that
ë1÷1 � : : :� ëk÷k � 0 (ëi 2 C):
Then for all i, using (14.13) we have
0 � hë1÷1 � : : :� ëk÷k , ÷ii � ëi:
Therefore ÷1, . . . , ÷k are linearly independent. j
We now relate inner products of characters to the spaces of CG-
homomorphisms which we constructed in Chapter 11.
14.24 Theorem
Let V and W be CG-modules with characters ÷ and ø, respectively.
Then
dim (HomCG (V , W )) � h÷, øi:
Proof We know from (14.14) that there are non-negative integers ci, di
(1 < i < k) such that
V � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk)
with ci factors Vi for each i, and
W � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk)
with di factors Vi for each i. By Proposition 11.2, for any i, j we have
dim (HomCG (Vi, Vj)) � äij:
Hence, using (11.5)(3) we see that
dim (HomCG (V , W )) �Xk
i�1
cidi:
Inner products of characters 145
On the other hand,
÷ �Xk
i�1
ci÷i and ø �Xk
i�1
di÷i
and so (14.13) implies that
h÷, øi �Xk
i�1
cidi:
The result follows. j
Decomposing CG-modules
It is sometimes of practical importance to be able to decompose a
given CG-module into a direct sum of CG-submodules, and we now
describe a process for doing this.
Once more we adopt Hypothesis 14.7:
CG � W1 � W2, where the CG-modules W1 and W2
have no common composition factor; and 1 � e1 � e2
with e1 2 W1, e2 2 W2.
Let V be any CG-module. We can write V � V1 � V2, where every
composition factor of V1 is a composition factor of W1 and every
composition factor of V2 is a composition factor of W2.
14.25 Proposition
With the above notation, for all v1 2 V1 and v2 2 V2 we have
v1e1 � v1, v2e1 � 0,
v1e2 � 0, v2e2 � v2:
Proof If v1 2 V1 then the function w2 ! v1w2 (w2 2 W2) is clearly a
CG-homomorphism from W2 to V1. Since W2 and V1 have no common
composition factor, we deduce the stated results just as in the proof of
Proposition 14.8. j
14.26 Proposition
If ÷ is an irreducible character of G, and V is any CG-module, then
VXg2G
÷(gÿ1)g
!
146 Representations and characters of groups
is equal to the sum of those CG-submodules of V which have character
÷ (where for r 2 CG, we de®ne Vr � fvr: v 2 Vg).
Proof Write
CG � U1 � : : :� Ur,
a direct sum of irreducible CG-submodules Ui. Let W1 be the sum of
those CG-submodules Ui which have character ÷, and let W2 be the
sum of the remaining CG-submodules Ui. Then the character of W1 is
m÷ where m � ÷(1), by Theorem 11.9. Also W1 and W2 satisfy
Hypothesis 14.7, and by Proposition 14.10, the element e1 of W1 is
given by
e1 � m
jGjXg2G
÷(gÿ1)g:
Let V1 be the sum of those CG-submodules of V which have
character ÷. Then Proposition 14.25 shows that Ve1 � V1. Clearly we
may omit the constant multiplier m/|G|, so
V1 � VXg2G
÷(gÿ1)g
!:
j
Once the irreducible characters of our group G are known, Proposi-
tion 14.26 provides a useful practical tool for ®nding CG-submodules
of a given CG-module V. The procedure is as follows:
(14.27) (1) Choose a basis v1, . . . , vn of V.
(2) For each irreducible character ÷ of G, calculate the vectors
vi(P
g2G÷(gÿ1)g) for 1 < i < n, and let V÷ be the sub-
space of V spanned by these vectors.
(3) Then V is the direct sum of the CG-modules V÷ as ÷ runs
over the irreducible characters of G. The character of V÷ is
a multiple of ÷.
We illustrate this method with a couple of simple examples. Some
more complicated uses of the method can be found in Chapter 32.
14.28 Examples
(1) Let G be any ®nite group and let V be any non-zero CG-module.
Taking ÷ to be the trivial character of G in Proposition 14.26, we see
that
Inner products of characters 147
VXg2G
g
!is the sum of all the trivial CG-submodules of V. For example, let
G � Sn and let V be the permutation module, with basis v1, . . . , vn
such that vig � vig for all i and all g 2 G. Then
VXg2G
g
!� sp (v1 � : : :� vn):
Hence V has a unique trivial CG-submodule.
(2) Let G be the subgroup of S4 which is generated by
a � (1 2 3 4) and b � (1 2)(3 4):
Then G � D8 (compare Example 1.5). Here is a list of the irreducible
characters ÷1, . . . , ÷5 of D8 (see Example 16.3(3)):
1 a a2 a3 b ab a2b a3b
÷1 1 1 1 1 1 1 1 1÷2 1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1÷3 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1÷4 1 ÿ1 1 ÿ1 ÿ1 1 ÿ1 1÷5 2 0 ÿ2 0 0 0 0 0
Let V be the permutation module for G, with basis v1, v2, v3, v4
such that vig � vig for all i and all g 2 G.
For 1 < i < 5, let
ei � ÷i(1)
8
Xg2G
÷i(gÿ1)g:
For example, e5 � 12(1 ÿ a2). Then
Ve1 � sp (v1 � v2 � v3 � v4),
Ve2 � 0,
Ve3 � 0,
Ve4 � sp (v1 ÿ v2 � v3 ÿ v4),
Ve5 � sp (v1 ÿ v3, v2 ÿ v4):
148 Representations and characters of groups
We have
V � Ve1 � Ve4 � Ve5,
and so we have expressed V as a direct sum of irreducible CG-
submodules whose characters are ÷1, ÷4 and ÷5, respectively.
You might like to check that
e1 � : : :� e5 � 1,
e2i � ei for 1 < i < 5,
eiej � 0 for i 6� j:
Compare these results with Corollary 14.9.
Note that the procedure described in (14.27) does not in general
enable us to write a given CG-module as a direct sum of irreducible
CG-submodules (since V÷ is not in general irreducible).
Summary of Chapter 14
1. The inner product of two functions W, ö from G to C is given by
hW, öi � 1
jGjXg2G
W(g)ö(g):
2. The irreducible characters ÷1, . . . , ÷k of G form an orthonormal
set; that is, h÷i, ÷ ji � äij for all i, j.
3. Every CG-module is determined by its character.
4. If ÷1, . . . , ÷k are the irreducible characters of G, and ø is any
character, then
ø � d1÷1 � : : :� dk÷k where di � hø, ÷ii:Each di is a non-negative integer. Also, ø is irreducible if and only
if kø, øl � 1.
Inner products of characters 149
Exercises for Chapter 14
1. Let G � S4. We shall see in Chapter 18 that G has characters ÷ and
ø which take the following values on the conjugacy classes:
Classrepresentative
1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
|CG(gi)| 24 4 3 8 4
÷ 3 ÿ1 0 3 ÿ1ø 3 1 0 ÿ1 ÿ1
Calculate h÷, ÷i, h÷, øi and hø, øi. Which of ÷ and ø is irreducible?
2. Let G � Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l, and let r1, r2,
r3 be the representations of G over C for which
ar1 �i 0
0 ÿi
!, br1 �
0 1
ÿ1 0
!,
ar2 �0 i
i 0
!, br2 �
0 ÿ1
1 0
!,
ar3 �ÿ1 0
0 1
!, br3 �
1 0
0 ÿ1
!:
Show that r1 and r2 are equivalent, but r3 is not equivalent to r1
or r2.
3. Suppose that r and ó are representations of G, and that for each g
in G there is an invertible matrix Tg such that
gó � Tÿ1g (gr)Tg:
Prove that there is an invertible matrix T such that for all g in G,
gó � Tÿ1(gr)T :
4. Suppose that ÷ is a non-zero, non-trivial character of G, and that
÷(g) is a non-negative real number for all g in G. Prove that ÷ is
reducible.
5. If ÷ is a character of G, show that
h÷reg, ÷i � ÷(1):
150 Representations and characters of groups
6. If ð is the permutation character of Sn, prove that
hð, 1Sni � 1:
(Hint: you may ®nd Exercise 11.4 relevant.)
7. Let ÷1, . . . , ÷k be the irreducible characters of the group G, and
suppose that
ø � d1÷1 � : : :dk÷k
is a character of G. What can you say about the integers di in the
cases kø, øl � 1, 2, 3 or 4?
8. Suppose that ÷ is a character of G and that for every g 2 G, ÷(g) is
an even integer. Does it follow that ÷ � 2ö for some character ö?
Inner products of characters 151
152
15
The number of irreducible characters
We devote this chapter to the theorem which states that the number of
irreducible characters of a ®nite group is equal to the number of
conjugacy classes of the group, and to some consequences of this
theorem. Together with the material from Chapter 14, the theorem
provides machinery for investigating characters which is used in the
remainder of the book.
Throughout, G is as usual a ®nite group.
Class functions
15.1 De®nition
A class function on G is a function ø: G! C such that ø(x) � ø(y)
whenever x and y are conjugate elements of G (that is, ø is constant
on conjugacy classes).
By Proposition 13.5(2), the characters of G are class functions on G.
The set C of all class functions on G is a subspace of the vector space
of all functions from G to C. A basis of C is given by those functions
which take the value 1 on precisely one conjugacy class and zero on
all other classes. Thus, if l is the number of conjugacy classes of G,
then
dim C � l:(15:2)
15.3 Theorem
The number of irreducible characters of G is equal to the number of
conjugacy classes of G.
Proof Let ÷1, . . . , ÷k be the irreducible characters of G, and let l be
the number of conjugacy classes of G. By Theorem 14.23, ÷1, . . . , ÷k
are linearly independent elements of C, so (15.2) implies that k < l.
In order to prove the reverse inequality l < k, we consider the
regular CG-module. If V1, : : : , Vk is a complete set of non-iso-
morphic irreducible CG-modules, we know from Theorem 8.7 that
CG � W1 � : : :� Wk ,
where for each i, Wi is isomorphic to a direct sum of copies of Vi.
Since CG contains the identity element 1, we can write
1 � f 1 � : : :� f k
with f i 2 Wi for 1 < i < k.
Now let z 2 Z(CG), the centre of CG. By Proposition 9.14, for each
i there exists ëi 2 C such that for all v 2 Vi,
vz � ëiv:
Hence wz � ëiw for all w 2 Wi, and in particular,
f iz � ëi f i (1 < i < k):
It follows that
z � 1z � ( f 1 � : : :� f k)z � f 1z� : : :� f kz
� ë1 f 1 � : : :� ëk f k :
This shows that Z(CG) is contained in the subspace of CG spanned by
f 1, : : : , f k . Since Z(CG) has dimension l by Proposition 12.22, we
deduce that l < k. This completes the proof that k � l. j
15.4 Corollary
The irreducible characters ÷1, : : : , ÷k of G form a basis of the vector
space of all class functions on G. Indeed, if ø is a class function, then
ø �Xk
i�1
ëi÷i
where ëi � kø, ÷il for 1 < i < k.
Proof Since ÷1, . . . , ÷k are linearly independent, they span a subspace
of C of dimension k. By (15.2), dim C � l, which is equal to k by
The number of irreducible characters 153
Theorem 15.3. Hence ÷1, . . . , ÷k span C, and so they form a basis of
C. The last part follows, using (14.13). j
Corollary 15.4 has the following useful consequence.
15.5 Proposition
Suppose that g, h 2 G. Then g is conjugate to h if and only if
÷(g) � ÷(h) for all characters ÷ of G.
Proof If g is conjugate to h then ÷(g) � ÷(h) for all characters ÷ of
G, by Proposition 13.5(2).
Conversely, suppose that ÷(g) � ÷(h) for all characters ÷. Then by
Corollary 15.4, ø(g) � ø(h) for all class functions ø on G. In
particular, this is true for the class function ø which takes the value 1
on the conjugacy class of g and takes the value 0 elsewhere. Then
ø(g) � ø(h) � 1, and so g is conjugate to h. j
15.6 Corollary
Suppose that g 2 G. Then g is conjugate to gÿ1 if and only if ÷(g) is
real for all characters ÷ of G.
Proof Since ÷(g) is real if and only if ÷(g) � ÷(gÿ1) (see Proposition
13.9(3)), the result follows immediately from Proposition 15.5. j
We conclude the chapter with an example illustrating some practical
methods of expressing characters and class functions of a group as
combinations of irreducible characters. As in previous examples, we
regard a character ÷ of G as a row vector, whose k entries are the
values of ÷ on the k conjugacy classes of G.
15.7 Example
We shall see in Section 18.4 that there is a certain group G of order
12 which has exactly six conjugacy classes with representatives
g1, . . . , g6 (where g1 � 1), and six irreducible characters ÷1, . . . , ÷6
given as follows:
154 Representations and characters of groups
gi g1 g2 g3 g4 g5 g6
|CG(gi)| 12 12 6 6 4 4
÷1 1 1 1 1 1 1÷2 1 ÿ1 ÿ1 1 i ÿi÷3 1 1 1 1 ÿ1 ÿ1÷4 1 ÿ1 ÿ1 1 ÿi i÷5 2 2 ÿ1 ÿ1 0 0÷6 2 ÿ2 1 ÿ1 0 0
Suppose we are given characters ÷ and ø of G as follows:
g1 g2 g3 g4 g5 g6
÷ 3 ÿ3 0 0 i ÿiø 4 0 0 4 0 0
Then it is easy to spot that
÷ � ÷2 � ÷6, ø � ÷1 � ÷2 � ÷3 � ÷4:
For example, the second entry in the row vector for ÷ is equal to
minus the ®rst entry. Inspecting the values of the irreducible characters
÷i, we see that ÷ must be a combination of ÷2, ÷4 and ÷6. The correct
answer now comes quickly to mind.
In fact, given any character ö of G whose degree is not large
compared with the degrees of the ÷i, it is not hard to use tactical
guesswork to express ö as a combination of the irreducible characters.
The reason for this is that the required coef®cients are known to be
non-negative integers, and the entries in the column corresponding to
g1 � 1 are positive integers (indeed, they are the degrees of the ÷i).
We suggest that you use the `guesswork method' to express the
following characters ë, ì of G as combinations of ÷1, . . . , ÷6:
g1 g2 g3 g4 g5 g6
ë 2 ÿ2 ÿ2 2 0 0ì 4 4 1 1 0 0
The number of irreducible characters 155
How do we cope with a class function or with a more dif®cult
character, like the following one?
g1 g2 g3 g4 g5 g6
ö 11 3 ÿ3 5 ÿ1 � 2i ÿ1 ÿ 2i
The answer is to use the inner product k , l. We know from Corollary
15.4 that the coef®cients ëi in the expression
ö � ë1÷1 � : : :� ë6÷6
are given by
ëi � hö, ÷ii (1 < i < 6):
Using Proposition 14.5(2), we calculate these inner products:
hö, ÷1i � 11 . 1
12� 3 . 1
12�ÿ3 . 1
6� 5 . 1
6� (ÿ1� 2i) . 1
4
� (ÿ1ÿ 2i) . 1
4� 1,
hö, ÷2i � 11 . 1
12� 3 . (ÿ1)
12� (ÿ3) . (ÿ1)
6� 5 . 1
6� (ÿ1� 2i) . (ÿi)
4
� (ÿ1ÿ 2i) . i
4� 3,
hö, ÷3i � 11 . 1
12� 3 . 1
12�ÿ3 . 1
6� 5 . 1
6� (ÿ1� 2i) . (ÿ1)
4
� (ÿ1ÿ 2i) . (ÿ1)
4� 2,
and similarly kö, ÷4l � 1, kö, ÷5l � 2 and kö, ÷6l � 0. Therefore
ö � ÷1 � 3÷2 � 2÷3 � ÷4 � 2÷5:
Summary of Chapter 15
1. The number of irreducible characters of a group is equal to the
number of conjugacy classes of the group.
156 Representations and characters of groups
2. The irreducible characters ÷1, . . . , ÷k of G form a basis of the
vector space of all class functions on G. If ø is a class function,
then
ø �Xk
i�1
ëi÷i where ëi � hø, ÷ii:
Exercises for Chapter 15
1. The three irreducible characters of S3 are ÷1, ÷2, ÷3:
Let ÷ be the class function on S3 with the following values:
Express ÷ as a linear combination of ÷1, ÷2 and ÷3. Is ÷ a character
of S3?
2. Let ø1, ø2 and ø3 be the class functions on S3 taking the following
values:
Express ø1, ø2 and ø3 as linear combinations of the irreducible
characters ÷1, ÷2 and ÷3 of S3.
1 (1 2) (1 2 3)
÷1 1 1 1÷2 1 ÿ1 1÷3 2 0 ÿ1
1 (1 2) (1 2 3)
÷ 19 ÿ1 ÿ2
1 (1 2) (1 2 3)
ø1 1 0 0ø2 0 1 0ø3 0 0 1
The number of irreducible characters 157
3. Suppose that G is the group of order 12 in Example 15.7, with
conjugacy class representatives g1, . . . , g6 and irreducible charac-
ters ÷1, . . . , ÷6 as in that example. Let ø be the class function on
G taking the following values:
Express ø as a linear combination of ÷1, . . . , ÷6. Is ø a character
of G?
4. Let G be a group of order 12.
(a) Show that G cannot have exactly 9 conjugacy classes. (Hint:
show that Z(G) cannot have order 6.)
(b) Using the solution to Exercise 11.2, prove that G has 4, 6 or 12
conjugacy classes. Find groups G in which each of these
possibilities is realized.
g1 g2 g3 g4 g5 g6
ø 6 0 3 ÿ3 ÿ1 ÿ i ÿ1 � i
158 Representations and characters of groups
159
16
Character tables and orthogonality relations
The irreducible characters of a ®nite group G are class functions, and
the number of them is equal to the number of conjugacy classes of G.
It is therefore convenient to record all the values of all the irreducible
characters of G in a square matrix. This matrix is called the character
table of G. The entries in a character table are related to each other in
subtle ways, many of which are encapsulated in the orthogonality
relations (Theorem 16.4). Much of the later material in the book will
be devoted to understanding character tables. The motivation for this is
Theorem 14.21, which tells us that every CG-module is determined by
its character. Thus, many problems in representation theory can be
solved by considering characters.
Character tables
16.1 De®nition
Let ÷1, : : : , ÷k be the irreducible characters of G and let g1, : : : , gk
be representatives of the conjugacy classes of G. The k 3 k matrix
whose ij-entry is ÷i(g j) (for all i, j with 1 < i < k, 1 < j < k) is
called the character table of G.
It is usual to number the irreducible characters and conjugacy classes
of G so that ÷1 � 1G, the trivial character, and g1 � 1, the identity
element of G. Beyond this, the numbering is arbitrary. Note that in the
character table, the rows are indexed by the irreducible characters of G
and the columns are indexed by the conjugacy classes (or, in practice,
by conjugacy class representatives).
16.2 Proposition
The character table of G is an invertible matrix.
Proof This follows immediately from the fact that the irreducible
characters of G, and hence also the rows of the character table, are
linearly independent (Theorem 14.23). j
16.3 Examples
(1) Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l. The irreducible
characters of G are given in Example 13.6(4). We take 1, a, b as
representatives of the conjugacy classes of G, and then the character
table of G is
(2) We can write down the character table of any ®nite abelian group
using Theorem 9.8. For example, the character table of C2 �ha: a2 � 1i is
and the character table of C3 � ka: a3 � 1l is
(3) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. You found all the
irreducible representations of G in Exercise 10.4. The conjugacy classes
1 a b
÷1 1 1 1÷2 1 1 ÿ1÷3 2 ÿ1 0
1 a
÷1 1 1÷2 1 ÿ1
1 a a2
÷1 1 1 1÷2 1 ù ù2 (ù � e2ði=3)÷3 1 ù2 ù
160 Representations and characters of groups
of G are given by (12.12), and representatives are 1, a2, a, b, ab.
Hence the character table of G is
1 a2 a b ab
÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1÷5 2 ÿ2 0 0 0
The character tables of all dihedral groups will be found in
Chapter 18.
The orthogonality relations
We have already seen many uses for the relations (14.13),
h÷r, ÷si � ärs,
among the irreducible characters ÷1, . . . , ÷k of G. These relations can
be expressed in terms of the rows of the character table, by writing
them as Xk
i�1
÷r(gi)÷s(gi)
jCG(gi)j � ärs
(see Proposition 14.5(2)). Similar relations exist between the columns
of the character table, and these are given by part (2) of our next
result.
16.4 Theorem
Let ÷1, . . . , ÷k be the irreducible characters of G, and let g1, : : : , gk
be representatives of the conjugacy classes of G. Then the following
relations hold for any r, s 2 {1, . . . , k}.
(1) The row orthogonality relations:Xk
i�1
÷r(gi)÷s(gi)
jCG(gi)j � ärs:
(2) The column orthogonality relations:Xk
i�1
÷i(gr)÷i(gs) � ärsjCG(gr)j:
Character tables and orthogonality relations 161
Proof The row orthogonality relations have already been proved. They
are recorded here merely for comparison with the column relations.
For 1 < s < k, let øs be the class function which satis®es
øs(gr) � ärs (1 < r < k):
By Corollary 15.4, øs is a linear combination of ÷1, . . . , ÷k , say
øs �Xk
i�1
ëi÷i (ëi 2 C):
We know that h÷i, ÷ ji � äij, so
ëi � høs, ÷ii � 1
jGjXg2G
øs(g)÷i(g):
Now øs(g) � 1 if g is conjugate to gs, and øs(g) � 0 otherwise; also
there are jGj=jCG(gs)j elements of G which are conjugate to gs, by
Theorem 12.8. Hence
ëi � 1
jGjXg2 gG
s
øs(g)÷i(g) � ÷i(gs)
jCG(gs)j :
Therefore
ärs � øs(gr) �Xk
i�1
ëi÷i(gr) �Xk
i�1
÷i(gr)÷i(gs)
jCG(gs)j ,
and the column orthogonality relations follow. j
16.5 Examples
We illustrate the column orthogonality relations.
(1) Let G � D6. We copy the character table of G from Example
16.3(1), and this time we record the order of the centralizer CG(gi)
next to each conjugacy class representative gi:
gi 1 a b|CG(gi)| 6 3 2
÷1 1 1 1÷2 1 1 ÿ1÷3 2 ÿ1 0
162 Representations and characters of groups
Consider the sumsP3
i�1 ÷i(g r)÷i(gs) for various cases:
r � 1, s � 2: 1 . 1� 1 . 1� 2 . (ÿ1) � 0;
r � 2, s � 2: 1 . 1� 1 . 1� (ÿ1) . (ÿ1) � 3;
r � 1, s � 3: 1 . 1� 1 . (ÿ1)� 2 . 0 � 0:
In each case, we read down columns r and s of the character table,
taking the products of the numbers which appear. The sum of the
products is 0 if r 6� s, and is the number at the top of the column (that
is, the order of the centralizer of gr) if r � s.
(2) Suppose we are given the following part of the character table of a
group G of order 12 which has exactly four conjugacy classes:
(where ù � e2ði=3). We shall use the column orthogonality relations to
determine the last row of the character table.
The entries in the ®rst column of the character table are the degrees
of the irreducible characters, so they are positive integers. By the
column orthogonality relations with r � s � 1, the sum of the squares
of these numbers is 12 (this also follows from Theorem 11.12). Hence
the last entry in the ®rst column is 3.
Let x denote the number at the foot of the second column. The
column orthogonality relation
X4
i�1
÷i(g1)÷i(g2) � 0
gives
1 . 1� 1 . 1� 1 . 1� 3x � 0:
Therefore x � ÿ1.
By considering the orthogonality relations between the ®rst column
and columns 3 and 4, we obtain the complete character table as
gi g1 g2 g3 g4
|CG( gi)| 12 4 3 3
÷1 1 1 1 1÷2 1 1 ù ù2
÷3 1 1 ù2 ù÷4
Character tables and orthogonality relations 163
follows:
Notice that the orthogonality relations hold between all pairs of
columns, although our calculation has used only those relations which
involve the ®rst column. For example,
X4
i�1
÷i(g2)÷i(g2) � 1 . 1� 1 . 1� 1 . 1� (ÿ1) . (ÿ1) � 4,
X4
i�1
÷i(g3)÷i(g3) � 1 . 1� ù . ù� ù2 . ù2 � 0 . 0 � 3,
X4
i�1
÷i(g3)÷i(g4) � 1 . 1� ù . ù2 � ù2 . ù� 0 . 0 � 0:
We shall see later that the character table which we have constructed
here is that of A4.
Those column orthogonality relations which involve the ®rst column
of the character table were proved in Chapter 13, since Theorem 13.19
and Proposition 13.20 give
Xk
i�1
di÷i(g) �jGj, if g � 1,
0, if g 6� 1,
8<:where di � ÷i(1). By taking the complex conjugate of each side of this
equation, we get
Xk
i�1
÷i(1)÷i(g) �jGj, if g � 1,
0, if g 6� 1,
8<:
gi g1 g2 g3 g4
|CG(gi)| 12 4 3 3
÷1 1 1 1 1÷2 1 1 ù ù2
÷3 1 1 ù2 ù÷4 3 ÿ1 0 0
164 Representations and characters of groups
and these are just the column orthogonality relations which involve the
®rst column.
Rows versus columns
Notice that in Example 16.5(2), where we were given three of the four
irreducible characters of G, we calculated the values of the last
character one at a time using the column orthogonality relations. An
alternative approach would have been to use the row orthogonality
relations h÷i, ÷4i � äi4 to obtain four equations in the four unknown
values ÷4(g j) (1 < j < 4). Although the calculation with the column
orthogonality relations was easier to perform, it is a fact that the
column orthogonality relations contain precisely the same information
as the row orthogonality relations, as we shall now show.
The character table of G is a k 3 k matrix, and we adjust the entries
÷i(g j) in this matrix to obtain another k 3 k matrix M, by letting the
ij-entry of M be
÷i(gj)
jCG(gj)j1=2:
Let M t denote the transpose of the complex conjugate of M.
Now the rs-entry in M M t isXk
i�1
÷r(gi)÷s(gi)
jCG(gi)j � ärs,
by the row orthogonality relations, so M M t � I . Indeed, the equation
M M t � I is just another way of expressing the row orthogonality
relations. On the other hand, the rs-entry in M t M is
1
jCG(gr)j1=2jCG(gs)j1=2
Xk
i�1
÷i(gr)÷i(gs) � ärs,
by the column orthogonality relations, so M t M � I .
Since the properties M t M � I and M M t � I of a square matrix M
are equivalent to each other, we see that the row and column
orthogonality relations are equivalent.
We could have used the above argument to deduce the column
orthogonality relations from the row ones. More importantly, the row
and column orthogonality relations encapsulate the same information,
Character tables and orthogonality relations 165
so when we are working with character tables, we can deduce exactly
the same results using either set of relations.
Summary of Chapter 16
Let G be a ®nite group with irreducible characters ÷1, . . . , ÷k and
conjugacy class representatives g1, . . . , gk.
1. The character table of G is the k 3 k matrix with ij-entry ÷i(g j).
2. The row orthogonality relations state that for all r, s,Xk
i�1
÷r(gi)÷s(gi)
jCG(gi)j � ärs:
3. The column orthogonality relations state that for all r, s,Xk
i�1
÷i(gr)÷i(gs) � ärsjCG(gr)j:
Exercises for Chapter 16
1. Write down the character table of C2 3 C2.
2. A certain group G of order 8 is known to have a total of ®ve
conjugacy classes, with representatives g1, . . . , g5, and four linear
characters ÷1, . . . , ÷4 taking the following values:
gi g1 g2 g3 g4 g5
|CG(gi)| 8 8 4 4 4
÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1
Find the complete character table of G.
3. There exists a group G of order 10 which has precisely four
conjugacy classes, with representatives g1, . . . , g4, and has irreduci-
ble characters ÷1, ÷2 as follows:
166 Representations and characters of groups
where á � (ÿ1�p5)=2 and â � (ÿ1ÿp5)=2.
Find the complete character table of G.
(Hint: ®rst ®nd the values of the remaining irreducible characters on
g1, then on g4 ± use Corollary 13.10.)
4. A certain group G has two columns of its character table as
follows:
where g1 � 1 and æ 2 C.
(a) Find æ.
(b) Find another column of the character table.
5. Let ÷1, . . . , ÷k be the irreducible characters of G. Show that
Z(G) � g 2 G:Xk
i�1
÷i(g)÷i(g) � jGj( )
:
6. Let G be a ®nite group with conjugacy class representatives
g1, : : : , gk and character table C. Show that det C is either real or
purely imaginary, and that
jdet Cj2 �Yk
i�1
jCG(gi)j:
Find �(det C) when G � C3.
gi g1 g2 g3 g4
|CG(gi)| 10 5 5 2
÷1 1 1 1 1÷2 2 á â 0
gi g1 g2
|CG(gi)| 21 7
÷1 1 1÷2 1 1÷3 1 1÷4 3 æ÷5 3 æ
Character tables and orthogonality relations 167
168
17
Normal subgroups and lifted characters
If N is a normal subgroup of the ®nite group G, and N 6� {1}, then
the factor group G=N is smaller than G. The characters of G=N should
therefore be easier to ®nd than the characters of G. In fact, we can use
the characters of G=N to get some of the characters of G, by a process
which is known as lifting. Thus, normal subgroups help us to ®nd
characters of G. In the opposite direction, it is also true that the
character table of G enables us to ®nd the normal subgroups of G; in
particular, it is easy to tell from the character table whether or not G
is simple.
The linear characters of G (i.e. the characters of degree 1) are
obtained by lifting the irreducible characters of G=N in the case where
N is the derived subgroup of G. (The derived subgroup is de®ned
below in De®nition 17.7.) The linear characters, in turn, can be used to
get new irreducible characters from a given irreducible character, in a
way which we shall describe.
Lifted characters
We begin by constructing a character of G from a character of G=N .
17.1 Proposition
Assume that N v G, and let ~÷ be a character of G=N . De®ne
÷: G! C by
÷(g) � ~÷(Ng) (g 2 G):
Then ÷ is a character of G, and ÷ and ~÷ have the same degree.
Proof Let ~r: G=N ! GL (n, C) be a representation of G=N with
character ~÷. The function r: G! GL (n, C) which is given by the
composition
g ! Ng ! (Ng)~r (g 2 G)
is a homomorphism from G to GL (n, C). Thus r is a representation of
G. The character ÷ of r satis®es
÷(g) � tr (gr) � tr ((Ng)~r) � ~÷(Ng)
for all g 2 G. Moreover, ÷(1) � ~÷(N), so ÷ and ~÷ have the same
degree. j
17.2 De®nition
If N v G and ~÷ is a character of G=N , then the character ÷ of G
which is given by
÷(g) � ~÷(Ng) (g 2 G)
is called the lift of ~÷ to G.
17.3 Theorem
Assume that N v G. By associating each character of G=N with its
lift to G, we obtain a bijective correspondence between the set of
characters of G=N and the set of characters ÷ of G which satisfy
N < Ker ÷. Irreducible characters of G=N correspond to irreducible
characters of G which have N in their kernel.
Proof If ~÷ is a character of G=N , and ÷ is the lift of ~÷ to G, then
~÷(N) � ÷(1). Also, if k 2 N then
÷(k) � ~÷(Nk) � ~÷(N ) � ÷(1),
so N < Ker ÷.
Now let ÷ be a character of G with N < Ker ÷. Suppose that
r: G! GL (n, C) is a representation of G with character ÷. If g1,
g2 2 G and Ng1 � Ng2 then g1 gÿ12 2 N, so (g1 gÿ1
2 )r � I, and hence
g1r � g2r. We may therefore de®ne a function ~r: G=N ! GL (n, C)
by
(Ng)~r � gr (g 2 G):
Then for all g, h 2 G we have
((Ng)(Nh))~r � (Ngh)~r � (gh)r � (gr)(hr)
� ((Ng)~r)((Nh)~r),
Normal subgroups and lifted characters 169
so ~r is a representation of G=N . If ~÷ is the character of ~r then
~÷(Ng) � ÷(g) (g 2 G):
Thus ÷ is the lift of ~÷.
We have now established that the function which sends each char-
acter of G=N to its lift to G is a bijection between the set of
characters of G=N and the set of characters of G which have N in
their kernel. It remains to show that irreducible characters correspond
to irreducible characters. To see this, let U be a subspace of Cn, and
note that
u(gr) 2 U for all u 2 U , u(Ng)~r 2 U for all u 2 U :
Thus, U is a CG-submodule of Cn if and only if U is a C(G=N )-
submodule of Cn. The representation r is therefore irreducible if and
only if the representation ~r is irreducible. Hence ÷ is irreducible if and
only if ~÷ is irreducible. j
If we know the character table of G=N for some normal subgroup N
of G, then Theorem 17.3 enables us to write down as many irreducible
characters of G as there are irreducible characters of G=N .
17.4 Example
Let G � S4 and
N � V4 � f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g,so that N v G (see Example 12.20). If we put a � N(1 2 3) and
b � N(1 2) then
G=N � ha, bi and a3 � b2 � N , bÿ1ab � aÿ1,
so G=N � D6. We know from Example 16.3(1) that the character table
of G=N is
N N (1 2) N(1 2 3)
~÷1 1 1 1~÷2 1 ÿ1 1~÷3 2 0 ÿ1
170 Representations and characters of groups
To calculate the lift ÷ of a character ~÷ of G=N , we note that
÷((1 2)(3 4)) � ~÷(N ) since (1 2)(3 4) 2 N ,
÷((1 2 3 4)) � ~÷(N (1 3)) since N (1 2 3 4) � N (1 3):
Hence the lifts of ~÷1, ~÷2, ~÷3 are ÷1, ÷2, ÷3, which are given by
1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0
Then ÷1, ÷2, ÷3 are irreducible characters of G, since ~÷1, ~÷2, ~÷3 are
irreducible characters of G=N .
Finding normal subgroups
The character table contains accessible information about the structure
of a group, as our next two propositions will demonstrate. First we
shall show how to ®nd all the normal subgroups of G, once the
character table of G is known. Recall that we can easily locate the
kernel of an irreducible character ÷ from the character table, since
Ker ÷ � fg 2 G: ÷(g) � ÷(1)g(see De®nition 13.13). Also Ker ÷ v G. Of course, any subgroup which
is the intersection of the kernels of irreducible characters is a normal
subgroup too. The following proposition shows that every normal
subgroup arises in this way.
17.5 Proposition
If N v G then there exist irreducible characters ÷1, : : : , ÷s of G
such that
N �\s
i�1
Ker ÷i:
Proof If g belongs to the kernel of each irreducible character of G,
then ÷(g) � ÷(1) for all characters ÷, so g � 1 by Proposition 15.5.
Hence the intersection of the kernels of all the irreducible characters of
G is {1}.
Normal subgroups and lifted characters 171
Now let ~÷1, . . . , ~÷s be the irreducible characters of G=N . By the
above observation, \s
i�1
Ker ~÷i � fNg:
For 1 < i < s, let ÷i be the lift to G of ~÷i. If g 2 Ker ÷i then
~÷i(N ) � ÷i(1) � ÷i(g) � ~÷i(Ng),
and so Ng 2 Ker ~÷i. Therefore if g 2T
Ker ÷i then Ng 2T
Ker ~÷i �{N}, and so g 2 N. Hence
N �\s
i�1
Ker ÷i: j
It is particularly easy to tell from the character table of G whether
or not G is simple:
17.6 Proposition
The group G is not simple if and only if
÷(g) � ÷(1)
for some non-trivial irreducible character ÷ of G, and some non-
identity element g of G.
Proof Suppose there is a non-trivial irreducible character ÷ such that
÷(g) � ÷(1) for some non-identity element g. Then g 2 Ker ÷, so
Ker ÷ 6� {1}. If r is a representation of G with character ÷, then
Ker ÷ � Ker r by Theorem 13.11(2). Since ÷ is non-trivial and irreduci-
ble, Ker r 6� G; hence Ker ÷ 6� G. Thus Ker ÷ is a normal subgroup of
G which is not equal to {1} of G, and so G is not simple.
Conversely, suppose that G is not simple, so that there is a normal
subgroup N of G with N 6� {1} and N 6� G. Then by Proposition 17.5,
there is an irreducible character ÷ of G such that Ker ÷ is not {1} or
G. As Ker ÷ 6� G, ÷ is non-trivial; and taking 1 6� g 2 Ker ÷, we have
÷(g) � ÷(1). j
Linear characters
Recall that a linear character of a group is a character of degree 1. We
shall show how to ®nd all linear characters of any group G, since the
172 Representations and characters of groups
®rst move in constructing the character table of G is often to write
down the linear characters. As a preliminary step, it is necessary to
determine the derived subgroup of G, which is de®ned in the following
way.
17.7 De®nition
For a group G, let G9 be the subgroup of G which is generated by all
elements of the form
gÿ1 hÿ1 gh (g, h 2 G):
Then G9 is called the derived subgroup of G.
We abbreviate gÿ1 hÿ1 gh as [g, h]. Thus
G9 � h[g, h]: g, h 2 Gi:
17.8 Examples
(1) If G is abelian then [g, h] � 1 for all g, h 2 G, so G9 � {1}.
(2) Let G � S3. Clearly [g, h] is always an even permutation, so
G9 < A3. If g � (1 2) and h � (2 3) then [g, h] � (1 2 3). Hence
G9 � h(1 2 3)i � A3.
We are going to show that G9 v G and that the linear characters of
G are the lifts to G of the irreducible characters of G=G9. One step is
provided by the following proposition.
17.9 Proposition
If ÷ is a linear character of G, then G9 < Ker ÷.
Proof Let ÷ be a linear character of G. Then ÷ is a homomorphism
from G to the multiplicative group of non-zero complex numbers.
Therefore, for all g, h 2 G,
÷(gÿ1 hÿ1 gh) � ÷(g)ÿ1÷(h)ÿ1÷(g)÷(h) � 1:
Hence G9 < Ker ÷. j
Next, we explore some group-theoretic properties of the derived
subgroup.
Normal subgroups and lifted characters 173
17.10 Proposition
Assume that N v G.
(1) G9 v G.
(2) G9 < N if and only if G=N is abelian. In particular, G=G9 is
abelian.
Proof (1) Note that for all a, b, x 2 G, we have
xÿ1(ab)x � (xÿ1ax)(xÿ1bx), and
xÿ1aÿ1x � (xÿ1ax)ÿ1:
Now G9 consists of products of elements of the form [g, h] and their
inverses. Therefore, to prove that G9 v G it is suf®cient by the ®rst
sentence to prove that xÿ1[g, h]x 2 G9 for all g, h, x 2 G. But
xÿ1[g, h]x � xÿ1 gÿ1 hÿ1 ghx
� (xÿ1 gx)ÿ1(xÿ1 hx)ÿ1(xÿ1 gx)(xÿ1 hx)
� [xÿ1 gx, xÿ1 hx]:
Therefore G9 v G.
(2) Let g, h 2 G. We have
ghgÿ1 hÿ1 2 N , Ngh � Nhg , (Ng)(Nh) � (Nh)(Ng):
Hence G9 < N if and only if G=N is abelian. Since we have proved
that G9 v G, we deduce that G=G9 is abelian. j
It follows from Proposition 17.10 that G9 is the smallest normal
subgroup of G with abelian factor group.
Given the derived subgroup G9, we can obtain the linear characters
of G by applying the next theorem.
17.11 Theorem
The linear characters of G are precisely the lifts to G of the irreducible
characters of G=G9. In particular, the number of distinct linear
characters of G is equal to jG=G9j, and so divides |G|.
Proof Let m � jG=G9j. Since G=G9 is abelian, Theorem 9.8 shows that
G=G9 has exactly m irreducible characters ~÷1, . . . , ~÷m, all of degree 1.
The lifts ÷1, . . . , ÷m of these characters to G also have degree 1, and
by Theorem 17.3 they are precisely the irreducible characters ÷ of G
174 Representations and characters of groups
such that G9 < Ker ÷. In view of Proposition 17.9, the characters
÷1, . . . , ÷m are therefore all the linear characters of G. j
17.12 Example
Let G � Sn. We shall show that G9 � An. If n � 1 or 2 then Sn is
abelian, so G9 � {1} � An. We proved that S93 � A3 in Example
17.8(2), so we assume that n > 4.
As Sn=An � C2, we have G9 < An by Proposition 17.10(2). If
g � (1 2), h � (2 3) and k � (1 2)(3 4), then
[g, h] � (1 2 3), [h, k] � (1 4)(2 3):
Since G9 v G, all the elements in (1 2 3)G and (1 4)(2 3)G belong to
G9. Therefore, by Theorem 12.15, G9 contains all 3-cycles and all
elements of cycle-shape (2, 2). But every product of two transpositions
is equal to the identity, a 3-cycle or an element of cycle-shape (2, 2);
and An consists of permutations, each of which is the product of an
even number of transpositions. Therefore An < G9. We have now
proved that G9 � An.
17.13 Example
We ®nd the linear characters of Sn (n > 2). From the last example, we
know that S9n � An. Since Sn=S9n � fAn, An(1 2)g � C2, the group
Sn=S9n has two linear characters ~÷1 and ~÷2, where
~÷1(An(1 2)) � 1,
~÷2(An(1 2)) � ÿ1:
Therefore by Theorem 17.11, Sn has exactly two linear characters ÷1,
÷2, which are given by
÷1 � 1Sn,
÷2(g) �1, if g 2 An,
ÿ1, if g =2 An:
(
Not only are the linear characters of G important in being irreducible
characters, but they can also be used to construct new irreducible
characters from old, as the next result shows.
Normal subgroups and lifted characters 175
17.14 Proposition
Suppose that ÷ is a character of G and ë is a linear character of G.
Then the product ÷ë, de®ned by
÷ë(g) � ÷(g)ë(g) (g 2 G)
is a character of G. Moreover, if ÷ is irreducible, then so is ÷ë.
Proof Let r: G! GL (n, C) be a representation with character ÷.
De®ne rë: G! GL (n, C) by
g(rë) � ë(g)(gr) (g 2 G):
Thus g(rë) is the matrix gr multiplied by the complex number ë(g).
Since r and ë are homomorphisms it follows easily that rë is a
homomorphism. The matrix g(rë) has trace ë(g) tr (gr), which is
ë(g)÷(g). Hence rë is a representation of G with character ÷ë.
Now for all g 2 G, the complex number ë(g) is a root of unity, so
ë(g)ë(g) � 1. Therefore
h÷ë, ÷ëi � 1
jGjXg2G
÷(g)ë(g)÷(g)ë(g)
� 1
jGjXg2G
÷(g)÷(g) � h÷, ÷i:
By Theorem 14.20, it follows that ÷ë is irreducible if and only if ÷ is
irreducible. j
The general case of a product of two characters will be discussed in
Chapter 19.
Summary of Chapter 17
1. Characters of G=N correspond to characters ÷ of G for which
N < Ker ÷. The character of G which corresponds to the character ~÷of G=N is the lift of ~÷, and is given by ÷(g) � ~÷(Ng) (g 2 G).
2. The normal subgroups of G can be found from the character table
of G.
3. The linear characters of G are precisely the lifts to G of the
irreducible characters of G=G9.
176 Representations and characters of groups
Exercises for Chapter 17
1. Let G � Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l.(a) Find the ®ve conjugacy classes of G.
(b) Find G9, and construct all the linear characters of G.
(c) Complete the character table of G.
Compare your table with the character table of D8 (Example
16.3(3)).
2. Let a and b be the following permutations in S7:
a � (1 2 3 4 5 6 7), b � (2 3 5)(4 7 6):
Let G � ka, bl. Check that
a7 � b3 � 1, bÿ1ab � a2:
(a) Show that G has order 21.
(b) Find the conjugacy classes of G.
(c) Find the character table of G.
3. Show that every group of order 12 has 3, 4 or 12 linear characters,
and hence cannot be simple.
4. A certain group G of order 12 has precisely six conjugacy classes,
with representatives g1, . . . , g6 (where g1 � 1), and has irreducible
characters ÷, ö with values as follows:
g1 g2 g3 g4 g5 g6
÷ 1 ÿi i 1 ÿ1 ÿ1ö 2 0 0 ÿ1 ÿ1 2
Use Proposition 17.14 to complete the character table of G. What
are the sizes of the conjugacy classes of G?
5. The character table of D8 is as shown (see Example 16.3(3)):
1 a2 a, a3 b, a2b ab, a3b
÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1÷5 2 ÿ2 0 0 0
Normal subgroups and lifted characters 177
Express each normal subgroup of D8 as an intersection of kernels
of irreducible characters, as in Proposition 17.5.
6. You are given that the group
T4n � ha, b: a2n � 1, an � b2, bÿ1ab � aÿ1ihas order 4n. (It is known as a dicyclic group.)
(a) Show that if å is any (2n)th root of unity in C, then there is a
representation of T4n over C which sends
a! å 0
0 åÿ1
� �, b! 0 1
å n 0
� �:
(b) Find all the irreducible representations of T4n.
7. For n > 1, the group
U6n � ha, b: a2n � b3 � 1, aÿ1ba � bÿ1ihas order 6n.
(a) Let ù � e2ði=3. Show that if å is any (2n)th root of unity in C,
then there is a representation of U6n over C which sends
a! 0 åå 0
� �, b! ù 0
0 ù2
� �:
(b) Find all the irreducible representations of U6n.
8. Let n be an odd positive integer. The group
V8n � ha, b: a2n � b4 � 1, ba � aÿ1bÿ1, bÿ1a � aÿ1bihas order 8n.
(a) Show that if å is any nth root of unity in C, then there is a
representation of V8n over C which sends
a! å 0
0 ÿåÿ1
� �, b! 0 1
ÿ1 0
� �:
(b) Find all the irreducible representations of V8n.
178 Representations and characters of groups
179
18
Some elementary character tables
We now illustrate the techniques we have presented so far by construct-
ing the character tables of several groups, including the groups S4 and
A4, and all dihedral groups.
18.1 The group S4
In Example 17.4, we produced three irreducible characters ÷1, ÷2, ÷3 of
S4 by lifting characters of the factor group S4=V4. We shall now use
Proposition 17.14, which deals with the product of a character with a
linear character, to complete the character table of S4.
Let ÷4 be the character
÷4(g) � jfix (g)j ÿ 1 (g 2 S4)
which is given in Proposition 13.24. By Proposition 17.14, the product
÷4÷2 is also a character of S4. The values of ÷2, ÷4 and ÷4÷2 are as
follows:
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4
÷2 1 ÿ1 1 1 ÿ1÷4 3 1 0 ÿ1 ÿ1÷4÷2 3 ÿ1 0 ÿ1 1
Note that
h÷4, ÷4i � 9
24� 1
4� 1
8� 1
4� 1,
so ÷4 is irreducible. The character ÷4÷2 is also irreducible, either by
using the same calculation or by quoting the result of Proposition
17.14. Let ÷5 � ÷4÷2. Since S4 has ®ve conjugacy classes, and we have
produced ®ve irreducible characters, we have now found the complete
character table of S4, as shown.
18.2 The group A4
Let G � A4, the alternating group of degree 4. Then |G| � 12, and G
has four conjugacy classes, with representatives
1, (1 2)(3 4), (1 2 3), (1 3 2)
(see Example 12.18(1)).
Let í be the character of A4 given by Proposition 13.24, so that
í(g) � |®x (g)| ÿ 1 for all g 2 A4. The values of í are as follows:
gi 1 (1 2)(3 4) (1 2 3) (1 3 2)|CG(gi)| 12 4 3 3
í 3 ÿ1 0 0
Note that
hí, íi � 9
12� 1
4� 1,
so í is an irreducible character of G of degree 3.
Since G has four irreducible characters, and the sum of the squares
of their degrees is 12, there must be exactly three linear characters of
G. Thus jG=G9j � 3 by Theorem 17.11. It is not dif®cult to con®rm
this by showing that
Character table of S4
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4
÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0÷4 3 1 0 ÿ1 ÿ1÷5 3 ÿ1 0 ÿ1 1
180 Representations and characters of groups
G9 � V4 � f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:
Now G=G9 � fG9, G9(1 2 3), G9(1 3 2)g � C3, and the character table
of G=G9 is
(where ù � e2ði=3). The lifts of ~÷1, ~÷2, ~÷3 to G, together with the
character ÷4 � í, give the complete character table of A4:
18.3 The dihedral groups
Let G be the dihedral group D2n of order 2n, with n > 3, so that
G � ha, b: an � b2 � 1, bÿ1ab � aÿ1i:We shall derive the character table of G.
Write å � e2ði=n. For each integer j with 1 < j , n=2, de®ne
Aj � å j 0
0 åÿ j
� �, Bj � 0 1
1 0
� �:
Check that
Anj � B2
j � I , Bÿ1j AjBj � Aÿ1
j :
It follows that by de®ning r j: G! GL(2, C) by
(arbs)r j � (Aj)r(Bj)
s (r, s 2 Z),
we obtain a representation r j of G for each j with 1 < j , n=2.
G9 G9(1 2 3) G9(1 3 2)
~÷1 1 1 1~÷2 1 ù ù2
~÷3 1 ù2 ù
Character table of A4
gi 1 (1 2)(3 4) (1 2 3) (1 3 2)|CG(gi)| 12 4 3 3
÷1 1 1 1 1÷2 1 1 ù ù2
÷3 1 1 ù2 ù÷4 3 ÿ1 0 0
Some elementary character tables 181
Each r j is an irreducible representation, either by the proof of
Example 5.5(2) or by applying the result of Exercise 8.4.
If i and j are distinct integers with 1 < i , n=2 and 1 < j , n=2,
then å i 6� å j and å i 6� åÿ j, so ari and ar j have different eigenvalues.
Therefore there is no matrix T with ari � Tÿ1(ar j)T, and so ri and r j
are not equivalent.
Let ø j be the character of r j. We have now constructed distinct
irreducible characters ø j of G, one for each j which satis®es
1 < j , n=2.
At this point it is convenient to consider separately the cases where
n is odd and where n is even.
Case 1: n odd
By (12.11) the conjugacy classes of D2n (n odd) are
f1g, far, aÿrg(1 < r < (nÿ 1)=2), fasb: 0 < s < nÿ 1g:Thus there are (n� 3)=2 conjugacy classes.
The (nÿ 1)=2 irreducible characters
ø1, ø2, : : : , ø(nÿ1)=2
each have degree 2. As G has (n � 3)/2 irreducible characters in all,
there are two more to be found.
Since kal v G and G=hai � C2, we obtain two linear characters ÷1,
÷2 of G by lifting the irreducible characters of G=hai to G. These
characters ÷1 and ÷2 are given by ÷1 � 1G and
÷2(g) � 1 if g � ar for some r,
ÿ1 if g � arb for some r:
�We have now found all the irreducible characters of D2n (n odd).
(Incidentally, we have proved that D92n � kal for n odd, in view of
Theorem 17.11.)
The character table of D2n (n odd) is therefore as follows (where
å � e2ði=n):
gi 1 ar (1 < r < (n ÿ 1)/2) b|CG(gi)| 2n n 2
÷1 1 1 1÷2 1 1 ÿ1ø j 2 å jr � åÿ jr 0(1 < j < (n ÿ 1)/2)
182 Representations and characters of groups
Case 2: n even
If n is even, say n � 2m, then the conjugacy classes of D2n, as
supplied by (12.12), are
f1g, famg, far, aÿrg(1 < r < mÿ 1), fasb: s eveng, fasb: s oddg:
Hence G has m � 3 irreducible characters, of which m ÿ 1 are given
by
ø1, ø2, : : : , ømÿ1:
To ®nd the remaining four irreducible characters, we ®rst note that
ha2i � fa j: j eveng is a normal subgroup of G and
G=ha2i � fha2i, ha2ia, ha2ib, ha2iabg
� C2 3 C2:
Therefore G has four linear characters ÷1, ÷2, ÷3, ÷4 (and G9 � ka2l).Since these linear characters are the lifts of the irreducible characters
of G/ka2l, they are easy to calculate, and their values appear in the
following complete character table of D2n (n even, n � 2m, å � e2ði=n).
gi 1 am ar (1 < r < m ÿ 1) b ab|CG(gi)| 2n 2n n 4 4
÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 (ÿ1)m (ÿ1)r 1 ÿ1÷4 1 (ÿ1)m (ÿ1)r ÿ1 1ø j 2 2(ÿ1) j å jr � åÿ jr 0 0(1 < j < m ÿ 1)
18.4 Another group of order 12
We shall now describe a non-abelian group G of order 12 which is not
isomorphic to either A4 or D12, and we shall construct the character
table of G. It is in fact known that every non-abelian group of order
Some elementary character tables 183
12 is isomorphic to A4, D12 or G, but we shall not prove this result
here.
Let a and b be the following permutations in S12:
a � (1 2 3 4 5 6)(7 8 9 10 11 12),
b � (1 7 4 10)(2 12 5 9)(3 11 6 8),
and let G � ka, bl, a subgroup of S12. Since a has order 6 and b =2 kal,the group G has at least 12 elements, namely
ar, arb (0 < r < 5):
Check that a and b satisfy
a6 � 1, a3 � b2, bÿ1ab � aÿ1:
It follows from these relations that every element of G has the form
arbs with 0 < r < 5, 0 < s < 1 as given above, and so |G| � 12.
The relations further imply that
CG(a) � hai, CG(a3) � G, CG(b) � f1, a3, b, a3bg:These, and similar facts, help us to ®nd the conjugacy classes of G,
which are tabulated below:
Conjugacy class Representative gi |CG(gi)|
{1} 1 12
{a3} a3 12
{a, aÿ1} a 6
{a2, aÿ2} a2 6
{b, a2b, a4b} b 4
{ab, a3b, a5b} ab 4
Therefore G has six irreducible characters.
Observe that ka2l � {1, a2, a4} v G, and
G=ha2i � fha2i, ha2ia, ha2ib, ha2iabg:Since ka2la � ka2lb2, we have G=ha2i � C4. By lifting the irreducible
characters of C4 to G, we obtain the linear characters ÷1, ÷2, ÷3, ÷4 of
G given below:
184 Representations and characters of groups
gi 1 a3 a a2 b ab|CG(gi)| 12 12 6 6 4 4
÷1 1 1 1 1 1 1÷2 1 ÿ1 ÿ1 1 i ÿi÷3 1 1 1 1 ÿ1 ÿ1÷4 1 ÿ1 ÿ1 1 ÿi i÷5 á1 á2 á3 á4 á5 á6
÷6 â1 â2 â3 â4 â5 â6
It remains to ®nd the values ár, âr taken by the last two irreducible
characters ÷5, ÷6. For this, we shall use the column orthogonality
relations, Theorem 16.4(2).
Observe that á1, â1 are the degrees of ÷5, ÷6, so they are positive
integers; also a3 is an element of order 2, so á2 and â2 are integers by
Corollary 13.10. By the column orthogonality relations applied to
columns 1 and 2, we have
4� á21 � â2
1 � 12,
4� á22 � â2
2 � 12,
á1á2 � â1â2 � 0:
Since á1, â1 are positive integers, the ®rst equation gives á1 � â1 � 2.
The other two equations then imply that á2 � ÿâ2 � �2. Since we
have not yet distinguished between ÷5 and ÷6, we may take á2 � 2 and
â2 � ÿ2.
For r . 2, the column orthogonality relations
X6
i�1
÷i(gr)÷i(g1) � 0 andX6
i�1
÷i(gr)÷i(g2) � 0
now give us two equations involving 2ár � 2âr and 2ár ÿ 2âr, respec-
tively, so we can solve them for ár and âr. Explicitly:
r � 3: 2á3 � 2â3 � 0, 4� 2á3 ÿ 2â3 � 0;
r � 4: 4� 2á4 � 2â4 � 0, 2á4 ÿ 2â4 � 0;
r � 5: 2á5 � 2â5 � 0, 2á5 ÿ 2â5 � 0;
r � 6: 2á6 � 2â6 � 0, 2á6 ÿ 2â6 � 0:
Some elementary character tables 185
Hence
á3 � ÿ1, â3 � 1,
á4 � ÿ1, â4 � ÿ1,
á5 � 0, â5 � 0,
á6 � 0, â6 � 0:
The complete character table of G is therefore as follows:
gi 1 a3 a a2 b abCG(gi)| 12 12 6 6 4 4
÷1 1 1 1 1 1 1÷2 1 ÿ1 ÿ1 1 i ÿi÷3 1 1 1 1 ÿ1 ÿ1÷4 1 ÿ1 ÿ1 1 ÿi i÷5 2 2 ÿ1 ÿ1 0 0÷6 2 ÿ2 1 ÿ1 0 0
We can deduce that G is not isomorphic to A4 or D12 from the fact
that the character table of G is different from those of A4 and D12.
It is instructive to note that we produced the last two irreducible
characters of G by simply using the orthogonality relations, without
constructing the corresponding CG-modules. This is typical of more
advanced calculations, and illustrates the fact that it is usually much
easier to construct an irreducible character of a group than to obtain an
irreducible representation. (In fact, it is not hard to construct the
representations of the above group G with characters ÷5 and ÷6 ± see
Exercise 17.6.)
Summary of Chapter 18
In this chapter we gave the character tables of various groups, as
follows.
1. Section 18.1: the group S4.
2. Section 18.2: the group A4.
3. Section 18.3: the dihedral groups.
186 Representations and characters of groups
Exercises for Chapter 18
1. Regard D8 as a subgroup of S4 permuting the four corners of a
square, as in Example 1.1(3). Let ð be the corresponding permuta-
tion character of D8. Find the values of ð on the elements of D8,
and express ð as a sum of irreducible characters.
2. Write down explicitly the character table of D12, and show that all
its entries are integers.
Use the character table to ®nd seven distinct normal subgroups of
D12. (Hint: use Proposition 17.5.)
3. Let G � T4n � ha, b: a2n � 1, an � b2, bÿ1ab � aÿ1i, as in
Exercise 17.6. Find the character table of G.
(Hint: use the result of Exercise 17.6. It is a good idea to do the
cases n odd and n even separately.)
4. Let G � U6n � ka, b: a2n � b3 � 1, aÿ1ba � bÿ1l, as in Exercise
17.7. Find the character table of G.
5. Let G � V8n � ha, b: a2n � b4 � 1, ba � aÿ1bÿ1, bÿ1a � aÿ1bi,with n odd, as in Exercise 17.8. Find the character table of G.
Some elementary character tables 187
188
19
Tensor products
The idea of multiplying a character of a group G by a linear character of
G was introduced at the end of Chapter 17, and it can be extended to
include the product of any pair of characters ÷ and ø. The value of the
product ÷ø on an element g of G is simply ÷(g)ø(g). It is therefore
straightforward to calculate the product, but a little ingenuity is required
in order to justify the conclusion that the product ÷ø is a character of G.
The plan is to take CG-modules V and W with characters ÷ and ørespectively, and to put them together to form a new CG-module, called
the tensor product of V and W, which has character ÷ø.
An important special case of the product ÷ø occurs when ÷ � ø, so
we consider the character ÷2, and more generally ÷3, ÷4, and so on. If
÷ is not linear, then the degrees of ÷, ÷2, . . . increase, and by taking
successive powers of ÷ we obtain arbitrarily many new characters.
Potentially, then, we have a chance of getting a large proportion of the
character table of G from just one non-linear character ÷ of G; and
indeed, products of characters provide a very good source of new
characters from given ones. We shall illustrate this by constructing the
character tables of S5 and S6.
At the end of the chapter, we apply tensor products in a different
way, to ®nd all the irreducible characters of a direct product G 3 H,
given those of G and H.
Tensor product spaces
Let V and W be vector spaces over C with bases v1, . . . , vm and
w1, : : : , wn, respectively. For each i, j with 1 < i < m, 1 < j < n, we
introduce a symbol vi wj. The tensor product space V W is de®ned
to be the mn-dimensional vector space over C with a basis given by
fvi wj: 1 < i < m, 1 < j < ng:Thus V W consists of all expressions of the formX
i, j
ëij(vi wj) (ëij 2 C):
For v 2 V and w 2 W with v �Pmi�1 ëivi and w �Pn
j�1 ì jwj
(ëi, ì j 2 C), we de®ne v w 2 V W by
v w �X
i, j
ëiì j(vi wj):
For example,
(2v1 ÿ v2) (w1 � w2)
� 2v1 w1 � 2v1 w2 ÿ v2 w1 ÿ v2 w2:
Do not be misled by the notation into believing that every element of
V W has the form v w, because this is not the case. For instance,
it is impossible to express
v1 w1 � v2 w2
in the form v w.
19.1 Proposition
(1) If v 2 V, w 2 W and ë 2 C, then
v (ëw) � (ëv) w � ë(v w):
(2) If x1, . . . , xa 2 V and y1, . . . , yb 2 W, thenXa
i�1
xi
!
Xb
j�1
yj
0@ 1A �Xi, j
xi yj:
Proof (1) Let v �Pmi�1 ëivi and w �Pn
j�1 ì jw j. Then
v (ëw) �X
i
ëivi
!
Xj
ëì jw j
!�X
i, j
ëëiì j(vi wj),
(ëv) w �X
i
ëëivi
!
Xj
ì jw j
!�X
i, j
ëëiì j(vi wj),
ë(v w) � ëX
i, j
ëiì j(vi wj) �X
i, j
ëëiì j(vi wj):
Tensor products 189
Therefore v (ëw) � (ëv) w � ë(v w).
The proof of part (2) is equally straightforward, and we leave it as
an exercise. j
Our construction of V W depended upon choosing a basis of V
and a basis of W at the beginning; the next proposition shows that
other bases of V and W work equally well.
19.2 Proposition
If e1, . . . , em is a basis of V and f1, . . . , fn is a basis of W, then the
elements in
fei f j: 1 < i < m, 1 < j < nggive a basis of V W.
Proof Write
vi �Xm
k�1
ëik ek , wj �Xn
l�1
ì jl f l (ëik , ì jl 2 C):
Then by Proposition 19.1, we have
vi wj �Xk, l
ëikì jl(ek f l):
Now the elements vi wj (1 < i < m, 1 < j < n) are a basis of
V W , and hence the mn elements ek fl (1 < k < m, 1 < l < n)
span V W. Since V W has dimension mn, it follows that the
elements ek fl are also a basis of V W.
Tensor product modules
We have introduced the tensor product of two vector spaces, so we are
now in a position to de®ne the tensor product of two CG-modules.
Let G be a ®nite group and let V and W be CG-modules with bases
v1, . . . , vm and w1, : : : , wn, respectively. We know that the elements
vi wj (1 < i < m, 1 < j < n)
give a basis of V W. The multiplication of vi wj by an element of
190 Representations and characters of groups
G is de®ned in the following simple way, which is then extended
linearly to a multiplication on the whole of V W.
19.3 De®nition
Let g 2 G. For all i, j, de®ne
(vi wj)g � vi g w j g
and, more generally, letXi, j
ëij(vi wj)
!g �
Xi, j
ëij(vig wjg)
for arbitrary complex numbers ëij.
19.4 Proposition
For all v 2 V, w 2 W and all g 2 G, we have
(v w)g � vg wg:
Proof Let v �Pmi�1 ëivi and w �Pn
j�1 ì jwj. Then
(v w)g �X
i, j
ëiì j(vi wj)
!g by Proposition 19:1
�X
i, j
ëiì j(vi g wjg)
�X
i
ëivi g
!
Xj
ì jwjg
!by Proposition 19:1
� vg wg: j
You should be warned that (v w)r 6� vr wr for most elements r in
CG. For example, consider what happens when r is a scalar multiple of g.
19.5 Proposition
The rule for multiplying an element of V W by an element of G,
given in De®nition 19.3, makes the vector space V W into a CG-
module.
Tensor products 191
Proof Let 1 < i < m, 1 < j < n, and g, h 2 G. Then
(vi wj)g � vi g wjg 2 V W ,
(vi wj)(gh) � vi(gh) wj(gh)
� (vi g)h (wjg)h
� (vi g wjg)h by Proposition 19:4
� ((vi wj)g)h,
(vi wj)1 � vi wj,
and Xi, j
ëij(vi wj)
!g �
Xi, j
ëij((vi wj)g):
Therefore all the conditions of Proposition 4.6 are ful®lled, and V W
is a CG-module. j
We now calculate the character of V W.
19.6 Proposition
Let V and W be CG-modules with characters ÷ and ø, respectively.
Then the character of the CG-module V W is the product character
÷ø, where
÷ø(g) � ÷(g)ø(g) for all g 2 G:
Proof Let g 2 G. By Proposition 9.11 we can choose a basis e1, . . . ,
em of V and a basis f1, . . . , fn of W such that
eig � ëiei (1 < i < m) and f jg � ì j f j (1 < j < n)
for some complex numbers ëi, ì j. Then
÷(g) �Xm
i�1
ëi, ø(g) �Xn
j�1
ì j:
Now for 1 < i < m and 1 < j < n,
(ei f j)g � eig f jg � ëiì j(ei f j),
and by Proposition 19.2, these vectors ei fj form a basis of V W.
Hence, if ö is the character of V W then
192 Representations and characters of groups
ö(g) �X
i, j
ëiì j �X
i
ëi
! Xj
ì j
!� ÷(g)ø(g),
as required. j
19.7 Corollary
The product of two characters of G is again a character of G.
19.8 Example
The character table of S4 was given in Section 18.1. We reproduce it
here, and calculate ÷3÷4 and ÷4÷4.
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4
÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0÷4 3 1 0 ÿ1 ÿ1÷5 3 ÿ1 0 ÿ1 1
÷3÷4 6 0 0 ÿ2 0÷4÷4 9 1 0 1 1
We see that
÷3÷4 � ÷4 � ÷5, and
÷4÷4 � ÷1 � ÷3 � ÷4 � ÷5:
Powers of characters
Corollary 19.7 shows that if ÷ is a character of G then so is ÷2, where
÷2 � ÷÷, the product of ÷ with itself. More generally, for every non-
negative integer n, we de®ne ÷ n by
÷ n(g) � (÷(g))n for all g 2 G:
Thus ÷0 � 1G. An inductive proof using Corollary 19.7 shows that ÷ n
is a character of G. When ÷ is a faithful character (that is, Ker � {1}),
the powers of ÷ carry a lot of information about the whole character
table of G, as can be seen from Theorem 19.10 below.
Tensor products 193
In the course of the proof of Theorem 19.10, we shall need the
following result concerning the so-called `Vandermonde matrix'.
(19.9) If á1, . . . , ár are distinct complex numbers, then the matrix
A �
1 á1 á21 : : : árÿ1
1
1 á2 á22 : : : árÿ1
2
: : : : : : :
1 ár á2r : : : árÿ1
r
0BBB@1CCCA
is invertible.
We ®rst sketch a proof of this result.
Suppose that x1, . . . , xr are indeterminates, and consider
Ä � det
1 x1 x21 : : : xrÿ1
1
1 x2 x22 : : : xrÿ1
2
: : : : : : :
1 xr x2r : : : xrÿ1
r
0BBB@1CCCA:
If i 6� j and xi � xj then two rows of the given matrix are equal, so
Ä � 0. It follows that Ä is divisible byYi , j
(xi ÿ xj) � (x1 ÿ x2)(x1 ÿ x3) : : : (x1 ÿ xr)
3 (x2 ÿ x3) : : : (x2 ÿ xr)
..
.
3 (xrÿ1 ÿ xr):
Now the coef®cient of xrÿ11 xrÿ2
2 : : :xrÿ1 in this product is 1: for in the
way we have displayed the product, the term xrÿ11 must come from all
the factors in the ®rst row, xrÿ22 must come from all the factors in the
second row, and so on. On the other hand, to obtain xrÿ11 xrÿ2
2 : : : xrÿ1
in the expansion of the determinant Ä, we must take xrÿ11 from the
®rst row of the matrix, xrÿ22 from the second row, and so on. Hence
the coef®cient of xrÿ11 xrÿ2
2 . . . xrÿ1 in Ä is �1. It follows that
Ä � �Yi , j
(xi ÿ xj):
194 Representations and characters of groups
To obtain (19.9), we substitute ái for xi (1 < i < r), and deduce that
the matrix A is invertible since its determinant is non-zero.
19.10 Theorem
Let ÷ be a faithful character of G, and suppose that ÷(g) takes
precisely r different values as g varies over all the elements of G. Then
every irreducible character of G is a constituent of one of the powers
÷0, ÷1, . . . , ÷ rÿ1.
Proof Let the r values taken by ÷ be á1, . . . , ár, and for 1 < i < r,
de®neGi � fg 2 G: ÷(g) � áig:
Take á1 � ÷(1), so that G1 � Ker ÷. As ÷ is faithful, G1 � {1}.
Now let ø be an irreducible character of G. We must show that
h÷ j, øi 6� 0 for some j with 0 < j < r ÿ 1.
For 1 < i < r, let
âi �Xg2Gi
ø(g),
and note that â1 � ø(1) 6� 0. Then for all j > 0,
h÷ j, øi � 1
jGjXg2G
(÷(g)) j ø(g) � 1
jGjXr
i�1
(ái)jâi:
Let A be the r 3 r matrix with ij-entry (ái)jÿ1, and let b be the row
vector which is given byb � (â1, : : : , âr):
Now A is invertible by (19.9), and b 6� 0 since â1 6� 0; hence bA 6� 0.
But the ( j� 1)th entry in the row vector bA is equal to jGjh÷ j, øi,and thus h÷ j, øi 6� 0 for some j with 0 < j < r ÿ 1, as we wished to
prove. j
19.11 Examples
(1) If G 6� {1} and ÷ is the regular character of G, then ÷(g) takes just
two different values (see Proposition 13.20), so Theorem 19.10 says
that every irreducible character of G is a constituent of 1G or ÷; we
know this already, by Theorem 10.5.
(2) Let G � S4, and refer to Example 19.8. Let ÷ � ÷4. Then ÷(g)
takes four different values. We have seen that
÷2 � ÷1 � ÷3 � ÷4 � ÷5
Tensor products 195
and we ®nd that
h÷3, ÷2i � 1:
Thus ÷0, ÷1, ÷2, ÷3 (indeed, in this case, just ÷2, ÷3) have between
them as constituents all the irreducible characters ÷1, . . . , ÷5 of G,
illustrating Theorem 19.10.
Decomposing ÷2
In view of Theorem 19.10, it is of some importance to be able to
decompose powers of a character ÷ into sums of irreducible characters.
We are going to provide a method for decomposing ÷2, the square of
÷. This special case is particularly useful in ®nding irreducible char-
acters, as we shall see.
Let V be a CG-module with character ÷. By Proposition 19.6, the
module V V has character ÷2. Let v1, . . . , vn be a basis of V, and
de®ne a linear transformation T: V V! V V by
(vi v j)T � v j vi for all i, j
and extending linearly ± that is,Xi, j
ëij(vi v j)
!T �
Xi, j
ëij(v j vi):
Check that for all v, w 2 V, we have
(v w)T � w v:
Hence T is independent of the choice of basis.
Now de®ne subsets of V V as follows:
S(V V ) � fx 2 V V : xT � xg,A(V V ) � fx 2 V V : xT � ÿxg,
Since T is linear, it is easy to see that S(V V) and A(V V) are
subspaces of V V (indeed, they are eigenspaces of T). The subspace
S(V V) is called the symmetric part of V V, and the subspace
A(V V ) is known as the antisymmetric part of V V.
19.12 Proposition
The subspaces S(V V) and A(V V) are CG-submodules of V V.
Also,V V � S(V V )� A(V V ):
196 Representations and characters of groups
Proof For all ëij 2 C and g 2 G,Xi, j
ëij(vi v j)
!Tg �
Xi, j
ëij(v j g vi g)
�X
i, j
ëij(vi g v j g)T
�X
i, j
ëij(vi v j)
!gT :
Therefore T is a CG-homomorphism from V V to itself. Hence, for
x 2 S(V V), y 2 A(V V) and g 2 G, we have
(xg)T � (xT )g � xg, and
(yg)T � (yT )g � ÿyg,
so xg 2 S(V V) and yg 2 A(V V). Thus S(V V) and A(V V)
are CG-submodules of V V.
If x 2 S(V V) \ A(V V) then x � xT � ÿx, so x � 0. Further, for
all x 2 V we have
x � 12(x� xT )� 1
2(xÿ xT ):
Since T 2 is the identity, 12(x� xT ) 2 S(V V ) and 1
2(xÿ xT ) 2
A(V V ). Therefore,
V V � S(V V )� A(V V ): j
Note that the symmetric part of V V contains all vectors which
have the form v w � w v with v, w 2 V, while the antisymmetric
part of V V contains all vectors of the form v w ÿ w v. We now
present bases of the symmetric and antisymmetric parts of V V
which consist of elements like these.
19.13 Proposition
Let v1, . . . , vn be a basis of V.
(1) The vectors vi v j � v j vi (1 < i < j < n) form a basis of
S(V V ). The dimension of S(V V ) is n(n� 1)=2.
(2) The vectors vi v j ÿ v j vi (1 < i , j < n) form a basis of
A(V V). The dimension of A(V V ) is n(nÿ 1)=2.
Tensor products 197
Proof Clearly the vectors vi v j � v j vi (1 < i < j < n) are linearly
independent elements of S(V V ), and the vectors vi v j ÿ v j vi
(1 < i , j < n) are linearly independent elements of A(V V ). Hence
dim S(V V ) > n(n� 1)=2, dim A(V V ) > n(nÿ 1)=2:
By Proposition 19.12,
dim S(V V ) � dim A(V V ) � dim V V � n2:
Hence the above inequalities are equalities, and the result follows. j
De®ne ÷S to be the character of the CG-module S(V V ), and ÷A
to be the character of the CG-module A(V V ). By Proposition 19.12,
÷2 � ÷S � ÷A:
The next result gives the values of the characters ÷S and ÷A.
19.14 Proposition
For g 2 G, we have
÷S(g) � 12(÷2(g)� ÷(g2)), and
÷A(g) � 12(÷2(g)ÿ ÷(g2)):
Proof By Proposition 9.11 we can choose a basis e1, . . . , en of V such
that ei g � ëiei (1 < i < n) for some complex numbers ëi. Then
(ei ej ÿ ej ei)g � ëië j(ei ej ÿ ej ei),
and hence from Proposition 19.13(2),
÷A(g) �Xi , j
ëië j:
Now ei g2 � ë2i ei, so ÷(g) �Pi ëi and ÷(g2) �Pi ë2
i . Therefore
÷2(g) � (÷(g))2 �X
i
ë2i � 2
Xi , j
ëië j � ÷(g2)� 2÷A(g):
Hence
÷A(g) � 12(÷2(g)ÿ ÷(g2)):
Also, ÷2 � ÷S � ÷A, which implies that
÷S(g) � ÷2(g)ÿ ÷A(g) � 12(÷2(g)� ÷(g2)): j
198 Representations and characters of groups
19.15 Example
Let G � S4. The character table of G is given in Example 19.8. Let
÷ � ÷4. The values of ÷, and the values of ÷S and ÷A, given by
Proposition 19.14, appear below.
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4
÷ 3 1 0 ÿ1 ÿ1÷S 6 2 0 2 0÷A 3 ÿ1 0 ÿ1 1
We ®nd that ÷S � ÷1 � ÷3 � ÷4 and ÷A � ÷5.
The techniques which we have developed so far give a useful method
for ®nding new irreducible characters of a group, given one or two
irreducible characters to start with. The strategy is simple:
(1) Given a character ÷, form ÷S and ÷A, and use inner products to
analyse ÷S and ÷A for new irreducible characters.
(2) If ø is a new character found in (1), then form øS and øA and
repeat.
We illustrate this strategy with two examples.
19.16 Example The character table of S5
Let G � S5, the symmetric group of degree 5. By Example 12.16(4), G
has conjugacy class representatives gi, conjugacy class sizes and
centralizer orders |CG(gi)| as follows:
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)Class size 1 10 20 15 30 20 24|CG(gi)| 120 12 6 8 4 6 5
Thus G has exactly seven irreducible characters.
(a) Linear characters By Example 17.13, G9 � A5 and G has exactly
two linear characters ÷1, ÷2, obtained by lifting the irreducible char-
acters of G=G9. We have
Tensor products 199
÷1 � 1G, and
÷2(g) � 1, if g is an even permutation,
ÿ1, if g is an odd permutation:
�(b) Permutation character Proposition 13.24 gives us a character ÷3
of G with values
÷3(g) � jfix (g)j ÿ 1 (g 2 G):
Observe that
h÷3, ÷3i � 42
120� 22
12� 12
6� (ÿ1)2
6� (ÿ1)2
5� 1:
Hence ÷3 is irreducible, by Theorem 14.20. Next, Proposition 17.14
shows that ÷4 � ÷3÷2 is also an irreducible character.
At this point we have the following portion of the character table of
G:
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)|CG(gi)| 120 12 6 8 4 6 5
÷1 1 1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1 ÿ1 1÷3 4 2 1 0 0 ÿ1 ÿ1÷4 4 ÿ2 1 0 0 1 ÿ1
(c) Tensor products We now use tensor products to construct the last
three irreducible characters of G.
Write ÷ � ÷3. By Proposition 19.14 the values of the characters ÷S
and ÷A are
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)|CG(gi)| 120 12 6 8 4 6 5
÷S 10 4 1 2 0 1 0÷A 6 0 0 ÿ2 0 0 1
Thus,
200 Representations and characters of groups
h÷A, ÷Ai � 36
120� 4
8� 1
5� 1,
and so ÷A is a new irreducible character, which we call ÷5.
Next,
h÷S , ÷1i � 10
120� 4
12� 1
6� 2
8� 1
6� 1,
h÷S , ÷3i � 40
120� 8
12� 1
6ÿ 1
6� 1, and
h÷S , ÷Si � 100
120� 16
12� 1
6� 4
8� 1
6� 3,
Therefore,
÷S � ÷1 � ÷3 � ø,
where ø is an irreducible character of degree 5. Let ÷6 � ø, so that
÷6 � ÷S ÿ ÷1 ÿ ÷3.
Finally, ÷7 � ÷6÷2 is a different irreducible character of degree 5.
We have now found all seven irreducible characters of S5. The
character table of S5 is as shown.
Character table of S5
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)|CG(gi)| 120 12 6 8 4 6 5
÷1 1 1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1 ÿ1 1÷3 4 2 1 0 0 ÿ1 ÿ1÷4 4 ÿ2 1 0 0 1 ÿ1÷5 6 0 0 ÿ2 0 0 1÷6 5 1 ÿ1 1 ÿ1 1 0÷7 5 ÿ1 ÿ1 1 1 ÿ1 0
19.17 Example The character table of S6
In this example, we use techniques similar to those of the previous
example to ®nd 8 of the 11 irreducible characters of the symmetric
group S6; we then ®nd the last three irreducible characters by using the
orthogonality relations.
Let G � S6, of order 720. For ease of printing, it is convenient to
label each conjugacy class by the cycle-shape of its elements. Using
Tensor products 201
this notation, the conjugacy class sizes and centralizer orders are as
follows (see Exercise 12.3):
Cycle-shape (1) (2) (3) (2,2) (4) (3,2) (5) (2,2,2) (3,3) (4,2) (6)Class size 1 15 40 45 90 120 144 15 40 90 120|CG(gi)| 720 48 18 16 8 6 5 48 18 8 6
Since G has 11 conjugacy classes, it has 11 irreducible characters.
(a) Linear characters As with all symmetric groups Sn (n > 2), the
derived subgroup is An, and we get exactly two linear characters ÷1
and ÷2, where
÷1 � 1G,
÷2(g) � 1, if g is even,
ÿ1, if g is odd
�(see Example 17.13).
(b) Permutation character and tensor products The function ÷3 which
is given by
÷3(g) � jfix (g)j ÿ 1 (g 2 G)
is a character of G, by Proposition 13.24. Let ÷ � ÷3. The values of ÷,
÷S and ÷A are as follows:
Class (1) (2) (3) (2,2) (4) (3,2) (5) (2,2,2) (3,3) (4,2) (6)|CG(gi)| 720 48 18 16 8 6 5 48 18 8 6
÷ � ÷3 5 3 2 1 1 0 0 ÿ1 ÿ1 ÿ1 ÿ1÷S 15 7 3 3 1 1 0 3 0 1 0÷A 10 2 1 ÿ2 0 ÿ1 0 ÿ2 1 0 1
We calculate that
h÷3, ÷3i � 1, h÷A, ÷Ai � 1,
h÷S , ÷1i � 1, h÷S , ÷3i � 1,
h÷S , ÷Si � 3:
202 Representations and characters of groups
Therefore ÷3 is irreducible; so is ÷4 � ÷3÷2. Also, ÷5 � ÷A is irreduci-
ble, as is ÷6 � ÷5÷2. Further,
÷S � ÷1 � ÷3 � ÷7,
where ÷7 is another irreducible character, of degree 9. Finally,
÷8 � ÷7÷2 is also irreducible.
The irreducible characters ÷1, . . . , ÷8 are recorded in the following
portion of the character table of G.
Class (1) (2) (3) (2,2) (4) (3,2) (5) (2,2,2) (3,3) (4,2) (6)|CG(gi)| 720 48 18 16 8 6 5 48 18 8 6
÷1 1 1 1 1 1 1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1 ÿ1 1 ÿ1 1 1 ÿ1÷3 5 3 2 1 1 0 0 ÿ1 ÿ1 ÿ1 ÿ1÷4 5 ÿ3 2 1 ÿ1 0 0 1 ÿ1 ÿ1 1÷5 10 2 1 ÿ2 0 ÿ1 0 ÿ2 1 0 1÷6 10 ÿ2 1 ÿ2 0 1 0 2 1 0 ÿ1÷7 9 3 0 1 ÿ1 0 ÿ1 3 0 1 0÷8 9 ÿ3 0 1 1 0 ÿ1 ÿ3 0 1 0
(c) Orthogonality relations We now use the column orthogonality
relations to complete the character table of G. It will be shown later
(Corollary 22.16) that all the entries in the character tables of all
symmetric groups are integers, but for the moment we know for certain
only that ÷(g) is an integer if g2 � 1 (see Corollary 13.10). It is
therefore convenient ®rst to concentrate on elements of order 2, so that
we can guarantee that the solutions to the equations which we deal
with are integers.
Let s denote the permutation (1 2) and t denote the permutation
(1 2)(3 4). Thus the conjugacy classes of s and t correspond to the
second and fourth columns of the character table, respectively, in the
ordering which we have adopted. From Corollary 13.10 we know that
÷(s) and ÷(t) are integers for all characters ÷ of G.
Ingeniously, we call the three irreducible characters of G which have
yet to be found ÷9, ÷10 and ÷11.
The column orthogonality relations give
X11
i�1
÷i(s)2 � 48:
Tensor products 203
Hence
÷9(s)2 � ÷10(s)2 � ÷11(s)2 � 2:
We can assume, without loss of generality, that
÷9(s)2 � ÷10(s)2 � 1, ÷11(s)2 � 0:
Now ÷9÷2 is an irreducible character, and is not equal to any of
÷1, . . . , ÷8. Moreover, since ÷9÷2(s) � ÿ÷9(s), we see that ÷9÷2 is not
÷9 or ÷11. Therefore,
÷9÷2 � ÷10:
Once more, we lose no generality in assuming that
÷9(s) � 1, ÷10(s) � ÿ1:
The plan is now to ®nd ÷i(1) and ÷i(t) for i � 9, 10, 11. That is, we
aim to evaluate the integers a, b, c, d, e, f in the following portion of
the character table:
The column orthogonality relations giveX11
i�1
÷i(1)÷i(s) � 0,X11
i�1
÷i(s)÷i(t) � 0,
X11
i�1
÷i(t)÷i(t) � 16,X11
i�1
÷i(1)÷i(t) � 0,
whence
aÿ b � 0, d ÿ e � 0,
d2 � e2 � f 2 � 2, ad � be� cf � 10:
The only solution to these equations in integers with a . 0 and b . 0
is
d � e � 1, f � 0, a � b � 5:
Element 1 s tClass (1) (2) (2,2)
÷9 a 1 d÷10 b ÿ1 e÷11 c 0 f
204 Representations and characters of groups
Finally, we ®nd that c � 16 by using the relation
X11
i�1
÷i(1)2 � 720:
The above portion of the character table is therefore
We can now determine the three unknown entries in each further
column, since the column orthogonality relations will give three
independent equations in these unknowns (as the above 3 3 3 matrix is
invertible). Having done these calculations, we ®nd that the complete
character table of S6 is as shown.
Direct products
We conclude the chapter by showing that tensor products can be used
to determine the character table of a direct product G 3 H, given the
character tables of G and H.
Let V be a CG-module, with basis v1, . . . , vm, and let W be a
Element 1 s tClass (1) (2) (2,2)
÷9 5 1 1÷10 5 ÿ1 1÷11 16 0 0
Character table of S6
Class (1) (2) (3) (2,2) (4) (3,2) (5) (2,2,2) (3,3) (4,2) (6)|CG(gi)| 720 48 18 16 8 6 5 48 18 8 6
÷1 1 1 1 1 1 1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1 ÿ1 1 ÿ1 1 1 ÿ1÷3 5 3 2 1 1 0 0 ÿ1 ÿ1 ÿ1 ÿ1÷4 5 ÿ3 2 1 ÿ1 0 0 1 ÿ1 ÿ1 1÷5 10 2 1 ÿ2 0 ÿ1 0 ÿ2 1 0 1÷6 10 ÿ2 1 ÿ2 0 1 0 2 1 0 ÿ1÷7 9 3 0 1 ÿ1 0 ÿ1 3 0 1 0÷8 9 ÿ3 0 1 1 0 ÿ1 ÿ3 0 1 0÷9 5 1 ÿ1 1 ÿ1 1 0 ÿ3 2 ÿ1 0÷10 5 ÿ1 ÿ1 1 1 ÿ1 0 3 2 ÿ1 0÷11 16 0 ÿ2 0 0 0 1 0 ÿ2 0 0
Tensor products 205
CH-module, with basis w1, . . . , wn. For all i, j with 1 < i < m and
1 < j < n, and all g 2 G, h 2 H, de®ne
(vi wj)(g, h) � vi g wjh
and extend this de®nition linearly to the whole of V W, that is, for
ëij 2 C, Xi, j
ëij(vi wj)
!(g, h) �
Xi, j
ëij(vi g wjh):
As in Proposition 19.4, we ®nd that
(v w)(g, h) � vg wh,
for all v 2 V, w 2 W. Then a proof similar to that of Proposition 19.5
shows that V W is a C(G 3 H)-module.
Let ÷ be the character of V and ø be the character of W. By the
proof of Proposition 19.6, the character of V W is ÷ 3 ø, where
(÷ 3 ø)(g, h) � ÷(g)ø(h) (g 2 G, h 2 H):
19.18 Theorem
Let ÷1, . . . , ÷a be the distinct irreducible characters of G and let
ø1, . . . , øb be the distinct irreducible characters of H. Then G 3 H
has precisely ab distinct irreducible characters, and these are
÷i 3 ø j (1 < i < a, 1 < j < b):
Proof For all i, j, k, l,
h÷i 3 ø j, ÷k 3 ÷ liG3 H � 1
jG 3 H jXg2Gh2H
÷i(g)ø j(h)÷k(g)ø l(h)
� 1
jGjXg2G
÷i(g)÷k(g)
!1
jH jXh2H
ø j(h)ø l(h)
!� h÷i, ÷kiGhø j, ø liH � äikä jl:
(Here the subscripts G 3 H, G and H indicate inner products of
characters of G 3 H, G and H, respectively.) Thus the ab characters
÷i 3 ø j are distinct and irreducible.
Next, note that for all g, x 2 G and h, y 2 H, we have
(x, y)ÿ1(g, h)(x, y) � (xÿ1 gx, yÿ1 hy):
206 Representations and characters of groups
Hence elements (g, h) and (g9, h9) of G 3 H are conjugate if and only
if the elements g and g9 are conjugate in G and the elements h and
h9 are conjugate in H. Consequently, if g1, . . . , ga are representatives
of the conjugacy classes of G and h1, . . . , hb are representatives of the
conjugacy classes of H, then the elements
(gi, hj) (1 < i < a, 1 < j < b)
are representatives of the conjugacy classes of G 3 H. In particular,
G 3 H has precisely ab conjugacy classes.
By Theorem 15.3, G 3 H has exactly ab irreducible characters, so
the irreducible characters ÷i 3 ø j which we have found must be all the
irreducible characters of G 3 H. j
19.19 Example The character table of S3 3 C2
The character table of S3 (� D6) is given in Example 16.3(1). We
reproduce it here, alongside the character table of C2.
The conjugacy classes of S3 3 C2 are represented by
(1, 1), ((1 2), 1), ((1 2 3), 1), (1, ÿ1), ((1 2), ÿ1), ((1 2 3), ÿ1),
and by Theorem 19.18, the character table of S3 3 C2 is as shown.
Character table of S3
gi 1 (1 2) (1 2 3)|CG(gi)| 6 2 3
÷1 1 1 1÷2 1 ÿ1 1÷3 2 0 ÿ1
Character table of C2
hi 1 ÿ1|CH (hi)| 2 2
ø1 1 1ø2 1 ÿ1
Character table of S3 3 C2
(gi, hj) (1, 1) ((1 2), 1) ((1 2 3), 1) (1, ÿ1) ((1 2), ÿ1) ((1 2 3), ÿ1)|CG3 H (gi, hj)| 12 4 6 12 4 6
÷1 3 ø1 1 1 1 1 1 1÷2 3 ø1 1 ÿ1 1 1 ÿ1 1÷3 3 ø1 2 0 ÿ1 2 0 ÿ1÷1 3 ø2 1 1 1 ÿ1 ÿ1 ÿ1÷2 3 ø2 1 ÿ1 1 ÿ1 1 ÿ1÷3 3 ø2 2 0 ÿ1 ÿ2 0 1
Tensor products 207
Compare the solution to Exercise 18.2, where we give the character
table of D12 (Exercise 1.5 shows that D12 � S3 3 C2).
Summary of Chapter 19
1. The product of any two characters of G is a character of G.
2. If ÷ is a character of G, then so are ÷S and ÷A, where
÷S(g) � 12(÷2(g)� ÷(g2)),
÷A(g) � 12(÷2(g)ÿ ÷(g2))
for all g 2 G.
3. The irreducible characters of G 3 H are those characters ÷ 3 ø,
where ÷ is an irreducible character of G and ø is an irreducible
character of H. The values of ÷ 3 ø are given by
(÷ 3 ø)(g, h) � ÷(g)ø(h)
for all g 2 G, h 2 H.
Exercises for Chapter 19
1. Let ÷, ø and ö be characters of the group G. Show that
h÷ø, öi � h÷, øöi � hø, ÷öi:2. Suppose that ÷ and ø are irreducible characters of G. Prove that
h÷ø, 1Gi � 1, if ÷ � ø,
0, if ÷ 6� ø:
�3. Let ÷ be a character of G which is not faithful. Show that there is
some irreducible character ø of G such that k÷ n, øl � 0 for all
integers n with n > 0.
(This shows that the hypothesis that ÷ is faithful cannot be dropped
from Theorem 19.10.)
4. In Example 20.13 of the next chapter we shall show that there exist
irreducible characters ÷ and ö of A5 which take the following
values:
208 Representations and characters of groups
1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)
÷ 5 ÿ1 1 0 0ö 3 0 ÿ1 (1�p5)=2 (1ÿp5)=2
Calculate the values of ÷S , ÷A, öS and öA. Express these characters
as linear combinations of the irreducible characters ø1, . . . , ø5 of
A5 which are given in Example 20.13.
5. A certain group G or order 24 has precisely seven conjugacy classes
with representatives g1, . . . , g7; further, G has a character ÷ with
values as follows:
gi g1 g2 g3 g4 g5 g6 g7
|CG(gi)| 24 24 4 6 6 6 6
÷ 2 ÿ2 0 ÿù2 ÿù ù ù2
where ù � e2ði=3. Moreover, g21, g2
2, g23, g2
4, g25, g2
6, g27 are con-
jugate to g1, g1, g2, g5, g4, g4, g5, respectively.
Find ÷S and ÷A, and show that both are irreducible.
By forming products of the irreducible characters found so far,
®nd the character table of G.
6. Write down the character table of D6 3 D6.
Tensor products 209
210
20
Restriction to a subgroup
In this chapter and the next, we are going to look at ways of relating the
representations of a group to the representations of its subgroups. Here, we
introduce the elementary idea of restricting a CG-module to a subgroup H
of G, and illustrate its use. The case where H is a normal subgroup of G is
of particular interest, and Clifford's Theorem 20.8 gives important informa-
tion in this case. We apply this result in the situation where H is of index 2
in G, which occurs, for example, when G � Sn and H � An.
Restriction
Let H be a subgroup of the ®nite group G. Then CH is a subset of CG. If
V is a CG-module, then V is also a CH-module, since properties (1)±(5) of
De®nition 4.2 certainly hold for all g, h 2 H if they hold for all g, h 2 G.
This simple way of converting a CG-module into a CH-module is known
as restricting from G to H. If V is a CG-module then we write the
corresponding CH-module as V # H , and call it the restriction of V to H.
The character of V # H is obtained from the character ÷ of V by
evaluating ÷ on the elements of H only. We write this character of H
as ÷ # H , and refer to it as the restriction of ÷ to H. More generally,
if f: G! C is any function, then f # H denotes the restriction of f to
H (so that ( f # H)(h) � f (h) for all h 2 H).
20.1 Example
Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. As in Example 4.5(1),
let V be the CG-module with basis v1, v2 for which
v1a � v2, v1b � v1,
v2a � ÿv1, v2b � ÿv2:
If H is the subgroup {1, a2, b, a2b} of G, then V # H is the CH-
module with basis v1, v2 for which
v1a2 � ÿv1, v1b � v1,
v2a2 � ÿv2, v2b � ÿv2:
The character ÷ of V is given by
g 1 a a2 a3 b ab a2b a3b
÷(g) 2 0 ÿ2 0 0 0 0 0
and the character ÷ # H of V # H is given by
If V is a CG-module and H is a subgroup of G, then
dim V � dim (V # H). However, it might be the case that V is an
irreducible CG-module while V # H is not an irreducible CH-module;
Example 20.1 illustrates this fact. On the other hand, if V # H is an
irreducible CH-module then V is an irreducible CG-module; for if U
is a CG-submodule of V, then U # H is a CH-submodule of V # H .
20.2 Example
Let G � S5 and let H be the subgroup A4 of G consisting of all even
permutations of {1, 2, 3, 4} ®xing 5. By 18.2, the character table of H
is
h 1 a2 b a2b
÷(h) 2 ÿ2 0 0
gi 1 (1 2)(3 4) (1 2 3) (1 3 2)jC H (gi)j 12 4 3 3
ø1 1 1 1 1ø2 1 1 ù ù2
ø3 1 1 ù2 ùø4 3 ÿ1 0 0
Restriction to a subgroup 211
(where ù � e2ði=3). The character table of G is given in Example
19.16, with irreducible characters labelled ÷1, . . . , ÷7.
For each i with 1 < i < 7, we calculate the character ÷i # H as a
sum of irreducible characters ø j. From Example 19.16 we see that
÷1 # H � ÷2 # H , ÷3 # H � ÷4 # H and ÷6 # H � ÷7 # H :
Therefore we need only consider ÷1 # H, ÷3 # H , ÷5 # H and ÷6 # H .
These have values
It is easy to spot that
÷1 # H � ø1,
÷3 # H � ø1 � ø4,
÷5 # H � 2ø4,
÷6 # H � ø2 � ø3 � ø4:
Constituents of a restricted character
To help us discuss the way in which a restricted character ÷ # H can
be expressed in terms of the irreducible characters of H, we introduce
the following notation.
20.3 De®nitions
The inner product k , lG is the inner product on the vector space of
functions from G to C which we have de®ned earlier, and k , l H is the
inner product on the vector space of functions from H to C, de®ned
similarly. Thus, if W1 and W2 are functions from G to C, then
hW1, W2iG � 1
jGjXg2G
W1(g)W2(g),
1 (1 2)(3 4) (1 2 3) (1 3 2)
÷1 # H 1 1 1 1÷3 # H 4 0 1 1÷5 # H 6 ÿ2 0 0÷6 # H 5 1 ÿ1 ÿ1
212 Representations and characters of groups
and if ö1 and ö2 are functions from H to C, then
hö1, ö2iH � 1
jH jXh2H
ö1(h)ö2(h):
If ÷ is a character of G and ø1, . . . , ør are the irreducible
characters of the subgroup H of G, then by Theorem 14.17,
÷ # H � d1ø1 � : : :� drør
for some non-negative integers d1, : : : , d r which are given by
di � h÷ # H , øiiH :
We say that øi is a constituent of ÷ # H if the coef®cient di in the
above expression is non-zero.
The next proposition shows that every irreducible character of H is a
constituent of the restriction of some irreducible character of G.
20.4 Proposition
Let H be a subgroup of G and let ø be a non-zero character of H.
Then there exists an irreducible character ÷ of G such that
h÷ # H , øiH 6� 0:
Proof Let ÷1, . . . , ÷k be the irreducible characters of G. Recall from
Theorem 13.19 and Proposition 13.20 that the regular character ÷reg of
G satis®es
÷reg(g) � jGj if g � 1,
0 if g 6� 1,
�and ÷reg �
Xk
i�1
÷i(1)÷i:
Now
0 6� jGjjH jø(1) � h÷reg # H , øiH �Xk
i�1
÷i(1)h÷i # H , øiH :
Therefore h÷i # H , øiH 6� 0 for some i. j
Suppose that we know the character table of G. In the light of
Proposition 20.4, we could hope to ®nd the character table of the
subgroup H by restricting the irreducible characters ÷ of G to H.
Unfortunately, it may be very dif®cult in practice to write down the
restrictions ÷ # H in terms of irreducible characters of H. The best
Restriction to a subgroup 213
chance of doing this occurs when the index jG: H j(� jGj=jH j) is
small, since the restrictions ÷ # H then do not have many constituents,
as the following result shows.
20.5 Proposition
Let H be a subgroup of G, let ÷ be an irreducible character of G, and
let ø1, : : : , ør be the irreducible characters of H. Then ÷ # H �d1ø1 � : : : � d rør, where the non-negative integers d1, . . . , d r
satisfy Xr
i�1
d2i < jG: H j:(20:6)
Moreover, we have equality in (20.6) if and only if ÷(g) � 0 for all
elements g of G which lie outside H.
Proof By Theorem 14.17, we haveXr
i�1
d2i � h÷ # H , ÷ # HiH � 1
jH jXh2H
÷(h)÷(h):
Also, since ÷ is irreducible,
1 � h÷, ÷iG � 1
jGjXg2G
÷(g)÷(g)
� 1
jGjXh2H
÷(h)÷(h)� K
� jH jjGjXr
i�1
d2i � K,
where K � (1=jGj)P g=2H÷(g)÷(g): Now K > 0, and K � 0 if and only
if ÷(g) � 0 for all g with g =2 H. The conclusions of the proposition
follow at once. j
We can say more about the constituents of ÷ # H in the case where
H is a normal subgroup of G. For example, we will see that all the
constituents of ÷ # H have the same degree.
The way to exploit the fact that H v G us revealed in the following
proposition.
214 Representations and characters of groups
20.7 Proposition
Suppose that H v G. Let V be an irreducible CG-module and U be an
irreducible CH-submodule of V # H. For every g 2 G let Ug �fug: u 2 Ug. Then
(1) The set Ug is an irreducible CH-submodule of V # H and
dim Ug � dim U .
(2) As a CH-module, V is a direct sum of some of the CH-
modules Ug.
(3) If g1, g2, g 2 G and Ug1 and Ug2 are isomorphic CH-modules,
then Ug1 g and Ug2 g are isomorphic CH-modules.
Proof (1) Clearly, Ug is a subspace of V; and since H v G we have
ghgÿ1 2 H for all h 2 H, so
(ug)h � u(ghgÿ1)g 2 Ug (u 2 U),
proving that Ug is a CH-submodule of V # H. Further, if W is a CH-
submodule of Ug, then Wgÿ1 is a CH-submodule of U; since U is
irreducible, Wgÿ1 � {0} or U, whence W � {0} or Ug. Therefore, Ug
is an irreducible CH-module, as claimed. Moreover, u! ug (u 2 U) is
an invertible linear transformation from U to Ug, so dim U � dim Ug.
(2) The sum of all the subspaces Ug with g 2 G is a CG-sub-
module of V. Therefore, since V is irreducible, we have
V �Xg2G
Ug:
Then by Proposition 7.12, V is a direct sum of some of the CH-
modules Ug.
(3) Now let ö be a CH-isomorphism from Ug1 to Ug2. De®ne
è : Ug1 g! Ug2 g by
è : wg ! (wö)g (w 2 Ug1):
Then è is clearly an isomorphism of vector spaces. Suppose that
h 2 H . Then gh � h9g for some h9 2 H, and
(wgh)è � (wh9g)è � (wh9ö)g � (wö)h9g
� (wö)gh � (wgè)h:
Therefore, è is a CH-isomorphism, and the proof of the proposition is
complete. j
Restriction to a subgroup 215
We now come the fundamental theorem on the restriction of a
character to a normal subgroup.
20.8 Clifford's Theorem
Suppose that H v G and that ÷ is an irreducible character of G. Then
(1) all the constituents of ÷ # H have the same degree; and
(2) if ø1, . . . , øm are the constituents of ÷ # H, then
÷ # H � e(ø1 � : : : � øm)
for some positive integer e.
Proof Let V be a CG-module with character ÷. Then it follows from
Proposition 20.7, parts (1) and (2), that all the constituents of ÷ # H
have the same degree.
Let e � h÷ # H , ø1i. Then V contains a CH-module X1 whose
character is eø1, and which is therefore a direct sum of e isomorphic
CH-modules, each having character ø1; say
X 1 � U1 � : : : � Ue:
By Proposition 20.7(3), if g 2 G then X1 g is a direct sum of
isomorphic CH-modules. On the other hand, V is a sum of CH-
modules of the form X1 g, by Proposition 20.7(2). Hence V has the
form
V � X 1 � : : : � X m
where each Xi is a direct sum of e isomorphic CH-modules, and
Xi 6� Xj if i 6� j. Therefore,
÷ # H � e(ø1 � : : : � øm): j
Our main applications of Clifford's Theorem will concern the case
where jG: H j � 2, but you might like to look at Corollary 22.14 in
Chapter 22 to see an advanced use of the theorem.
Normal subgroups of index 2
We are shortly going to give more precise information about the
constituents of ÷ # H when H is a normal subgroup of G of index 2
(that is, jG: H j � 2). Examples where this happens are G � Sn,
216 Representations and characters of groups
H � An, or G � D2n � ha, b: an � b2 � 1, bÿ1ab � aÿ1i, H � hai. In
fact, if H is a subgroup of index 2 in G, then H must be normal in G
(see Exercise 1.10).
When H is a normal subgroup of index 2 in G, the character tables
of G and H are closely related. We describe this relationship in
(20.13) below, and we then illustrate the results by ®nding the character
table of A5 from that of S5 (which we have already obtained in
Example 19.16).
20.9 Proposition
Suppose that H is a normal subgroup of index 2 in G, and let ÷ be an
irreducible character of G. Then either
(1) ÷ # H is irreducible, or
(2) ÷ # H is the sum of two distinct irreducible characters of H of the
same degree.
Proof If ø1, . . . , ør are the irreducible characters of H, then by
Proposition 20.5,
÷ # H � d1ø1 � : : :� drør,
wherePr
i�1d2i < 2. Since d1, : : : , d r are non-negative integers, we
deduce that either ÷ # H � øi for some i, or ÷ # H � øi � ø j for some
i, j with i 6� j. In the latter case, øi and ø j have the same degree, by
Clifford's Theorem 20.8 j
For practical purposes, it is often desirable to have more details
about the two cases in Proposition 20.9, and we shall supply these
next.
Since G=H � C2, we may lift the non-trivial linear character of
G=H to obtain a linear character ë of G which satis®es
ë(g) �1 if g 2 H ,
ÿ1 if g =2 H :
(Note that for all irreducible characters ÷ of G, ÷ and ÷ë are irreducible
characters of the same degree (see Proposition 17.14). Also,
÷ # H � ÷ë # H , since ë(h) � 1 for all h 2 H.
20.10 Proposition
Suppose that H is a normal subgroup of index 2 in G, and that ÷ is
Restriction to a subgroup 217
an irreducible character of G. Then the following three conditions are
equivalent:
(1) ÷ # H is irreducible;
(2) ÷(g) 6� 0 for some g 2 G with g =2 H;
(3) the characters ÷ and ÷ë of G are not equal.
Proof We use Proposition 20.5; since jG: H j � 2, ÷ # H is irreducible
if and only if the inequality in (20.6) is strict, and this happens if and
only if ÷(g) 6� 0 for some g 2 G with g =2 H. Thus (1) is equivalent to
(2).
To see that (2) is equivalent to (3), observe that
÷ë(g) �÷(g) if g 2 H ,
ÿ÷(g) if g =2 H ,
(
so ÷(g) 6� 0 for some g with g =2 H if and only if ÷ë 6� ÷. j
According to Proposition 20.9, if H is a normal subgroup of G of
index 2, and ÷ is an irreducible character of G, then ÷ # H is the sum
of one or two irreducible characters of H. In the next proposition we
consider the ®rst possibility.
20.11 Proposition
Suppose that H is a normal subgroup of index 2 in G, and that ÷ is
an irreducible character of G for which ÷ # H is irreducible. If ö is an
irreducible character of G which satis®es
ö # H � ÷ # H ,
then either ö � ÷ or ö � ÷ë.
Proof We have
(÷� ÷ë)(g) �2÷(g) if g 2 H ,
0 if g =2 H :
(
Hence
218 Representations and characters of groups
h÷� ÷ë, öiG � 1
jGjXg2H
2÷(g)ö(g)
� 1
jH jXg2H
÷(g)ö(g)
� h÷ # H , ö # HiH :
Now k÷ # H, ö # Hl H � 1, since ÷ # H is irreducible and
ö # H � ÷ # H. Therefore k÷ � ÷ë, ölG � 1, and so either ö � ÷ or
ö � ÷ë. j
Finally, we look at the case where ÷ # H is reducible.
20.12 Proposition
Suppose that H is a normal subgroup of index 2 in G, and that ÷ is
an irreducible character of G for which ÷ # H is the sum of two
irreducible characters of H, say ÷ # H � ø1 � ø2. If ö is an irreduci-
ble character of G such that ö # H has ø1 or ø2 as a constituent,
then ö � ÷.
Proof In view of Proposition 20.10, ÷(g) � 0 for all g with g =2 H.
Therefore,
hö, ÷iG � 1
jGjXg2G
ö(g)÷(g) � 1
jGjXg2H
ö(g)÷(g)
� 12hö # H , ÷ # HiH :
If ö # H has ø1 or ø2 as a constituent, then hö # H , ÷ # HiH 6� 0, so
kö, ÷lG 6� 0, and hence ö � ÷. j
We summarize our results on subgroups of index 2, by explaining
how to list the irreducible characters of H on the assumption that we
know the character table of G.
(20.13) Let H be a normal subgroup of index 2 in the group G.
(1) Each irreducible character ÷ of G which is non-
zero somewhere outside H restricts to be an
irreducible character of H. Such characters of G
occur in pairs (÷ and ÷ë) which have the same
restriction to H (Propositions 20.10, 20.11).
Restriction to a subgroup 219
(2) If ÷ is an irreducible character of G which is zero
everywhere outside H, then ÷ restricts to be the
sum of two distinct irreducible characters of H of
the same degree. The two characters of H which
we get from ÷ in this way come from no other
irreducible character of G (Propositions 20.9,
20.10, 20.12).
(3) Every irreducible character of H appears among
those obtained by restricting irreducible characters
of G, as in parts (1) and (2) (Proposition 20.4).
In case (2) of (20.13), extra work is needed to calculate the values
taken by the two constituents of ÷ # H . Fortunately, in practice case (1)
occurs more frequently than case (2).
20.14 Example The character table of A5
Write H � A5, and note that H is a normal subgroup of index 2 in
the group S5. The conjugacy classes of H and their sizes are given in
Example 12.18(2), and the irreducible characters ÷1, . . . , ÷7 of S5 can
be found in Example 19.16.
Observe that ÷1, ÷3 and ÷6 are non-zero somewhere outside H, so by
(20.13)(1), ÷1 # H , ÷3 # H and ÷6 # H are irreducible characters of H.
Call them ø1, ø2 and ø3, respectively. Also, ÷5(g) � 0 for all g =2 H,
so by (20.13)(2), ÷5 # H � ø4 � ø5 where ø4 and ø5 are distinct
irreducible characters of H of degree 3. Note that ÷2 # H � ÷1 # H ,
÷4 # H � ÷3 # H and ÷7 # H � ÷6 # H, and hence ø1, . . . , ø5 are the
distinct irreducible characters of H by (20.13)(3).
The results we have obtained so far have been deduced from our
summary (20.13) of facts about characters of subgroups of index 2.
They can also be veri®ed by calculating inner products. We have
established that the character table of H is
gi 1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)|CG(gi)| 60 3 4 5 5
ø1 1 1 1 1 1ø2 4 1 0 ÿ1 ÿ1ø3 5 ÿ1 1 0 0ø4 3 á2 á3 á4 á5
ø5 3 â2 â3 â4 â5
220 Representations and characters of groups
We use the column orthogonality relations to calculate the unknowns
ái and âi. The values of ái � âi for 2 < i < 5 are given by noting that
ø4 � ø5 � ÷5 # H (or by using the column orthogonality relations for
column 1 and column i). We get
á2 � â2 � 0, á3 � â3 � ÿ2, á4 � â4 � á5 � â5 � 1:
Using Proposition 12.13, we see that each element of A5 is conjugate
to its inverse. Hence by Proposition 13.9(4), all the numbers in the
character table are real. By the orthogonality relation for column i with
itself (2 < i < 5), we obtain
3 � 3� á22 � â2
2,
4 � 2� á23 � â2
3,
5 � 2� á24 � â2
4 � 2� á25 � â2
5:
Hence
á2 � â2 � 0, á3 � â3 � ÿ1,
and we ®nd that á4 and â4 are the solutions of the quadratic
x2 ÿ xÿ 1 � 0. Since we have not yet distinguished between ø4 and
ø5, we may take
á4 � 12(1�p5), â4 � 1
2(1ÿp5):
Similarly, á5 and â5 are 12(1 � p5). Since ø4 6� ø5, we have
á5 � 12(1ÿp5), â5 � 1
2(1�p5):
Thus the character table of A5 is as shown.
where á � 12(1 � p5), â � 1
2(1 ÿ p5).
Character table of A5
gi 1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)|CG(gi)| 60 3 4 5 5
ø1 1 1 1 1 1ø2 4 1 0 ÿ1 ÿ1ø3 5 ÿ1 1 0 0ø4 3 0 ÿ1 á âø5 3 0 ÿ1 â á
Restriction to a subgroup 221
Proposition 17.6 allows us to deduce from the character table that
A5 is a simple group.
Summary of Chapter 20
Assume throughout that H is a subgroup of G.
1. If ÷ is a character of G, then ÷ # H is a character of H. The values
of ÷ # H are given by
(÷ # H)(h) � ÷(h)
for all h 2 H . In particular, ÷ and ÷ # H have the same degree.
2. The number of irreducible constituents of ÷ # H is bounded above
by jG : H j. Indeed, if ø1, : : : , ør are the irreducible characters of
H, and ÷ # H � d1ø1 � : : :� d rør, thenXr
i�1
d2i < jG : H j:
3. If H v G and ÷ is an irreducible character of G, then all the
constituents of ÷ # H have the same degree.
Exercises for Chapter 20
1. Let G � S4 and let H be the subgroup k(1 2 3 4), (1 3)l of G.
(a) Show that H � D8.
(b) For each irreducible character ÷ of G (given in Section 18.1),
express ÷ # H as a sum of irreducible characters of H.
2. Use the restrictions of the irreducible characters of S6, given in
Example 19.17, to ®nd the character table of A6.
(The seven conjugacy classes of A6 can be found by consulting the
solutions to Exercises 12.3 and 12.4.)
3. Let G be a group with an abelian subgroup H of index n. Prove
that ÷(1) < n for every irreducible character ÷ of G.
4. Suppose that G is a group with a subgroup H of index 3, and let ÷be an irreducible character of G. Prove that
h÷ # H , ÷ # HiH � 1, 2 or 3:
Give examples to show that each possibility can occur.
222 Representations and characters of groups
5. It is known that the complete list of degrees of the irreducible
characters of S7 is
1, 1, 6, 6, 14, 14, 14, 14, 15, 15, 20, 21, 21, 35, 35:
Also, A7 has nine conjugacy classes. Find the complete list of
degrees of the irreducible characters of A7.
Restriction to a subgroup 223
224
21
Induced modules and characters
Throughout this chapter we assume that H is a subgroup of the ®nite
group G. We saw in the last chapter that restriction gives a simple way
of converting a CG-module into a CH-module. Much more subtle than
this is the process of induction, which constructs a CG-module from a
given CH-module, and induction is the main concern of this chapter.
As H is smaller than G, it is usually the case that CH-modules are
easier to understand and construct than CG-modules, so induction can
often give us an important handle on the representations of a group if
we know some representations of its subgroups. We shall see many
applications of this method in later chapters.
Before describing the process of induction, we require some results
which connect CH-homomorphisms with CG-homomorphisms.
CH-homomorphisms and CG-homomorphisms
Let U be a CH-submodule of the regular CH-module CH. If r 2 CG,
then
W: u! ru (u 2 U )
de®nes a CH-homomorphism from U to CG, since for all s 2 CH,
(us)W � rus � (uW)s. We now prove the striking fact that every CH-
homomorphism from U to CG has this simple form.
21.1 Proposition
Assume that H < G, and let U be a CH-submodule of CH. If W is a
CH-homomorphism from U to CG, then there exists r 2 CG such that
uW � ru for all u 2 U :
Proof By Maschke's Theorem 8.1, there is a CH-submodule W of CH
such that CH � U � W. De®ne ö: CH! CG by
ö: u� w! uW (u 2 U , w 2 W ):
Then ö is easily seen to be a CH-homomorphism. Let r � 1ö. For
u 2 U,uW � uö � (1u)ö � (1ö)u � ru,
and so W is of the required form. j
We give two corollaries of Proposition 21.1, the ®rst of which is just
the case H � G of the proposition.
21.2 Corollary
Let U be a CG-submodule of CG. Then every CG-homomorphism from
U to CG has the form
u! ru (u 2 U )
for some r 2 CG.
21.3 Corollary
Let U and V be CG-submodules of CG. Then the following two
statements are equivalent:
(1) U \ V � {0};
(2) there exists r 2 CG such that for all u 2 U, v 2 V,
ru � u and rv � 0:
Proof Assume that U \ V � {0}. Then the sum U � V is a direct sum,
so
u� v! u (u 2 U , v 2 V )
is a function; moreover, it is a CG-homomorphism from U � V to CG
(see Proposition 7.11). Therefore by Corollary 21.2, there exists
r 2 CG such that for all u 2 U, v 2 V,
r(u� v) � u:
Then ru � u if u 2 U, and rv � 0 if v 2 V.
Conversely, assume that for some r 2 CG we have ru � u and
rv � 0 for all u 2 U, v 2 V. If x 2 U \ V then rx � x and rx � 0, and
so x � 0. Consequently U \ V � {0}. j
Induced modules and characters 225
Induction from H to G
For any subset X of CG, we write X(CG) for the subspace of CG
which is spanned by all the elements xg with x 2 X, g 2 G. That is,
X (CG) � sp fxg: x 2 X , g 2 Gg:Clearly, X(CG) is then a CG-submodule of CG.
Remember that H is a subgroup of G, so CH is a subset of CG.
21.4 De®nition
Assume that H is a subgroup of G. Let U be a CH-submodule of
CH, and let U " G denote the CG-module U(CG). Then U " G is
called the CG-module induced from U.
21.5 Example
Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l, and let H � kal, a cyc-
lic subgroup of G of order 3. Let ù � e2ði=3, and de®ne
W0 � sp (1� a� a2),
W1 � sp (1� ù2a� ùa2),
W2 � sp (1� ùa� ù2a2):
These are CH-submodules of CH (see Example 10.8(1)). Clearly,
W0 " G � sp (1� a� a2, b� ab� a2b),
W1 " G � sp (1� ù2a� ùa2, b� ù2ab� ùa2b),
W2 " G � sp (1� ùa� ù2a2, b� ùab� ù2a2b):
Recall from Example 10.8(2) that
CG � U1 � U2 � U3 � U4,
a direct sum of irreducible CG-modules Ui, where
U1 � sp (1� a� a2 � b� ab� a2b),
U2 � sp (1� a� a2 ÿ bÿ abÿ a2b),
U3 � sp (1� ù2a� ùa2, b� ù2ab� ùa2b),
U4 � sp (1� ùa� ù2a2, b� ùab� ù2a2b):
Thus
W0 " G � U1 � U2, W1 " G � U3, W2 " G � U4:
226 Representations and characters of groups
In particular, W0 " G is reducible, while W1 " G and W2 " G are
irreducible.
We now show that isomorphic CH-modules give isomorphic induced
CG-modules.
21.6 Proposition
Assume that H < G. Suppose that U and V are CH-submodules of CH
and that U is CH-isomorphic to V. Then U " G is CG-isomorphic to
V " G.
Proof Let W: U! V be a CH-isomorphism. By Proposition 21.1, there
exists r 2 CG such that uW � ru for all u 2 U, and also there exists
s 2 CG such that vWÿ1 � sv for all v 2 V. Consequently
sru � u and rsv � v for all u 2 U , v 2 V :
If a 2 U " G then a is a linear combination of elements ug
(u 2 U, g 2 G), so ra is a linear combination of elements rug, and
hence ra 2 V " G. Therefore
ö: a! ra (a 2 U " G)
is a function from U " G to V " G. Moreover, ö is a CG-homomorph-
ism, as (aö)g � rag � (ag)ö (a 2 U " G, g 2 G). Since sru � u and
rsv � v for all u 2 U, v 2 V, we have
sra � a, rsb � b for all a 2 U " G, b 2 V " G:
Hence the function
b! sb (b 2 V " G)
is the inverse of ö. Therefore ö is a CG-isomorphism, proving that
U " G is CG-isomorphic to V " G. j
The next proposition and its corollary enable us to de®ne the
induced module U " G for an arbitrary CH-module U.
21.7 Proposition
Assume that U and V are CH-submodules of CH with U \ V � {0}.
Then (U " G) \ (V " G) � f0g.
Induced modules and characters 227
Proof By Corollary 21.3, there exists r 2 CG such that ru � u and
rv � 0 for all u 2 U, v 2 V. Then for all u 2 U, v 2 V and all g 2 G,
rug � ug and rvg � 0:
Since U " G is spanned by elements of the form ug (u 2 U, g 2 G),
this implies that
ru9 � u9 for all u9 2 U " G,
and similarly,
rv9 � 0 for all v9 2 V " G:
Therefore (U " G) \ (V " G) � f0g by Corollary 21.3. j
21.8 Corollary
Let U be a CH-submodule of CH, and suppose that
U � U1 � : : :� Um,
a direct sum of CH-submodules Ui. Then
U " G � (U1 " G)� : : :� (Um " G):
Proof We prove this by induction on m. It is trivial for m � 1. Now
U � U1 � V, where V � U2 � : : : � Um. The de®nition of U " G
implies that
U " G � (U1 " G)� (V " G):
Therefore by Proposition 21.7,
U " G � (U1 " G)� (V " G):
By induction, V " G � (U2 " G)� : : : � (Um " G), and hence, using
(2.10), we obtain
U " G � (U1 " G)� : : :� (Um " G),
as required. j
We can now de®ne the induced module U " G for an arbitrary CH-
module U (where U is not necessarily a CH-submodule of CH).
21.9 De®nition
Let U be a CH-module. Then (by Theorems 8.7 and 10.5),
U � U1 � : : :� Um
228 Representations and characters of groups
for certain irreducible CH-submodules Ui of CH. De®ne U " G to be
the following (external) direct sum:
U " G � (U1 " G)� : : :� (Um " G):
Proposition 21.6 and Corollary 21.8 ensure that this de®nition is
consistent with De®nition 21.4.
We emphasize that the de®nition of the induced module U " G in
the case where U is a CH-submodule of CH is a natural one:
U " G � U (CG):
We shall always prove results for general induced modules U " G by
®rst dealing with the special case where U is a CH-submodule of CH,
and then applying the fact (which is immediate from De®nition 21.9)
that
(U1 � : : :� Um) " G � (U1 " G)� : : :� (Um " G):(21:10)
Our ®rst major result on general induced modules is known as `the
transitivity of induction'.
21.11 Theorem
Suppose that H and K are subgroups of G such that H < K < G. If U
is a CH-module, then
(U " K) " G � U " G:
Proof Assume ®rst that U is a CH-submodule of CH. Then U(CK) is
spanned by elements of the form
uk (u 2 U , k 2 K):
Therefore, (U(CK))(CG) is spanned by elements of the form
ukg (u 2 U , k 2 K, g 2 G):
Since K < G, it follows that (U(CK))(CG) � U(CG). That is,
(U " K) " G � U " G:(21:12)
Now let U be an arbitrary CH-module. Then
U � U1 � : : :� Um
for certain irreducible CH-submodules Ui of CH. By (21.10),
U " K � (U1 " K)� : : :� (Um " K):
Induced modules and characters 229
Therefore
(U " K) " G � (U1 " K) " G � : : :� (Um " K) " G
by (21:10)
� (U1 " G)� : : :� (Um " G) by (21:12)
� U " G by Definition 21.9. j
Induced characters
21.13 De®nition
If ø is the character of a CH-module U, then the character of the
induced CG-module U " G is denoted by ø " G, and is called the
character induced from ø.
The next example illustrates an important relationship between
induced characters and restrictions of characters.
21.14 Example
Let G � S5 and let H be the subgroup A4 of G, as in Example 20.2.
We showed in that example that if ÷1, . . . , ÷7 are the irreducible
characters of G (given in Example 19.16), and ø1, . . . , ø4 are the
irreducible characters of H (given in 18.2) then
÷1 # H � ø1,
÷2 # H � ø1,
÷3 # H � ø1 � ø4,
÷4 # H � ø1 � ø4,
÷5 # H � 2ø4,
÷6 # H � ø2 � ø3 � ø4,
÷7 # H � ø2 � ø3 � ø4:
By Theorem 14.17, the coef®cients which appear here are the values of
h÷i # H , ø jiH for appropriate i, j. We record these coef®cients in a
230 Representations and characters of groups
matrix whose ij-entry is h÷i # H , ø jiH :
1CCCCCCCCA
0BBBBBBBB@
ø1 ø2 ø3 ø4
÷1 1 0 0 0
÷2 1 0 0 0
÷3 1 0 0 1
÷4 1 0 0 1
÷5 0 0 0 2
÷6 0 1 1 1
÷7 0 1 1 1
The rows of this matrix tell us how the irreducible characters of G
restrict to H. For example, row 3 gives
÷3 # H � 1 . ø1 � 0 . ø2 � 0 . ø3 � 1 . ø4:
Remarkably, it turns out that the columns of the matrix tell us how
the irreducible characters of H induce to G. To be precise, the seven
integers in column 1 give
ø1 " G � 1 . ÷1 � 1 . ÷2 � 1 . ÷3 � 1 . ÷4 � 0 . ÷5 � 0 . ÷6 � 0 . ÷7:
Similarly,ø2 " G � ø3 " G � ÷6 � ÷7, and
ø4 " G � ÷3 � ÷4 � 2÷5 � ÷6 � ÷7:
Thus the ij-entry in our matrix, which we already know to be equal to
h÷i # H , ø jiH , is also equal to h÷i, ø j " GiG.
In fact, it is true that
h÷, ø " GiG � h÷ # H , øiH
for all characters ÷ of G and ø of H, and we devote the next section
to a proof of this result, which is known as the Frobenius Reciprocity
Theorem.
The Frobenius Reciprocity Theorem
Before proving the theorem, we need the following preliminary result.
21.15 Proposition
Assume that H < G. Let U be a CH-submodule of CH, and let V be a
CG-submodule of CG. Then the vector spaces
HomCG (U " G, V ) and HomC H (U , V # H)
have equal dimensions.
Induced modules and characters 231
Proof Suppose that W 2 HomCG (U " G, V ). Then by Corollary 21.2,
there is an element r 2 CG such that
sW � rs for all s 2 U " G:
De®ne W: U ! CG to be the restriction of W to U; that is,
uW � ru for all u 2 U :
Then W 2 HomC H (U , V # H). Thus the function
W! W
is a linear transformation from HomCG (U " G, V ) to
HomC H (U , V # H). We shall show that this linear transformation is
invertible.
Let ö 2 HomC H (U , V # H). Then by Proposition 21.1, there exists
r 2 CG such that uö � ru for all u 2 U. De®ne W from U " G to CG
by
sW � rs (s 2 U " G):
Then W 2 HomCG (U " G, V ). Moreover, ö � W. Therefore the function
W! W is surjective.
Finally, note that if r1, r2 2 CG and r1u � r2u for all u 2 U, then
r1s � r2s for all s 2 U " G, since s is a linear combination of
elements ug with u 2 U, g 2 G. Hence the function W! W is injective,
and so it is an invertible linear transformation from HomCG(U " G, V )
to HomC H (U , V # H). These two vector spaces therefore have the
same dimension, as required. j
21.16 The Frobenius Reciprocity Theorem
Assume that H < G. Let ÷ be a character of G and let ø be a
character of H. Then
hø " G, ÷iG � hø, ÷ # HiH :
Proof First assume that the characters ÷ and ø are irreducible. Then
there is a CH-submodule U of CH which has character ø, and there
is a CG-submodule V of CG which has character ÷. By Theorem
14.24, we have
hø " G, ÷iG � dim (HomCG (U " G, V )), and
hø, ÷ # HiH � dim (HomC H (U , V # H)):
232 Representations and characters of groups
From Proposition 21.15, we therefore deduce that
hø " G, ÷iG � hø, ÷ # HiH(21:17)
in the special case we are considering, namely when ÷ and ø are
irreducible.
For the general case, let ÷1, . . . , ÷k be the irreducible characters of
G and let ø1, . . . , øm be the irreducible characters of H. Then for
some integers di, ej we have
÷ �Xk
i�1
di÷i and ø �Xm
j�1
ejø j:
Therefore
hø " G, ÷iG �Xm
j�1
ejø j " G,Xk
i�1
di÷i
* +G
�Xm
j�1
Xk
i�1
ejdihø j " G, ÷iiG
�Xm
j�1
Xk
i�1
ejdihø j, ÷i # HiH by (21:17)
�Xm
j�1
ejø j,Xk
i�1
di÷i # H
* +H
� hø, ÷ # HiH :
This completes the proof of the Frobenius Reciprocity Theorem. j
21.18 Corollary
If f is a class function on G, and ø is a character of H, then
hø " G, f iG � hø, f # HiH :
Proof This follows at once from the Frobenius Reciprocity Theorem,
since by Corollary 15.4, the characters of G span the vector space of
class functions on G. j
The values of induced characters
We now show how to evaluate induced characters. Let ø be a character
of the subgroup H of G, and for convenience of notation, de®ne the
Induced modules and characters 233
function _ø: G! C by
_ø(g) �ø(g) if g 2 H ,
0 if g =2 H :
(
21.19 Proposition
The values of the induced character ø " G are given by
(ø " G)(g) � 1
jH jXy2G
_ø(yÿ1 gy)
for all g 2 G.
Proof Let f: G! C be the function given by
f (g) � 1
jH jXy2G
_ø(yÿ1 gy) (g 2 G):
We aim to prove that f � ø " G.
If w 2 G then
f (wÿ1 gw) � 1
jH jXy2G
_ø(yÿ1wÿ1 gwy) � f (g)
since wy runs through G as y runs through G. Therefore f is a class
function, and so by Corollary 15.4, it is suf®cient to show that
h f , ÷iG � hø " G, ÷iG for all irreducible characters ÷ of G.
Let ÷ be an irreducible character of G. Then
h f , ÷iG � 1
jGjXg2G
f (g)÷(g)
� 1
jGj1
jH jXg2G
Xy2G
_ø(yÿ1 gy)÷(g):
Put x � yÿ1 gy. Then
h f , ÷iG � 1
jGj1
jH jXx2G
Xy2G
_ø(x)÷(yxyÿ1):
� 1
jH jXx2H
ø(x)÷(x)
234 Representations and characters of groups
since _ø(x) � 0 if x =2 H, and ÷(yxyÿ1) � ÷(x) for all y 2 G. Therefore
h f , ÷iG � hø, ÷ # HiH
and so by the Frobenius Reciprocity Theorem,
h f , ÷iG � hø " G, ÷iG:
This was the equation required to show that f � ø " G, so the proof is
complete. j
21.20 Corollary
If ø is a character of the subgroup H of G, then the degree of ø " G
is given by
(ø " G)(1) � jGjjH jø(1):
Proof This follows immediately by evaluating (ø " G)(1) using Prop-
osition 21.19. Alternatively, the stated degree of ø " G can be found
using just the de®nition of induced modules (see Exercise 21.3). j
For practical purposes, a formula for the values of induced characters
different from that given in Proposition 21.19 is more useful, and we
shall derive this next (it is given in Proposition 21.23 below).
For x 2 G, de®ne the class function f Gx on G by
f Gx (y) � 1 if y 2 xG
0 if y =2 xG
�(y 2 G):
Thus f is the characteristic function of the conjugacy class xG.
21.21 Proposition
If ÷ is a character of G and x 2 G, then
h÷, f Gx iG �
÷(x)
jCG(x)j :
Induced modules and characters 235
Proof We have
h÷, f Gx iG �
1
jGjXg2G
÷(g) f Gx (g)
� 1
jGjXg2xG
÷(g) � jxGjjGj ÷(x)
� ÷(x)
jCG(x)j by Theorem 12:8: j
Note that a result similar to Proposition 21.21 was used in the proof
of Theorem 16.4.
If H < G and h 2 H then h H � hG; but if g 2 G then gG may
contain 0, 1, 2 or more conjugacy classes of H. To put this another
way, we have:
(21.22) Suppose that x 2 G.
(1) If no element of xG lies in H, then f Gx # H � 0.
(2) If some element of xG lies in H, then there are elements
x1, . . . , xm 2 H such that
f Gx # H � f H
x1� : : :� f H
xm:
Statement (2) just says that H \ xG breaks up into m conjugacy
classes of H, with representatives x1, : : : , xm.
21.23 Proposition
Let ø be a character of the subgroup H of G, and suppose that x 2 G.
(1) If no element of xG lies in H, then (ø " G)(x) � 0.
(2) If some element of xG lies in H, then
(ø " G)(x) � jCG(x)j ø(x1)
jCH (x1)j � : : :�ø(xm)
jCH (xm)j� �
,
where x1, . . . , xm 2 H and f Gx # H � f H
x1� . . . � f H
xm(as in (21.22)).
Proof By Proposition 21.21 and Corollary 21.18, we have
(ø " G)(x)
jCG(x)j � hø " G, f Gx iG � hø, f G
x # HiH :
236 Representations and characters of groups
If no element of xG lies in H, then f Gx # H � 0, and hence
(ø " G)(x) � 0. And if some element of xG lies in H, and
f Gx # H � f H
x1� : : : � f H
xmas in (21.22)(2), then
(ø " G)(x)
jCG(x)j � hø, f Hx1� : : :� f H
xmiH
� hø, f Hx1iH � : : :� hø, f H
xmiH
� ø(x1)
jCH (x1)j � : : :�ø(xm)
jCH (xm)j by Proposition 21:21:
The result follows. j
21.24 Example
Let G � S4 and let H � ka, bl, where
a � (1 2 3 4), b � (1 3):
Then H � D8, since a4 � b2 � 1 and bÿ1ab � aÿ1. By (12.12), the
conjugacy classes of H are
f1g,
fa2 � (1 3)(2 4)g,
fa � (1 2 3 4), a3 � (1 4 3 2)g,
fb � (1 3), a2b � (2 4)g,
We have
f G1 # H � f H
1 , f G(1 3) # H � f H
(1 3), f G(1 2 3) # H � 0,
f G(1 2)(3 4) # H � f H
(1 3)(2 4) � f H(1 2)(3 4), and
f G(1 2 3 4) # H � f H
(1 2 3 4):
For example, the statement
f G(1 2)(3 4) # H � f H
(1 3)(2 4) � f H(1 2)(3 4)
records the fact that the G-conjugacy class (1 2)(3 4)G contains exactly
two H-conjugacy classes, with representatives (1 3)(2 4) and (1 2)(3 4).
Induced modules and characters 237
The orders of the centralizers of the elements of H are as follows:
h 1 (1 3)(2 4) (1 2 3 4) (1 3) (1 2)(3 4)|CG(h)| 24 8 4 4 8|CH (h)| 8 8 4 4 4
Suppose that ø is a character of H. Then according to Proposition
21.23, we have
(ø " G)(1) � 24ø(1)
8,
(ø " G)((1 3)) � 4ø((1 3))
4,
(ø " G)((1 2 3)) � 0,
(ø " G)((1 2)(3 4)) � 8ø((1 3)(2 4))
8� ø((1 2)(3 4))
4
� �,
(ø " G)((1 2 3 4)) � 4ø((1 2 3 4))
4:
Referring to Example 16.3(3) for the irreducible characters ÷1, . . . , ÷5
of H � D8, we therefore have
1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
÷1 " G 3 1 0 3 1÷2 " G 3 ÿ1 0 ÿ1 1÷3 " G 3 1 0 ÿ1 ÿ1÷4 " G 3 ÿ1 0 3 ÿ1÷5 " G 6 0 0 ÿ2 0
In the next example, we use induced characters to ®nd the character
table of a group of order 21.
21.25 Example (cf. Exercise 17.2)
De®ne permutations a, b in S7 by
a � (1 2 3 4 5 6 7), b � (2 3 5)(4 7 6)
238 Representations and characters of groups
and let G be the subgroup ka, bl of S7. Check that
a7 � b3 � 1, bÿ1ab � a2:
It follows from these relations that the elements of G are all of the
form aibj with 0 < i < 6, 0 < j < 2. Also, G has order 21.
We aim to ®nd the character table of G. First we ®nd the conjugacy
classes. Since hai < CG(a), 7 divides jCG(a)j; and since b =2 CG(a),
jCG(a)j, 21. Hence jCG(a)j � 7, and similarly jCG(b)j � 3. Using this,
we see that the conjugacy classes of G are
f1g,fa, a2, a4g,fa3, a5, a6g,faib: 0 < i < 6g,faib2: 0 < i < 6g:
We take 1, a, a3, b and b2 to be representatives of the conjugacy
classes. Notice that G has exactly ®ve irreducible characters.
Since kal v G and G=hai � C3, we obtain three linear characters ÷1,
÷2, ÷3 of G as the lifts of the linear characters of G=hai. Their values
are shown below:
g 1 a a3 b b2
|CG(g)| 21 7 7 3 3
÷1 1 1 1 1 1÷2 1 1 1 ù ù2
÷3 1 1 1 ù2 ù
where ù � e2ði=3.
Let H � kal. We shall obtain the last two irreducible characters of G
by inducing linear characters of H. Let ç � e2ði=7. For 1 < k < 6, there
is a character øk of H given by
øk(a j) � ç jk (0 < j < 6):
To use the formula in Proposition 21.23 for calculating øk " G, note
that
f Ga # H � f H
a � f Ha2 � f H
a4
Induced modules and characters 239
since no two of the elements a, a2, a4 are conjugate in H. Hence by
Proposition 21.23,
(ø1 " G)(a) � ç� ç2 � ç4
and similarly
(ø1 " G)(a3) � ç3 � ç5 � ç6, (ø1 " G)(1) � 3:
And since no G-conjugate of b or b2 lies in H,
(ø1 " G)(b) � (ø1 " G)(b2) � 0:
Similarly
(ø3 " G)(1) � 3, (ø3 " G)(a) � ç3 � ç5 � ç6,
(ø3 " G)(a3) � ç� ç2 � ç4, and (ø3 " G)(b) � (ø3 " G)(b2) � 0:
Thus if we write ÷4 � ø1 " G and ÷5 � ø3 " G, then the values of
÷4 and ÷5 are
1 a a3 b b2
÷4 3 ç � ç2 � ç4 ç3 � ç5 � ç6 0 0÷5 3 ç3 � ç5 � ç6 ç � ç2 � ç4 0 0
Now ÷4 # H � ø1 � ø2 � ø4 and ÷5 # H � ø3 � ø5 � ø6. Therefore
÷4 6� ÷5, since ø1, . . . , ø6 are linearly independent.
We now calculate that
h÷4, ÷4iG � 9
21� 2
7� 2
7� 0
3� 0
3� 1,
and similarly h÷5, ÷5iG � 1. Thus ÷4 and ÷5 are our last two irreducible
characters, and the character table of G is as shown.
Character table of ha, b: a7 � b3 � 1, bÿ1ab � a2i
g 1 a a3 b b2
|CG(gi)| 21 7 7 3 3
÷1 1 1 1 1 1÷2 1 1 1 ù ù2
÷3 1 1 1 ù2 ù÷4 3 ç � ç2 � ç4 ç3 � ç5 � ç6 0 0÷5 3 ç3 � ç5 � ç6 ç � ç2 � ç4 0 0
240 Representations and characters of groups
Summary of Chapter 21
Assume that H is a subgroup of G.
1. For each CH-module U, an induced CG-module U " G can be
de®ned. If U is a CH-module of CH, then U " G is simply U(CG).
2. If ø is a character of H then the induced character ø " G is given
by
(ø " G)(g) � 1
jH jXy2G
_ø(yÿ1 gy):
In particular, the degree of ø:G is |G:H|ø(1).
3. If no element of gG lies in H, then
(ø " G)(g) � 0:
If some element of gG lies in H, then
(ø " G)(g) � jCG(g)j ø(x1)
jCH (x1)j � : : :�ø(xm)
jCH (xm)j� �
where f Gg # H � f H
x1� : : : � f H
xm.
4. The Frobenius Reciprocity Theorem states that
hø " G, ÷iG � hø, ÷ # HiH ,
where ø is a character of H and ÷ is a character of G.
Exercises for Chapter 21
1. Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l and let H be the
subgroup ka2, bl. De®ne U to be the 1-dimensional subspace of CH
spanned by
1ÿ a2 � bÿ a2b:
(a) Check that U is a CH-submodule of CH.
(b) Find a basis of the induced CG-module U " G.
(c) Write down the character of the CH-module U and the char-
acter of the CG-module U " G. Is U " G irreducible?
2. Let G � S4 and let H be the subgroup k(1 2 3)l � C3.
(a) If ÷1, . . . , ÷5 are the irreducible characters of G, as given in
Induced modules and characters 241
Section 18.1, work out the restrictions ÷i # H (1 < i < 5) as
sums of the irreducible characters ø1, ø2, ø3 of C3.
(b) Calculate the induced characters ø j " G (1 < j < 3) as sums of
the irreducible characters ÷i of G.
3. Show direct from the de®nition that if H < G and ø is a character
of H, then
(ø " G)(1) � jGjjH jø(1):
4. Let H be a subgroup of G, let ø be a character of H, and let ÷ be
a character of G. Prove that
(ø(÷ # H)) " G � (ø " G)÷:
(Hint: consider the inner product of each side with an arbitrary
irreducible character of G, and use the Frobenius Reciprocity
Theorem.)
5. Let G � S7 and let H � ka, bl, where
a � (1 2 3 4 5 6 7), b � (2 3 5)(4 7 6),
as in Example 21.25. Let ö and ø be the irreducible characters of
H which are given by
gi 1 a a3 b b2
|CH (gi)| 21 7 7 3 3
ö 1 1 1 1 1ø 3 ç � ç2 � ç4 ç3 � ç5 � ç6 0 0
where ç � e2ði=7 (see Example 21.25).
You are given that jCG(a)j � 7 and jCG(b)j � 18. Calculate the
values of the induced characters ö " G and ø " G.
6. Suppose that H is a subgroup of G, and let ÷1, . . . , ÷k be the
irreducible characters of G. Let ø be an irreducible character of H.
Show that the integers d1, : : : , d k , which are given by ø " G
� d1÷1 � : : : dk÷k , satisfyXk
i�1
d2i < jG : H j:
(Compare Proposition 20.5.)
242 Representations and characters of groups
7. Suppose that H is a normal subgroup of index 2 in G, and let ø be
an irreducible character of H. Discover and prove results for ø " G
which are similar to those presented in Chapter 20 for the restriction
of irreducible characters of G to H.
Induced modules and characters 243
244
22
Algebraic integers
Among the properties of characters which may be regarded as funda-
mental, perhaps the most opaque is that which states that the degree of
an irreducible character of a ®nite group G must divide the order of
G. This is one of several results which we shall prove in this chapter,
using algebraic integers. Most of the results concern arithmetic proper-
ties of character values. We discuss properties of a group element
g 2 G which ensure that ÷(g) is an integer for all characters ÷ of G.
And we prove some useful congruence properties; for example, if p is
a prime number and g 2 G is an element of order pr for some r, then
÷(g) � ÷(1) mod p for any character ÷ of G for which ÷(g) is an
integer.
Algebraic integers
22.1 De®nition
A complex number ë is an algebraic integer if and only if ë is an
eigenvalue of some matrix, all of whose entries are integers.
Thus, for ë to be an algebraic integer, we require that
det (Aÿ ëI) � 0
for some square matrix A with integer entries. Equivalently, for the
same matrix A, we have
uA � ëu
for some non-zero row vector u.
We remark that ë is an algebraic integer if and only if ë is a root of
a polynomial of the form
x n � anÿ1x nÿ1 � : : :� a1x� a0
where a0, : : : , anÿ1 are integers (see Exercise 22.7). In fact, alge-
braic integers are usually de®ned in this way.
22.2 Examples
(1) Every integer n is an algebraic integer, since n is an eigenvalue of
the 1 3 1 matrix (n).
(2)p
2 is an algebraic integer, since it is an eigenvalue of the
matrix0 1
2 0
� �:
(3) If ë is an algebraic integer, then so are ÿë and the complex
conjugate ë of ë. To see this, note that if A is an integer matrix and u
is a row vector with uA � ëu, then
u(ÿA) � (ÿë)u and uA � ëu,
where u is the row vector obtained from u by replacing each entry by
its complex conjugate.
(4) Let A be the n 3 n matrix given by
A �
0 0 0 : : : 0 0 1
1 0 0 : : : 0 0 0
0 1 0 : : : 0 0 0
: : :0 0 0 : : : 1 0 0
0 0 0 : : : 0 1 0
0BBBBBB@
1CCCCCCA:Suppose that ù is an nth root of unity, and let u be the row vector
(1, ù, ù2, . . . , ùnÿ1). Then
uA � (ù, ù2, : : : , ùnÿ1, 1) � ùu:
This shows that every nth root of unity is an algebraic integer.
22.3 Theorem
If ë and ì are algebraic integers, then ëì and ë � ì are also
algebraic integers.
Proof There exist square matrices A and B, all of whose entries are
integers, and non-zero row vectors u and v, such that
uA � ëu, vB � ìv:
Suppose that A is an m 3 m matrix and B is an n 3 n matrix.
Algebraic integers 245
Let e1, . . . , em be a basis of Cm and f 1, : : : , f n be a basis of Cn.
Then the vectors ei fj (1 < i < m, 1 < j < n) form a basis of the
tensor product space V � Cm Cn. De®ne an endomorphism A B of
V by
(ei f j)(A B) � eiA f jB (1 < i < m, 1 < j < n),
extending linearly (that is, (P
ëij(ei f j))(A B) �Pëij(eiA f jB)).
It is easy to check as in the proof of Proposition 19.4 that for all
vectors x 2 Cm, y 2 Cn, we have
(x y)(A B) � xA yB:
Hence(u v)(A B) � uA vB � ëu ìv � ëì(u v):
Therefore ëì is an eigenvalue of A B. Since the matrix of A B
relative to the basis ei fj (1 < i < m, 1 < j < n) has integer entries,
it follows that ëì is an algebraic integer.
Let Im and In denote the identity m 3 m and n 3 n matrices,
respectively. Then
(u v)(A In � Im B) � uA vIn � uIm vB
� ëu v� u ìv
� (ë� ì)(u v),
and we deduce as above that ë � ì is an algebraic integer. j
Theorem 22.3 shows that the set of all algebraic integers forms a
subring of C. The next result provides a link between algebraic integers
and characters.
22.4 Corollary
If ÷ is a character of G and g 2 G, then ÷(g) is an algebraic integer.
Proof By Proposition 13.9, ÷(g) is a sum of roots of unity. Each root
of unity is an algebraic integer, by Example 22.2(4), so any sum of
roots of unity is an algebraic integer by Theorem 22.3. Hence ÷(g) is
an algebraic integer. j
22.5 Proposition
If ë is both a rational number and an algebraic integer, then ë is an
integer.
246 Representations and characters of groups
Proof Suppose that ë is a rational number which is not an integer. We
shall show that ë is not an algebraic integer, which is enough to
establish the proposition.
Write ë � r=s, where r and s are coprime integers and s 6� �1. Let
p be a prime number which divides s. For every n 3 n matrix A of
integers, the entries of sA ÿ rI which are not on the diagonal are
divisible by s, and hence also by p. Therefore
det (sAÿ rI) � (ÿr)n � mp
for some integer m. As p does not divide r (since r and s are
coprime), we deduce that det (sA ÿ rI) 6� 0. Thus
det (Aÿ ëI) � 1
s
� �n
det (sAÿ rI) 6� 0,
and hence ë is not an algebraic integer. j
The next result is an immediate consequence of Corollary 22.4 and
Proposition 22.5.
22.6 Corollary
Let ÷ be a character of G and let g 2 G. If ÷(g) is a rational number,
then ÷(g) is an integer.
In passing, note that we have, as a special case of Proposition 22.5,
the well known result thatp
2 is irrational. (Example 22.2(2) shows
thatp
2 is an algebraic integer.)
The degree of every irreducible character divides |G|
To prepare for the proof that |G| is divisible by the degree of each
irreducible character of G, we establish two preliminary lemmas. Recall
from De®nition 12.21 that if C is a conjugacy class of G, then
C �Xx2C
x 2 CG:
22.7 Lemma
Suppose that g 2 G and that C is the conjugacy class of G which
contains g. Let U be an irreducible CG-module, with character ÷. Then
Algebraic integers 247
uC � ëu for all u 2 U ,
where
ë � jGjjCG(g)j
÷(g)
÷(1):
Proof Since C lies in the centre of CG (see Proposition 12.22), we
know by Proposition 9.14 that there exists ë 2 C such that uC � ëu for
all u 2 U; that is,
uXx2C
x � ëu for all u 2 U :
Consequently if B is a basis of U, thenXx2C
[x]B � ëI :
Taking the traces of both sides of this equation, we obtainXx2C
÷(x) � ë÷(1),
and since ÷ is constant on the conjugacy class C, this yields
jCj÷(g) � ë÷(1):
Thus ë � jCj÷(g)=÷(1). As |C| � |G:CG(g)| by Theorem 12.8, the result
follows. j
22.8 Lemma
Let r �P g2G á gg 2 CG, where each á g is an integer. Suppose that u
is a non-zero element of CG such that
ur � ëu,
where ë 2 C. Then ë is an algebraic integer.
Proof Let g1, . . . , gn be the elements of G. Then for 1 < i < n, we
have
gir �Xn
j�1
aijgj
248 Representations and characters of groups
for certain integers aij. (In fact, aij � á g where g � gÿ1i gj.) The
statement that ur � ëu (with u 6� 0) says that ë is an eigenvalue of the
integer matrix A � (aij). Therefore ë is an algebraic integer. j
22.9 Example
Let G � Cn � kx: xn � 1l, and de®ne
u � 1� ùxÿ1 � ù2xÿ2 � : : :� ùnÿ1x 2 CG,
where ù is an nth root of unity. Then
ux � ùu
and so Lemma 22.8 con®rms that ù is an algebraic integer.
Notice that this example is just a reworking of Example 22.2(4).
22.10 Corollary
If ÷ is an irreducible character of G and g 2 G, then
ë � jGjjCG(g)j
÷(g)
÷(1)
is an algebraic integer.
Proof Let U be an irreducible CG-submodule of CG with character ÷,
and let C be the sum of the elements in the conjugacy class of G
which contains g. Then uC � ëu for all u 2 U, by Lemma 22.7.
Therefore ë is an algebraic integer, by Lemma 22.8. j
22.11 Theorem
If ÷ is an irreducible character of G, then ÷(1) divides |G|.
Proof Let g1, . . . , gk be representatives of the conjugacy classes of G.
Then for all i, both
jGjjCG(gi)j
÷(gi)
÷(1)and ÷(gi)
are algebraic integers, by Corollaries 22.10 and 22.4. Hence by
Theorem 22.3, Xk
i�1
jGjjCG(gi)j
÷(gi)÷(gi)
÷(1)
Algebraic integers 249
is an algebraic integer. This algebraic integer is equal to jGj=÷(1), by
the row orthogonality relations, Theorem 16.4(1). As jGj=÷(1) is a
rational number. Proposition 22.5 now implies that jGj=÷(1) is an
integer. That is, ÷(1) divides |G|. j
22.12 Examples
(1) If p is a prime number and G is a group of order pn for some n,
then ÷(1) is a power of p for all irreducible characters ÷ of G.
In particular, if jGj � p2 then ÷(1) � 1 for all irreducible characters
÷. (Note that ÷(1) , p, as the sum of the squares of the degrees of the
irreducible characters is equal to |G|.) Hence, using Proposition 9.18,
we recover the well known result that groups of order p2 are abelian.
(2) Let G be a group of order 2 p, where p is prime. By Theorem
22.11, the degree of each irreducible character of G is 1 or 2 (it
cannot be p for the reason noted in (1) above). By Theorem 17.11, the
number of linear characters of G divides |G|. Hence, either the degrees
of the irreducible characters of G are all 1, or they are
1, 1, 2, : : : , 2 (with ( pÿ 1)=2 degrees 2):
(3) If G � Sn then every prime p which divides the degree of an
irreducible character of G also divides n!, and hence satis®es p < n.
Theorem 22.11 also has the following interesting consequences
concerning irreducible characters of simple groups. (Recall that a group
G is simple if it has no normal subgroups apart from {1} and G.)
22.13 Corollary
No simple group has an irreducible character of degree 2.
Proof Suppose that G is a simple group which has an irreducible
character ÷ of degree 2. Let r: G! GL(2, C) be a representation of
G with character ÷. Since Ker r v G and G is simple, we have
Ker r � f1g, and so r is injective.
First, observe that G is non-abelian, by Proposition 9.5. Hence
G9 6� 1, and so G9 � G as G is simple. Therefore by Theorem 17.11,
G has no non-trivial linear characters. But g! det (gr) is a linear
character of G (see Exercise 13.7(a)), and this implies that
det (gr) � 1 for all g 2 G:
250 Representations and characters of groups
Now G has even order, by Theorem 22.11. Therefore G contains an
element x of order 2 (see Exercise 1.8).
Consider the 2 3 2 matrix xr. As r is injective, xr has order 2; and
by Proposition 9.11, there is a 2 3 2 matrix T such that Tÿ1(xr)T is a
diagonal matrix with diagonal entries �1. Since det (xr) � 1, we
conclude that
Tÿ1(xr)T � ÿ1 0
0 ÿ1
� �:
Thus
xr � T (ÿI)Tÿ1 � ÿI :
Consequently (xr)(gr) � (gr)(xr) for all g 2 G. As r is injective, this
means that xg � gx for all g 2 G, and hence
hxi v G:
This is a contradiction, as G is simple. j
Our next result again shows that information about character degrees
can sometimes be used to learn about the structure of a group. This
time, we assume that every irreducible character of G has degree a
power of a prime p, and we deduce that G has an abelian normal p-
complement N; that is, N is an abelian normal subgroup of G, and jN jis coprime to p, while jG: N j is a power of p.
22.14 Corollary
Suppose that p is a prime and the degree of every irreducible character
of G is a power of p. Then G has an abelian normal p-complement. In
particular, G is not simple unless G has prime order.
Proof The result is correct if G is abelian (see Theorem 9.6), so we
assume that G is non-abelian.
Theorems 11.12 and 17.11 give us the equation
jGj � jG=G9j �X
÷(1)2,
the sum being over the irreducible characters ÷ of G for which
÷(1) . 1.
Since G is non-abelian, ÷(1) . 1 for some irreducible character ÷ of
G. Then ÷(1) is divisible by p, by our hypothesis, so p divides |G| by
Algebraic integers 251
Theorem 22.11; and we deduce from the equation above that p divides
the order of the abelian group G=G9. Since every ®nite abelian group
is isomorphic to a direct product of cyclic groups, it follows that G=G9
has a subgroup of index p. Hence G has a normal subgroup H of
index p.
Let ø be an irreducible character of H. Then h÷ # H , øi 6� 0 for
some irreducible character ø of H, by Proposition 20.4. Next, Clif-
ford's Theorem 20.8 shows that ø(1) divides ÷(1), so ø(1) is a power
of p. We may now apply induction on |G| to deduce that H has an
abelian normal p-complement N.
We have that |N| is coprime to p and jG : N j is a power of p, so it
remains to prove that N v G.
Suppose that g 2 G and the order of g is coprime to p. Then
g 2 H , since otherwise p divides the order of gH which in turn
divides the order of g; a similar argument shows that g 2 N. Hence N
consists of those elements of G whose order is coprime to p, and from
this fact it follows easily that N v G.
Finally, assume that G is simple, so either N � {1} or N � G. If
N � f1g then G is a p-group so Z(G) 6� {1} (see Exercise 12.7);
because G is simple, we have Z(G) � G, so G is abelian. On the other
hand, if N � G then G is again abelian. But an abelian simple group
has prime order, by Exercise 1.1. Therefore, G has prime order. j
A condition which ensures that ÷(g) is an integer
In Theorem 22.16 below we give a group-theoretic condition on an
element g of G which implies that ÷(g) is an integer for every
character ÷ of G. This result implies, for example, that for all n, every
entry in the character table of Sn is an integer (see Corollary 22.17).
Bearing in mind the dif®culties we encountered in constructing the
character tables of Sn for small values of n (we reached n � 6 in
Example 19.17), Theorem 22.16 is evidently a useful result.
Before proving Theorem 22.16, we require a preliminary lemma
concerning roots of unity. If a and b are positive integers, then we
denote their highest common factor by (a, b). Also, for integers d and
n, we write d|n to denote the fact that d divides n.
22.15 Lemma
If ù is an nth root of unity, then
252 Representations and characters of groups
X1<i<n,(i,n)�1
ùi
is an integer.
Proof We prove the result by induction on n. It is trivial for n � 1.
Also, if ù � 1 then the result is immediate. So suppose that ù is an
nth root of unity and ù 6� 1. Then ù is a root of the polynomial
(xn ÿ 1)=(xÿ 1) � x nÿ1 � : : :� x� 1: ThereforeP
ni�1 ù
i � 0.
Now we partition the sumP
ni�1 ù
i according to the highest common
factor d of i and n:
0 �Xn
i�1
ùi �Xdjn
X1<i<n(i,n)�d
ùi �Xdjn
X1< j<n=d,( j,n=d)�1
ùdj:
If djn then ùd is an (n=d)th root of unity, and if in addition d . 1,
then by our induction hypothesis,X1< j<n=d,( j,n=d)�1
ùdj 2 Z:
It follows that X1<i<n,(i,n)�1
ùi �Xn
i�1
ùi ÿXdjn,
d . 1
X1< j<n=d,( j,n=d)�1
ùdj 2 Z,
as required. j
22.16 Theorem
Let g be an element of order n in G. Suppose that g is conjugate to gi
for all i with 1 < i < n and (i, n) � 1. Then ÷(g) is an integer for all
characters ÷ of G.
Proof Let V be a CG-module with character ÷ of degree m. By
Proposition 9.11, there is a basis B of V such that
[g]B �ù1
. ..
ùm
0B@1CA0
0
Algebraic integers 253
where ù1, . . . , ùm are nth roots of unity. For 1 < i < n, the matrix
[gi]B has diagonal entries ùi1, . . . , ùi
m, and so
÷(gi) � ùi1 � : : :� ùi
m:
Therefore by Lemma 22.15, X1<i<n,(i,n)�1
÷(gi) 2 Z:
As g is conjugate to gi if 1 < i < n and (i, n) � 1, we have
÷(gi) � ÷(g) for such i, and hence
s÷(g) 2 Z,
where s is the number of integers i with 1 < i < n and (i, n) � 1.
Consequently ÷(g) is a rational number, and so ÷(g) is an integer by
Corollary 22.6. j
We remark that using Galois theory it is possible to prove the
converse of Theorem 22.16, namely that if ÷(gi) 2 Z for all characters
÷ of G, then g is conjugate to gi whenever i is coprime to n.
22.17 Corollary
All the character values of symmetric groups are integers.
Proof If g 2 Sn and i is coprime to the order of g, then the permuta-
tions g and gi have the same cycle-shape, and hence are conjugate by
Theorem 12.15. The result now follows from Theorem 22.16. j
The p9-part of a group element
The rest of the chapter is devoted to some important congruence
properties of character values. For example, one particularly useful
consequence of our results is that if p is a prime number, g is an
element of G of order pr for some r, and ÷ is a character of G such
that ÷(g) 2 Z, then ÷(g) � ÷(1) mod p.
Before going into the character theory, we need to de®ne the p9-part
of a group element. The de®nition will emerge from the following
lemma.
254 Representations and characters of groups
22.18 Lemma
Let p be a prime number and let g 2 G. Then there exist x, y 2 G such
that
(1) g � xy � yx,
(2) the order of x is a power of p, and
(3) the order of y is coprime to p.
Moreover, the elements x and y of G which satisfy conditions (1)±(3)
are unique.
Proof Let the order of g be upv, where u, v 2 Z and (u, p) � 1. Then
there exist integers a, b such that
au� bpv � 1:
Put x � gau and y � gbpv. Then
xy � yx � gau�bpv � g,
xpv � gaupv � 1,
yu � gbupv � 1:
Hence the order of x is a power of p and the order of y divides u, so
is coprime to p. Therefore x and y satisfy conditions (1)±(3).
Now suppose that x9, y9 2 G also satisfy (1)±(3); that is, g �x9y9 � y9x9, the order of x9 is a power of p and the order of y9 is
coprime to p. We must show that x � x9 and y � y9.
We have
x9g � x9y9x9 � gx9,
so x9 commutes with g, hence also with gau � x. Since both x and x9
have order a power of p, it follows that xÿ1x9 has order a power of p.
Similarly, y9 commutes with y and y(y9)ÿ1 has order coprime to p.
Finally, xy � g � x9y9, so
xÿ1x9 � y(y9)ÿ1:
If z � xÿ1x9 � y(y9)ÿ1, then we have shown that the order of z is both
a power of p and coprime to p. Therefore z � 1, and so x � x9 and
y � y9, as required. j
Algebraic integers 255
22.19 De®nition
We call the element y which appears in Lemma 22.18 the p9-part of
g.
We extract the following statement from the proof of Lemma 22.18.
(22.20) Let the order of g be upv, where u, v 2 Z and
(u, p) � 1, and choose integers a, b with au � bpv � 1.
Then the p9- part of g is gbpv.
For example, if p � 2 and g has order 6, then the p9-part of g is
gÿ2; the expression g � xy in Lemma 22.18 has x � g3, y � gÿ2.
A little ring theory
To prepare for our main result on congruence properties of character
values, we need a few basic facts about a certain subring of C in
which all our character values will lie.
Let n be a positive integer and let æ � e2ði=n. De®ne Z[æ] to be the
subring of C generated by Z and æ; that is,
Z[æ] � f f (æ): f (x) 2 Z[x]g:Clearly, every element of Z[æ] is an integer combination of the powers
1, æ, æ2, . . . , ænÿ1, so in fact
Z[æ] � f f (æ): f (x) 2 Z[x] of degree < nÿ 1g:Now let p be a prime number and let
pZ[æ] � fpr: r 2 Z[æ]g,a principal ideal of Z[æ].
22.21 Proposition
There are only ®nitely many ideals I of Z[æ] which contain pZ[æ].
Proof Consider the factor ring Z[æ]= pZ[æ]. By de®nition, this has as
its elements all the cosets pZ[æ] � r with r 2 Z[æ]. Every such coset
contains an element of the form
a0 � a1æ� : : :� anÿ1ænÿ1, with ai 2 Z, 0 < ai < pÿ 1 for all i:
As there are only ®nitely many such elements, we conclude that
Z[æ]=pZ[æ] is ®nite. The ideals of Z[æ] which contain pZ[æ] are in
256 Representations and characters of groups
bijective correspondence with the ideals of Z[æ]=pZ[æ] (the corre-
spondence being I ! I= pZ[æ]). Therefore there are only ®nitely many
such ideals, and the proof is complete. j
We deduce from Proposition 22.21 that there is a maximal ideal of
P of Z[æ] which contains pZ[æ]; that is, P is a proper ideal which is
contained in no larger proper ideal of Z[æ]. (A proper ideal of Z[æ] is
an ideal which is not equal to Z[æ].)
We now prove two easy results about the maximal ideal P.
22.22 Proposition
If r, s 2 Z[æ] and rs 2 P, then either r 2 P or s 2 P. In particular, if
rn 2 P for some positive integer n, then r 2 P.
Proof Assume that rs 2 P and r =2 P. We must show that s 2 P.
Since r =2 P, the ideal rZ[æ] � P of Z[æ] strictly contains P. As P is
maximal, we therefore have
rZ[æ]� P � Z[æ]:
Consequently, there exist a 2 Z[æ], b 2 P such that
1 � ra� b:
Then
s � rsa� sb:
As rs 2 P and b 2 P, it follows that s 2 P, as required.
For the last statement of the proposition, assume that r n 2 P. Since
r n � rr nÿ1, this implies that either r 2 P or r nÿ1 2 P. Repeating this
argument, we conclude that r 2 P. j
22.23 Proposition
We have P \ Z � pZ.
Proof Let m 2 P \ Z. If p Bj m then there are integers a, b with
am� bp � 1; but this implies that 1 2 P, which is false, since
P 6� Z[æ]. Thus pjm, which establishes that P \ Z � pZ. Since p 2 P,
we also have pZ � P \ Z. j
Algebraic integers 257
Congruences
At last we are ready to prove our results on congruences of character
values. Let G be a group of order n and let æ � e2ði=n. The ring Z[æ]
is of interest because all the character values of G lie in Z[æ] (see
Proposition 9.11). As in the previous section, let p be a prime number
and let P be a maximal ideal of Z[æ] containing pZ[æ].
22.24 Theorem
Let g 2 G and let y be the p9- part of g. If ÷ is any character of G,
then
÷(g)ÿ ÷(y) 2 P:
Proof Suppose that g has order m � upv, where u, v 2 Z and
(u, p) � 1. Choose integers a, b with au � bpv � 1. Then y � gbpv(see
(22.20)).
The orders of g and of y divide n � |G|, so ÷(g) and ÷(y) are both
sums of nth roots of unity, and hence lie in Z[æ].
Now let ù be an mth root of unity (so ù 2 Z[æ] as mjn). Then
ù � ùau�bpv, and so
ù pv � ùaupv. ùbp2v � ùbp2v
:
Consider the number (ù ÿ ùbpv) pv
. By the Binomial Theorem,
(ùÿ ùbpv
) pv � ù pv ÿ pvù pvÿ1
ùbpv � : : :� pv
r
� �ù pvÿ r
ùrbpv
� : : :� (ÿ1) pv
ùbp2v
:
For 0 , r , pv, the binomial coef®cient ( pv
r ) is divisible by p. Hence
(ùÿ ùbpv
) pv � ù pv � (ÿ1) pv
ùbp2v � pá,
where á 2 Z[æ]. Moreover, since ù pv � ùbp2v, we have
ù pv � (ÿ1) pv
ùbp2v �0, if p 6� 2,
2ù pv, if p � 2,
(so it follows that
(ùÿ ùbpv
) pv 2 pZ[æ]:
Thus (ù ÿ ùbpv) pv
lies in the maximal ideal P. Application of Proposi-
tion 22.22 now forces
258 Representations and characters of groups
(ùÿ ùbpv
) 2 P:(22:25)
By Proposition 9.11, there are mth roots of unity ù1, . . . , ùd such
that
÷(g) � ù1 � : : :� ùd and ÷(y) � ùbpv
1 � : : :� ùbpv
d :
Then
÷(g)ÿ ÷(y) � (ù1 ÿ ùbpv
1 )� : : :� (ùd ÿ ùbpv
d ),
which, by (22.25), lies in P. j
22.26 Corollary
Let p be a prime number. Suppose that g 2 G and that y is the p9- part
of g. If ÷ is a character of G such that ÷(g) and ÷(y) are both
integers, then
÷(g) � ÷(y) mod p:
Proof As ÷(g) and ÷(y) are both integers, Theorem 22.24 and Proposi-
tion 22.23 give
÷(g)ÿ ÷(y) 2 P \ Z � pZ:
Therefore ÷(g) � ÷(y) mod p. j
22.27 Corollary
Let p be a prime number. Suppose that g 2 G and the order of g is a
power of p. If ÷ is a character of G such that ÷(g) 2 Z, then
÷(g) � ÷(1) mod p:
Proof As the order of g is a power of p, the p9-part of g is 1, so the
result is immediate from Corollary 22.26. j
Notice that Corollary 13.10 is the special case of Corollary 22.27 in
which g has order 2.
We shall use the congruence results 22.24±22.27 extensively in our
character calculations in Chapters 25±7. For the moment, we just
illustrate the results with reference to some character tables which we
already know.
Algebraic integers 259
22.28 Example
Recall from Example 20.14 that the character table of A5 is as shown.
where á � (1�p5)=2, â � (1ÿp5)=2.
If g � (1 2 3) then Corollary 22.26 implies that ÷(g) � ÷(1) mod 3
whenever ÷(g) 2 Z. Thus the corresponding entries in columns 1 and 2
of the character table are congruent modulo 3, as can be seen by
inspecting the table. Similarly the entries in columns 1 and 3 are
congruent modulo 2. Also
÷i((1 2 3 4 5)) � ÷i(1) mod 5 for i � 1, 2, 3:
However, ÷4((1 2 3 4 5)) � á =2 Z. We illustrate Theorem 22.24 for this
value. If we take p � 5 and g � (1 2 3 4 5), then the p9-part of g is
1, and
÷4(g)ÿ ÷4(1) � áÿ 3 � 12(1�p5ÿ 6)
� p5 . 12(1ÿp5) � â
p5:
Put æ � e2ði=60, and let P be a maximal ideal of Z[æ] containing 5Z[æ].
Then (p
5)2 2 P, sop
5 2 P by Proposition 22.22. Since â 2 Z[æ] (see
Proposition 9.11), we have âp
5 2 P. That is,
÷4(g)ÿ ÷4(1) 2 P:
This illustrates Theorem 22.24.
Summary of Chapter 22
1. Character values are algebraic integers.
2. The degree of every irreducible character of G divides |G|.
Character table of A5
1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)
÷1 1 1 1 1 1÷2 4 1 0 ÿ1 ÿ1÷3 5 ÿ1 1 0 0÷4 3 0 ÿ1 á â÷5 3 0 ÿ1 â á
260 Representations and characters of groups
3. If g is conjugate to gi for all integers i which are coprime to the
order of g, then ÷(g) is an integer, for all characters ÷.
4. Let p be a prime number. If g 2 G and y is the p9-part of g, then
÷(g) � ÷(y) mod p for all characters ÷ of G such that ÷(g) and ÷(y)
are integers.
Exercises for Chapter 22
1. Let G be a group of order 15. Use Theorems 11.12, 17.11 and
22.11 to show that every irreducible character of G has degree 1.
Deduce that G is abelian.
2. Prove that the number of conjugacy classes in a group of order 16
is 7, 10 or 16.
3. Let p and q be prime numbers with p . q, and let G be a non-
abelian group of order pq.
(a) Find the degrees of all the irreducible characters of G.
(b) Show that |G9| � p.
(c) Show that q divides p ÿ 1 and that G has q� (( pÿ 1)=q)
conjugacy classes.
4. Let G be a group and let ö be a character of G such that
ö(g) � ö(h) for all non-identity elements g and h of G.
(a) Show that ö � a1G � b÷reg for some a, b 2 C.
(b) Show that a � b and a � b|G| are integers.
(c) Show that if ÷ is a non-trivial irreducible character of G, then
b÷(1) is an integer.
(d) Deduce that both a and b are integers.
5. Suppose that G is a group of odd order. This exercise shows that
the only irreducible character ÷ of G such that ÷ � ÷ is the trivial
character.
(a) Prove that if g 2 G and g � gÿ1, then g � 1.
(b) Now let ÷ be an irreducible character of G with ÷ � ÷. Prove
that
h÷, 1Gi � 1
jGj (÷(1)� 2á),
where á is an algebraic integer.
(c) Deduce that ÷ � 1G.
Algebraic integers 261
6. It is often possible to calculate the character table from limited
arithmetic information about the group G. This exercise illustrates
this point with the group G � S5.
A certain group G of order 120 has exactly seven conjugacy
classes, and contains an element g of order 5 such that jCG(g)j � 5.
Moreover, g, g2, g3 and g4 are conjugate in G.
(a) Show that for every irreducible character ÷ of G, ÷(g) is 0, 1
or ÿ1.
(b) Use Corollary 22.27 to deduce that G has two irreducible
characters of degree 5.
(c) Find ÷(1) and ÷(g) for all irreducible characters ÷ of G.
(d) You are given that all entries in the character table of G are
integers, and that the conjugacy classes of G have represent-
atives g1, . . . , g7 with orders and centralizer orders as follows:
Using Corollary 22.26 and the column orthogonality relations, ®nd
the character table of G.
7. Prove that a complex number ë is an algebraic integer if and only if
ë is a root of a polynomial of the form
x n � anÿ1x nÿ1 � : : :� a1x� a0,
where each ar (0 < r < n ÿ 1) is an integer.
gi g1 g2 g3 g4 g5 g6 g7
Order of gi 1 2 2 3 4 6 5|CG(gi)| 120 12 8 6 4 6 5
262 Representations and characters of groups
263
23
Real representations
Since Chapter 9, we have always taken our representations to be over
the ®eld C of complex numbers. However, some results in representa-
tion theory work equally well for the ®eld R of real numbers. There is
a subtle interplay between representations over C and representations
over R, which we shall explore in this chapter.
Often, characters of CG-modules are real-valued, and the ®rst main
result of the chapter describes the number of real-valued irreducible
characters of G.
Let r be a representation of G. If all the matrices gr (g 2 G) have
real entries, then of course the character of r is real-valued. However,
the converse is not true: it can happen that the character of r is real-
valued, while there is no representation ó equivalent to r with all the
matrices gó having real entries. Various criteria for whether or not a
character corresponds to a representation over R lead us to the
remarkable Frobenius±Schur Count of Involutions. This is used in the
last section to prove a famous result of Brauer and Fowler concerning
centralizers of involutions in ®nite simple groups.
The material in this chapter is perhaps at a slightly more advanced
level than that in the rest of the book, and is not used in the ensuing
chapters, which consist largely of the calculation of character tables
and applications of character theory. Nevertheless, the subject of real
representations not only is elegant and interesting, but also gives
delicate information about characters which often comes into play in
more dif®cult calculations.
Real characters
An element g of the ®nite group G is said to be real if g is conjugate
to gÿ1; and if g is real, then the conjugacy class gG is also said to be
real. Notice that if a conjugacy class is real, then it contains the
inverse of each of its elements, since (gÿ1)G � fxÿ1: x 2 gGg.On the other hand, a character ÷ of G is real if ÷(g) is real for all
g 2 G. Thus for example, the conjugacy class {1} of G is real, and the
trivial character of G is real.
23.1 Theorem
The number of real irreducible characters of G is equal to the number
of real conjugacy classes of G.
Proof Let X denote the character table of G, and let X denote the
complex conjugate of the matrix X.
For every irreducible character ÷ of G, the complex conjugate ÷ is
also an irreducible character (see Proposition 13.15), so X can be
obtained from X by permuting the rows. Hence there is a permutation
matrix P such that
PX � X
(see Exercise 4.4).
For every conjugacy class gG of G, the entries in the column of X
which corresponds to gG are the complex conjugates of the entries in
the column of X which corresponds to (gÿ1)G. Therefore X can be
obtained from X be permuting the columns, and so there is a permuta-
tion matrix Q such that
XQ � X
By Proposition 16.2, X is invertible. Therefore
Q � Xÿ1 X � Xÿ1 PX :
Consequently P and Q have the same trace, by Proposition 13.2. Since
the trace of a permutation matrix is equal to the number of points
®xed by the corresponding permutation, we have
the number of real irreducible characters of G is tr (P),
andthe number of real conjugacy classes of G is tr (Q):
As these numbers are equal, the result is proved. j
Part of the following corollary was obtained by a different method in
Exercise 22.5.
264 Representations and characters of groups
23.2 Corollary
The group G has a non-trivial real irreducible character if and only if
the order of G is even.
Proof If G has odd order, then no non-identity element of G is real
(see the solution to Exercise 23.1). Therefore by Theorem 23.1, the
only real character of G is the trivial character.
If G has even order, then by Exercise 1.8, G has an element g of
order 2. Hence G has at least two real conjugacy classes, {1} and gG,
and so G has at least two real irreducible characters by Theorem
23.1. j
Characters which can be realized over R
Let ÷ be a character of the group G. We say that ÷ can be realized
over R if there is a representation r: G! GL (n, C) with character ÷,
such that all the entries in each matrix gr (g 2 G) are real. This is the
same as saying that there is some CG-module V with character ÷, and
there is a basis v1, . . . , vn of V, such that for all g 2 G and 1 < i < n,
vi g is a linear combination of v1, . . . , vn with real coef®cients.
23.3 Examples
(1) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let ÷ be the
irreducible character of G of degree 2 (see Example 16.3(3)). Then ÷can be realized over R, since
ar � 0 1
ÿ1 0
� �, br � ÿ1 0
0 1
� �provides a representation r of G with character ÷ such that all the
matrices gr (g 2 G) have real entries.
(2) Let G � Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l, and let ÷ be the
irreducible character of G of degree 2 (see Exercise 17.1). The values
of ÷ are as follows:
1 a2 a b ab
÷ 2 ÿ2 0 0 0
Real representations 265
Thus ÷ is real. In fact, ÷ cannot be realized over R, but it is at the
moment unclear how to prove this. (We shall eventually establish this
in Example 23.18(3) below.)
Although every character which can be realized over R is perforce a
real character, Example 23.3(2) tells us that the converse is false.
RG-modules
Recall that in Chapter 4 we de®ned an FG-module, where F is R or
C. Thus an RG-module is a vector space over R, with a multiplication
by elements of G satisfying the conditions of De®nition 4.2. In this
section we shall study the relationship between RG-modules and CG-
modules.
23.4 Examples
Let V be a 2-dimensional vector space over R, with basis v1, v2.
(1) V becomes an RD8-module if we de®ne
v1a � v2, v1b � ÿv1,
v2a � ÿv1, v2b � v2
(compare Example 23.3(1)).
(2) V becomes an RC3-module, where C3 � kx: x3 � 1l, if we de®ne
v1x � v2,
v2x � ÿv1 ÿ v2:
(This gives the representation x! 0 1
ÿ1 ÿ1
� �of Exercise 3.2.)
Every RG-module can easily be converted into a CG-module.
Simply take a basis v1, . . . , vn of the RG-module, and consider the
vector space over C with basis v1, . . . , vn. This new vector space is
clearly a CG-module (with vi g de®ned as before). The construction is
even easier to understand in terms of representations: if r:
G! GL (n, R) is a representation then for each g 2 G, the matrix grhas its entries in R, and hence also in C. Therefore we obtain a
representation r: G! GL (n, C). Notice that a character ÷ of G can be
266 Representations and characters of groups
realized over R if and only if there exists an RG-module with
character ÷.
Rather more subtle than this is the construction of an RG-module from
a given CG-module. Let V be a CG-module with basis v1, . . . , vn,
and let g 2 G. There exist complex numbers z jk such that
v j g �Xn
k�1
zjkvk (1 < j < n):
Now let VR be the vector space over R with basis
v1, : : : , vn, iv1, : : : , ivn:
Write z jk � xjk � iyjk with xjk , yjk 2 R. We de®ne a multiplication of
VR by g by putting
v j g �Xn
k�1
(xjkvk � yjk(ivk)), and(23:5)
(iv j)g �Xn
k�1
(ÿyjkvk � xjk(ivk)) (1 < j < n),
and extending linearly to de®ne vg for all v 2 VR. In this way we
de®ne vg for all v 2 VR and all g 2 G. Regarding v j as an element of
the CG-module V, we have
(v j g)h � v j(gh) for all g, h 2 G, 1 < j < n:
It follows easily that, regarding v j and iv j as elements of VR, we have
(v j g)h � v j(gh) and ((iv j)g)h � (iv j)(gh):
Hence using Proposition 4.6, we see that (23.5) makes VR into an RG-
module.
If ÷ is the character of V, then
÷(g) �Xn
k�1
zkk :
The character of VR, evaluated at g, is
2Xn
k�1
xkk � ÷(g)� ÷(g):
Hence the character of VR is ÷ � ÷.
We summarize the basic properties of VR in the next proposition.
Real representations 267
23.6 Proposition
Let V be a CG-module with character ÷.
(1) The RG-module VR has character ÷ � ÷; in particular, dim VR
� 2 dim V.
(2) If V is an irreducible CG-module and VR is a reducible RG-
module, then ÷ can be realized over R.
Proof We have already proved part (1).
For part (2), suppose that V is an irreducible CG-module and VR is
a reducible RG-module. Then by part (1), VR � U � W where U is an
RG-module with character ÷ and W is an RG-module with character ÷.
Thus there is an RG-module, namely U, with character ÷, and so ÷ can
be realized over R. j
23.7 Examples
(1) Let G � C3 � kx: x3 � 1l, and let V be the 1-dimensional CG-
module with basis v1 such that
v1x � 12(ÿ1� i
p3)v1
(note that 12(ÿ1� i
p3) � e2ði=3). Then VR has basis v1, iv1, and with
respect to this basis, x is represented by the matrix
ÿ1=2p
3=2
ÿp3=2 ÿ1=2
� �:
(2) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let V be the 2-
dimensional CG-module with basis v1, v2 such that
v1a � iv1, v1b � v2,
v2a � ÿiv2, v2b � v1:
Then VR has basis v1, v2, v3, v4, where v3 � iv1, v4 � iv2. With
respect to this basis, we obtain the representation r, where
ar �0 0 1 0
0 0 0 ÿ1
ÿ1 0 0 0
0 1 0 0
0BB@1CCA, br �
0 1 0 0
1 0 0 0
0 0 0 1
0 0 1 0
0BB@1CCA:
The subspace of VR which is spanned by v1 � v4 and v2 � v3 is an
RG-submodule. Therefore the character of V can be realized over R,
268 Representations and characters of groups
by Proposition 23.6(2). In fact, we already know this from Example
23.3(1).
Bilinear forms
The question of whether or not a given character can be realized over
R turns out to be related to the existence of a certain bilinear form on
the corresponding CG-module.
Let V be a vector space over F, where F is R or C. A bilinear form
â on V is a function which associates with each ordered pair (u, v) of
vectors in V an element â(u, v) of F, and which has the following
properties:
â(ë1u1 � ë2u2, v) � ë1â(u1, v)� ë2â(u2, v),
â(u, ë1v1 � ë2v2) � ë1â(u, v1)� ë2â(u, v2),
for all u, v, u1, u2, v1, v2 2 V and ë1, ë2 2 F. (Thus for ®xed u, v, the
functions x! â(x, v) and y! â(u, y) are both linear ± hence the term
bilinear.)
The bilinear form â is symmetric if
â(u, v) � â(v, u) for all u, v 2 V :
And the bilinear form â is skew-symmetric if
â(u, v) � ÿâ(v, u) for all u, v 2 V :
If V is an FG-module, then a bilinear form â on V is said to be G-
invariant if
â(ug, vg) � â(u, v) for all u, v 2 V and g 2 G:
Our next result shows that every RG-module has a G-invariant
symmetric bilinear form with a strong positivity property. A similar
result for CG-modules was given in Exercise 8.6.
23.8 Theorem
If V is an RG-module, then there exists a G-invariant symmetric
bilinear form â on V such that
â(v, v) . 0 for all non-zero v 2 V :
Real representations 269
Proof Let v1, . . . , vn be a basis of V. For u �Pnj�1ë jv j, v �Pn
j�1ì jv j 2 V with ë j, ì j 2 R, de®ne
ã(u, v) �Xn
j�1
ë jì j:
Then ã is a symmetric bilinear form on V. Moreover, for non-zero
v 2 V,
ã(v, v) �Xn
j�1
ì2j . 0:
Now let
â(u, v) �Xx2G
ã(ux, vx) (u, v 2 V ):
Again, â is a symmetric bilinear form on V, and â(v, v) . 0 for all
non-zero v 2 V.
If g 2 G, then gx runs through G as x runs through G, and hence
â(ug, vg) �Xx2G
ã(ugx, vgx) � â(u, v):
Therefore â is G-invariant and the theorem is proved. j
23.9 Proposition
Let V be an RG-module and let â be a G-invariant bilinear form on V.
If U is an RG-submodule of V, then so is
W � fw 2 V : â(u, w) � 0 for all u 2 Ug:
Proof It is easy to see that W is a subspace of V. Now let w 2 W and
g 2 G. For all u 2 U, we have ugÿ1 2 U, so
â(u, wg) � â(ugÿ1, wggÿ1) � â(ugÿ1, w) � 0:
Thus wg 2 W, so W is an RG-submodule of V. j
23.10 Proposition
Suppose that â is a G-invariant symmetric bilinear form on the RG-
module V, and that there exist u, v 2 V with â(u, u) . 0 and
â(v, v) , 0. Then V is a reducible RG-module.
270 Representations and characters of groups
Proof Theorem 23.8 supplies us with a G-invariant symmetric bilinear
form â1 on V such that
â1(w, w) . 0 for all non-zero w 2 V :
By a general result on bilinear forms (see Exercise 23.7), there is a
basis v1, . . . , vn of V such that
â1(vi, v j) � â(vi, v j) � 0 if i 6� j,
andâ1(vi, vi) � 1 for all i,
â(v1, v1) . 0,
â(v2, v2) , 0:
Let â(v1, v1) � x, and de®ne ã by
ã(u, v) � â1(u, v)ÿ 1
xâ(u, v) (u, v 2 V ):
Since â and â1 are G-invariant symmetric bilinear forms on V, so is ã.
But for all v �Pni�1ëivi 2 V (ëi 2 R), we have
ã(v, v1) � ë1ã(v1, v1) � 0:
Therefore, if we de®ne
W � fw 2 V : ã(v, w) � 0 for all v 2 Vg,then W is non-zero, and is an RG-submodule of V by Proposition 23.9.
Moreover,
ã(v2, v2) � 1ÿ 1
xâ(v2, v2) . 0,
so W 6� V. Therefore V is a reducible RG-module. j
We can now relate bilinear forms to the question of whether or not
a given character of G can be realized over R.
23.11 Theorem
Let ÷ be an irreducible character of G. The following two conditions
are equivalent:
(1) ÷ can be realized over R;
(2) there exists a CG-module V with character ÷, and a non-zero G-
invariant symmetric bilinear form on V.
Real representations 271
Proof We ®rst show that (2) implies (1). Let V be a CG-module with
character ÷, and suppose that â is a non-zero G-invariant symmetric
bilinear form on V. There exist u, v 2 V with â(u, v) � â(v, u) 6� 0.
Sinceâ(u� v, u� v) � â(u, u)� â(v, v)� 2â(u, v),
there exists w 2 V with â(w, w) 6� 0. Let â(w, w) � z and v1 � zÿ1=2w.
Thenâ(v1, v1) � 1:
Extend v1 to a basis v1, . . . , vn of V. Then v1, . . . , vn, iv1, . . . , ivn
is a basis of the RG-module VR.
De®ne a function W from VR to V by
W:Xn
j�1
ë jv j �Xn
j�1
ì j(iv j)!Xn
j�1
(ë j � iì j)v j (ë j, ì j 2 R):
Then W is a bijection, and for all w1, w2, v 2 VR, all ë 2 R and all
g 2 G, we have
(w1 � w2)W � w1W� w2W,(23:12)
(ëv)W � ë(vW),
(vg)W � (vW)g:
Now de®ne a function ~â on ordered pairs of elements of VR by
~â(u, v) � the real part of â(uW, vW) (u, v 2 VR):
You can readily check, using the properties (23.12), that ~â is a G-
invariant symmetric bilinear form on VR. Notice that
~â(v1, v1) � 1 and ~â(iv1, iv1) � ÿ1:
Therefore VR is a reducible RG-module, by Proposition 23.10. It now
follows from Proposition 23.6(2) that ÷ can be realized over R. This
establishes that (2) implies (1) in the theorem.
Conversely, suppose that ÷ can be realized over R, and let U be an
RG-module with character ÷. By Theorem 23.8, there is a non-zero G-
invariant symmetric bilinear form ã on U. Let u1, : : : , un be a basis
of U, and let V be the vector space over C with basis u1, : : : , un. As
explained earlier, V is a CG-module (with ui g de®ned as for U).
De®ne ã̂ on V by
ã̂Xn
j�1
ë ju j,Xn
k�1
ìk uk
!�Xn
j�1
Xn
k�1
ë jìkã(u j, uk)
272 Representations and characters of groups
(where ë j, ìk 2 C). Then ã̂ is a non-zero G-invariant symmetric
bilinear form on the CG-module V, and V has character ÷. Thus (1)
implies (2), and the proof of the theorem is complete. j
The indicator function
We now associate with each irreducible character ÷ of G a certain
number, called the indicator of ÷, which is always 0, 1 or ÿ1. We shall
see later that this number tells us whether or not ÷ can be realized
over R.
Observe that
h÷2, 1Gi � 1
jGjXg2G
÷(g)÷(g) � h÷, ÷i:
Therefore, for irreducible characters ÷, we have
h÷2, 1Gi �0, if ÷ is not real,
1, if ÷ is real:
(Let V be a CG-module with character ÷. Recall from Chapter 19 that
÷2 is the character of the CG-module V V, and
÷2 � ÷S � ÷A,
where ÷S is the character of the symmetric part of V V, and ÷A is
the character of the antisymmetric part of V V. Hence if
h÷2, 1Gi � 1, then precisely one of ÷S and ÷A has 1G as a constituent.
23.13 De®nition
If ÷ is an irreducible character of G, then we de®ne the indicator é÷ of
÷ by
é÷ �0, if ÷ is not a constituent of ÷S or ÷A,
1, if 1G is a constituent of ÷S ,
ÿ1, if 1G is a constituent of ÷A:
8>><>>:We call é the indicator function on the set of irreducible characters of
G. Note that é÷ 6� 0 if and only if ÷ is real.
The next result gives a signi®cant property of the indicator function,
relating it to the internal structure of the group G.
Real representations 273
23.14 Theorem
For all x 2 G, X÷
(é÷)÷(x) � jfy 2 G: y2 � xgj,
where the sum is over all irreducible characters ÷ of G.
Proof De®ne a function W: G! C by
W(x) � jfy 2 G: y2 � xgj (x 2 G):
Note that W is a class function on G, since for g 2 G we have
y2 � x, (gÿ1 yg)2 � gÿ1xg:
Therefore by Corollary 15.4, W is a linear combination of the irreduci-
ble characters of G.
The de®nition of é÷ gives
é÷ � h÷S ÿ ÷A, 1Gi
� 1
jGjXg2G
÷(g2) by Proposition 19:14
� 1
jGjXx2G
Xg2G: g2�x
÷(g2)
� 1
jGjXx2G
W(x)÷(x)
� hW, ÷i:Therefore, W �P(é÷)÷, and the result follows. j
23.15 Example
Let G � S3. The character table of G is
1 (1 2) (1 2 3)
÷1 1 1 1÷2 1 ÿ1 1÷3 2 0 ÿ1
274 Representations and characters of groups
Using Proposition 19.14 we calculate that é÷ � 1 for each irreducible
character ÷ of G, soP
(é÷)÷ � ÷1 � ÷2 � ÷3, which takes the following
values:
Sure enough, in accordance with Theorem 23.14, four elements of G
square to be 1, namely 1, (1 2), (1 3) and (2 3); no elements square to
be (1 2); and one element, (1 3 2), squares to be (1 2 3).
Back to reality
We now relate the indicator function to the previous material on
bilinear forms. Using this, we show that the indicator of an irreducible
character determines whether or not it can be realized over R, and
deduce the Frobenius±Schur Count of Involutions.
23.16 Theorem
Let V be an irreducible CG-module with character ÷.
(1) There exists a non-zero G-invariant bilinear form on V if and
only if é÷ 6� 0.
(2) There exists a non-zero G-invariant symmetric bilinear form on
V if and only if é÷ � 1.
(3) There exists a non-zero G-invariant skew-symmetric bilinear form
on V if and only if é÷ � ÿ1.
Proof In this proof we regard C as a 1-dimensional vector space over
C, and de®ne a multiplication of C by elements of G by
ëg � ë (ë 2 C, g 2 G):
In this way, C becomes a trivial CG-module.
(1) Suppose that é÷ 6� 0. Then 1G is a constituent of ÷2, and hence
the CG-module V V has a trivial CG-submodule. By Proposition 8.8,
there is a non-zero CG-homomorphism from V V onto this trivial
CG-submodule, and hence there is a non-zero CG-homomorphism Wfrom V V onto the trivial CG-module C.
1 (1 2) (1 2 3)
÷1 � ÷2 � ÷3 4 0 1
Real representations 275
Now de®ne â by
â(u, v) � (u v)W (u, v 2 V ):
Then â is a non-zero bilinear form on V, and for u, v 2 V and g 2 G,
we have
â(ug, vg) � (ug vg)W � ((u v)g)W
� ((u v)W)g � (u v)W � â(u, v):
Thus â is G-invariant.
Conversely, suppose that there is a non-zero G-invariant bilinear form
â on V. Let v1, : : : , vn be a basis of V, so that vi v j
(1 < i < n, 1 < j < n) form a basis of V V. De®ne W: V V! C by
putting
(vi v j)W � â(vi, v j) (1 < i < n, 1 < j < n)
and extending linearly to the whole of V V. For g 2 G, we have
((vi v j)g)W � (vi g v j g)W � â(vi g, v j g)
� â(vi, v j) as â is G-invariant
� (vi v j)W:
Hence W is a non-zero CG-homomorphism from V V onto the trivial
CG-module C. Thus, by Proposition 10.1, V V has a trivial CG-
submodule. If follows that ÷2 has the trivial character 1G as a constitu-
ent, and therefore é÷ 6� 0.
(2) Suppose that é÷ � 1. Then 1G is a constituent of ÷S , which is the
character of the CG-module S(V V), the symmetric part of V V. As
in (1), it follows by Proposition 8.8 that there is a non-zero CG-
homomorphism W from S(V V) onto the trivial CG-module C. De®ne
â(u, v) � (u v� v u)W (u, v 2 V ):
Then â is a non-zero G-invariant symmetric bilinear form on V.
Conversely, suppose that there exists a non-zero G-invariant sym-
metric bilinear form â on V. Let v1, . . . , vn be a basis of V, and de®ne
W: S(V V)! C by putting
(vi v j � v j vi)W � â(vi, v j) (1 < i, j < n)
and extending linearly. Since â is symmetric, W is well-de®ned; and Wis a non-zero CG-homomorphism from S(V V) onto the trivial CG-
276 Representations and characters of groups
module C. Hence ÷S has the trivial character 1G as a constituent, and
so é÷ � 1.
(3) The proof of (3) is very similar to that of (2), and is omitted. j
We can now relate real representations of G to involutions in G,
where by an involution we mean an element of order 2.
23.17 Corollary (The Frobenius±Schur Count of Involutions)
For each irreducible character ÷ of G, we have
é÷ �0, if ÷ is not real,
1, if ÷ can be realized over R,
ÿ1, if ÷ is real, but ÷ cannot be realized over R:
8>><>>:Moreover, for all x 2 G,X
÷
(é÷)÷(x) � jfy 2 G: y2 � xgj,
where the sum is over all irreducible characters ÷ of G. In particular,X÷
(é÷)÷(1) � 1� t,
where t is equal to the number of involutions in G.
Proof When we de®ned the indicator function, we showed that é÷ 6� 0
if and only if ÷ is real. And Theorems 23.11 and 23.16(2) show that ÷can be realized over R if and only if é÷ � 1. This proves that é÷ is
determined as in the statement of the corollary.
The expression forP
÷(é÷)÷(x) was obtained in Theorem 23.14.
Putting x � 1, we see thatP
÷(é÷)÷(1) is equal to the number of
elements y of G satisfying y2 � 1. These elements are just the
involutions in G, together with the identity, so the number of them is
precisely 1 � t. j
We conclude with some examples illustrating the use of Corollary
23.17.
23.18 Examples
(1) Let ÷ be a linear character. Then é÷ � 1 if ÷ is real, and é÷ � 0 if
÷ is non-real.
Real representations 277
For an abelian group, the Frobenius±Schur Count of Involutions
shows that the number of real irreducible (linear) characters is equal to
the number of real conjugacy classes, since in this case g is conjugate
to gÿ1 if and only if g2 � 1. This special case of Theorem 23.1 can
be proved directly without much dif®culty (see Exercise 23.2).
(2) We know that all the irreducible characters of D8 �ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l can be realized over R (see Example
23.3(1), and note that all four linear characters are real). Thus é÷ � 1
for all irreducible characters ÷ of D8, and soX÷
(é÷)÷(1) � 1� 1� 1� 1� 2 � 6:
The ®ve involutions in D8 which are predicted by the Frobenius±
Schur Count of Involutions are a2, b, ab, a2b and a3b.
(3) In the group Q8 � ka, b: a4 � 1, a2 � b2, bÿ1ab � aÿ1l, there is just
one involution, namely a2. Now é÷ � 1 for each of the four linear
characters, and X÷
(é÷)÷(1) � 2
by the Frobenius±Schur Count of Involutions. Therefore, if ø is the
irreducible character of degree 2, then éø � ÿ1. In particular ø cannot
be realized over R.
(4) The symmetric group S4 has ten elements whose square is 1, namely
the identity, the six elements which are conjugate to (1 2), and the three
elements which are conjugate to (1 2)(3 4). Since the degrees 1, 1, 2, 3,
3 of the irreducible characters of S4 sum to be 10 (see Section 18.1), we
see that all the characters of S4 can be realized over R.
The Brauer±Fowler Theorem
We now apply Corollary 23.17 to give a proof of a famous theorem of
Brauer and Fowler.
23.19 Brauer±Fowler Theorem
Let n be a positive integer. Then there exist only ®nitely many non-
isomorphic ®nite simple groups containing an involution with central-
izer of order n.
278 Representations and characters of groups
Despite its fairly elementary proof, this result is of great historical
importance in ®nite group theory. It led Brauer to propose the
programme of determining, for each ®nite group C, all the simple
groups G possessing an involution u with CG(u) � C. This programme
was the start of the modern attempt to classify all ®nite simple groups,
which was ®nally completed in the early 1980s. For further information
about this, see the book by D. Gorenstein listed in the Bibliography.
In Exercise 10 at the end of the chapter you are asked to carry out
Brauer's programme in the case where C � C2. This should not trouble
you too much. In Chapter 30, Theorem 30.8, you will ®nd a much
more sophisticated case, in which C � D8.
For proof of Theorem 23.19 we require two preliminary lemmas.
23.20 Lemma
If a1, . . . , an are real numbers, thenP
a2i >
Pai� �2=n.
Proof This follows from the Cauchy±Schwarz inequality
kvk kwk > jv:wj, taking v � (a1, . . . , an) and w � (1, . . . , 1). j
23.21 Lemma
Let G be a group of even order m, and let t be the number of
involutions in G (so t . 0 by Exercise 8 of Chapter 1). Write
a � (mÿ 1)=t. Then G contains a non-identity element x such that
jG : CG(x)j < a2.
Proof By Corollary 23.17, we have
t <X÷
÷(1)
where the sum is over all non-trivial irreducible characters ÷ of G.
Writing k for the number of irreducible characters of G, we deduce
using Lemma 23.20 and Theorem 11.12 that
t2 < (X÷
÷(1))2 < (k ÿ 1)X
÷(1)2 � (k ÿ 1)(mÿ 1),
and hence mÿ 1 < (k ÿ 1)(mÿ 1)2=t2 � (k ÿ 1)a2. Now k ÿ 1 is
the number of non-identity conjugacy classes of G. If every non-
identity conjugacy class has size more then a2, then (k ÿ 1)a2 .
Real representations 279
jGj ÿ 1 � mÿ 1, a contradiction. Therefore some non-identity class xG
has size at most a2. Then jG : CG(x)j < a2, giving the result. j
Proof of Theorem 23.19 Suppose G is simple and contains an involu-
tion u such that jCG(u)j � n. Let jGj � m, and let t be the number of
involutions in G. By Proposition 12.6, every element of the conjugacy
class uG is an involution, and hence
t > juGj � jG : CG(u)j � m=n:
Therefore (mÿ 1)=t , n, and so by Lemma 23.21, there is a non-
identity element x 2 G such that jG : CG(x)j, n2.
Let H � CG(x). If H � G then x lies in Z(G), the centre of G,
which is a normal subgroup of G. Since G is simple it follows that
G � Z(G), so G is abelian and therefore G � C2.
Now suppose that H 6� G. Write r � jG : H j, so r , n2. By Exer-
cise 9 at the end of the chapter, there is a non-trivial homomorphism èfrom G to the symmetric group Sr. As G is simple, the normal
subgroup Ker è � f1g. Thus G is isomorphic to a subgroup of Sr,
hence of Sn2 . In particular, given n, there are only ®nitely many
possibilities for G. j
Summary of Chapter 23
1. The number of real irreducible characters of G is equal to the
number of real conjugacy classes of G.
Let é be the indicator function, and let ÷ be an irreducible
character of G.
2: é÷ �
0, if ÷ is non-real,
1, if there exists an RG-module U with character ÷,
with character ÷
ÿ1, if ÷ is real, but there does not exist an RG-module:
8>>>><>>>>:3: é÷ �
0, if 1G is not a constituent of ÷S or ÷A,
1, if 1 G is a constituent of ÷S ,
ÿ1, if 1G is a constituent of ÷A:
8><>:4.P
÷(é÷)÷(1) � |{g 2 G: g2 � 1}|.
280 Representations and characters of groups
Exercises for Chapter 23
1. Prove that if G is a group of odd order then no non-identity
element of G is real.
2. Let G be a ®nite abelian group. Use the description of the
irreducible characters of G, given in Theorem 9.8, to prove directly
that the number of real irreducible characters of G is equal to the
number of elements g in G for which g2 � 1.
3. Let G � D2n and consult Section 18.3 for the character table of G.
How many elements g of G satisfy g2 � 1? Deduce that é÷ � 1 for
all irreducible characters ÷ of G.
4. Let r be an irreducible representation of degree 2 of a group G,
and let ÷ be the character of r. Prove that ÷A(g) � det (gr) for all
g 2 G. Deduce that é÷ � ÿ1 if and only if det (gr) � 1 for all
g 2 G.
5. Let G � T4n � ha, b: a2n � 1, an � b2, bÿ1ab � aÿ1i, as in Exer-
cise 17.6. Let V be a 2-dimensional vector space over C with basis
v1, v2 and let å be a (2n)th root of unity in C with å 6� �1.
Exercise 17.6 shows that V becomes an irreducible CG-module if
we de®ne
v1a � åv1, v1b � v2,
v2a � åÿ1v2, v2b � å nv1:
Let ÷ be the character of this CG-module V.
(a) Note that å n � �1. Use Exercise 4 to show that é÷ � 1 if
å n � 1 and é÷ � ÿ1 if å n � ÿ1.
(b) Let â be the bilinear form on V for which
â(v1, v1) � â(v2, v2) � 0,
â(v1, v2) � 1, â(v2, v1) � å n:
Prove that the bilinear form â is G-invariant, and use Theorem
23.16 to provide a second proof that é÷ � 1 if å n � 1 and
é÷ � ÿ1 if å n � ÿ1.
(c) Prove that an is the only element of order 2 in T4n.
(d) Use the character table of G, which appears in the solution to
Exercise 18.3, to ®nd é÷ for each irreducible character ÷ of G.
Check that
Real representations 281
X÷
(é÷)÷(1) � 2,
in agreement with the Frobenius±Schur Count of Involutions.
6. Prove that if ÷ is an irreducible character of a group G, and
é÷ � ÿ1, then ÷(1) is even.
(Hint: the solution uses a well known result about skew-symmetric
bilinear forms.)
7. Suppose that V is a vector space over R and that â1 and â are
symmetric bilinear forms on V. Assume that â1(w, w) . 0 for all
non-zero w in V. Prove that there is a basis e1, : : : , en of V such
that
â1(ei, ei) � 1 for all i, and
â1(ei, ej) � â(ei, ej) � 0 for all i 6� j:
8. Schur's Lemma is crucial for the development of the theory of
CG-modules. This exercise indicates the extent to which results
like Schur's Lemma hold for RG-modules.
Let V and W be irreducible RG-modules.
(a) Prove that if W: V! W is an RG-homomorphism then either Wis an RG-isomorphism or vW � 0 for all v 2 V.
(b) Prove that if W: V! V is an RG-isomorphism and V remains
irreducible as a CG-module, then W � ë1V for some real num-
ber ë.
. (c) Give an example of a group G, an irreducible RG-module V
and an RG-homomorphism W: V! V which is not a multiple
of 1V .
9. Let G be a group with a subgroup H of index n. Let Ù be the set
of n right cosets Hx of H in G. For g 2 G, de®ne a function
r g :Ù! Ù by (Hx)r g � Hxg for all x 2 G.
Prove that r g is a permutation of Ù, and that the function r : g! r g
is a homomorphism from G to the symmetric group on Ù.
Show that the kernel of r isT
x2Gxÿ1Hx.
Deduce that if a group G has a subgroup H of index n, then
there is a homomorphism G! Sn with kernel contained in H.
10. Suppose that G is a ®nite group containing and involution t with
CG(t) � C2. Prove that |G : G9| � 2. Deduce that if G is simple,
then G � C2.
282 Representations and characters of groups
283
24
Summary of properties of character tables
In this short chapter we present no new results, but instead we gather
together from previous chapters various properties which are helpful
when we try to ®nd the character table of a particular group. In the
next four chapters we shall calculate several character tables in detail.
Usually we begin by working out the conjugacy classes and centrali-
zer orders of our given ®nite group G. The size of the character table
is determined by the number k of conjugacy classes of G; the character
table is then a k 3 k matrix, with columns indexed by the conjugacy
classes of G (the ®rst column corresponding to the conjugacy class
{1}), and with rows indexed by the irreducible characters of G.
When doing calculations, we commonly come across a new character
÷, which may or may not be irreducible. We can then calculate h÷, ÷i,which is given by
h÷, ÷i � 1
jGjXg2G
÷(g)÷(g):
The character ÷ is irreducible if and only if h÷, ÷i � 1 (see Theorem
14.20). If ÷ is reducible then we calculate h÷, ÷ii for each of the
irreducible characters ÷i which we already know, and then
÷ÿX
i
h÷, ÷ii÷i
will also be a character. We can thus determine whether ÷ is a linear
combination of the irreducible characters we already know; and if it is
not, then we can obtain from ÷ a linear combination of irreducible
characters, all of which are new.
We have developed a number of methods for producing characters ÷
on which to perform such calculations. For example, every subgroup of
Sn has a permutation character (see (13.22)); the product of two
characters is again a character (Proposition 19.6); given a character øwe can form the symmetric and antisymmetric parts of its square, øS
and øA (see Proposition 19.14); and if H is a subgroup of G then we
can restrict characters of G to H, and induce characters of H to G.
These and other properties of characters are summarized in the follow-
ing list.
Properties of characters
Assume that ÷1, . . . , ÷k are the irreducible characters of G.
(1) (Example 13.8(3)) There is a (trivial) character ÷ of G which is
given by
÷(g) � 1 for all g 2 G:
(2) (Theorem 17.11) The group G has precisely jG=G9j linear char-
acters. These are the characters ÷ given by
÷(g) � ø(gG9) (g 2 G)
as ø varies over the irreducible (linear) characters of G=G9.
(3) (Theorem 17.3) As a generalization of (2), if N v G and ø is an
irreducible character of G=N , then we get an irreducible character
÷ of G which is given by
÷(g) � ø(gN ) (g 2 G)
(÷ is the lift of ø). This method gives precisely those irreducible
characters of G which have N contained in their kernel.
(4) (Theorem 19.18) If G � G1 3 G2 then all the irreducible charac-
ters ÷ of G are given by
÷(g1, g2) � ö1(g1)ö2(g2) (g1 2 G1, g2 2 G2),
as öi varies over the irreducible characters of Gi (i � 1, 2).
(5) (Proposition 13.24) If G is a subgroup of Sn, then the function
í: G! C de®ned by
í(g) � jfix (g)j ÿ 1 (g 2 G)
is a character of G.
284 Representations and characters of groups
(6) (Theorems 11.12 and 22.11) The entries ÷i(1) (1 < i < k) in the
®rst column of the character table of G are positive integers, and
satisfy
Xk
i�1
÷i(1)2 � jGj:
Moreover, each integer ÷i(1) divides |G|.
(7) (Row orthogonality relations, Theorem 16.4(1)) For all i, j, we
have h÷i, ÷ ji � äij.
(8) (Column orthogonality relations, Theorem 16.4(2)) For all g,
h 2 G, we have
Xk
i�1
÷i(g)÷i(h) �jCG(g)j, if g and h are conjugate,
0, otherwise:
(
(9) (Exercise 13.5) If ÷ is an irreducible character of G and z 2 Z(G),
then there exists a root of unity å such that for all g 2 G,
÷(zg) � å÷(g):
(10) (Proposition 13.9(2)) If g is an element of order n in G, and ÷ is
a character of G, then ÷(g) is a sum of nth roots of unity.
Moreover, |÷(g)| < ÷(1).
(11) (Proposition 13.9(3, 4)) If g 2 G and ÷ is a character of G, then
÷(gÿ1) � ÷(g):
In particular, if g is conjugate to gÿ1 then ÷(g) is real for all
characters ÷ of G.
(12) (Corollary 15.6) If g 2 G and g is not conjugate to gÿ1, then
÷(g) is non-real for some character ÷ of G.
(13) (Theorem 22.16) Let g 2 G. If g is conjugate to gi for all positive
integers i which are coprime to the order of g, then ÷(g) is an
integer for all characters ÷ of G.
(14) (Corollary 22.26) Suppose that p is a prime number and that y is
the p9-part of the element g of G. If ÷ is a character of G such
that ÷(g) and ÷(y) are both integers, then
÷(g) � ÷(y) mod p:
Summary of properties of character tables 285
In particular, if the order of g is a power of p, then
÷(g) � ÷(1) mod p:
(15) (Proposition 13.15) If ÷ is an irreducible character of G, then so
is ÷, where
÷(g) � ÷(g) (g 2 G):
(16) (Theorem 23.1) The number of real irreducible characters of G is
equal to the number of real conjugacy classes of G.
(17) (Proposition 17.14) If ÷ is an irreducible character of G and ë is
a linear character of G, then ֑ is an irreducible character of G,
where
÷ë(g) � ÷(g)ë(g) (g 2 G):
(18) (Proposition 19.6) If ÷ and ø are characters of G, then so is the
product ÷ø, where
÷ø(g) � ÷(g)ø(g) (g 2 G):
(19) (Proposition 19.14) If ÷ is a character of G, then so are ÷S and
÷A, where for all g 2 G,
÷S(g) � 12(÷2(g)� ÷(g2)),
÷A(g) � 12(÷2(g)ÿ ÷(g2)):
(20) (De®nition 21.13, Proposition 21.23) If H is a subgroup of G and
ø is a character of H, then ø " G is a character of G, with
values given by Proposition 21.23.
(21) (Chapter 20) If H is a subgroup of G and ø is a character of G,
then ø # H is a character of H, where
(ø # H)(h) � ø(h) (h 2 H):
We have seen that the character table of a group G gives group-
theoretic information about G. For example, the ®rst column determines
|G| and jG=G9j (by (6) and (2)). We can see from the character table
whether or not G is simple (Proposition 17.6); indeed, we can ®nd all
the normal subgroups of G (Proposition 17.5). Two important normal
subgroups are G9 and Z(G); these can be determined in the following
ways. The derived subgroup G9 consists of those elements g in G
which satisfy ÷(g) � 1 for all linear characters ÷ of G. The centre
Z(G) can be found by noting which elements g of G satisfy
286 Representations and characters of groups
P÷(g)÷(g) � |G|, the sum being over all irreducible characters ÷ of G.
In Chapter 30 we shall see some more impressive results about
subgroups of G, which can be deduced from the character table.
As a ®nal remark, it is of course true that isomorphic groups have
the same character table; however, the converse is false: in Exercise
17.1 we gave examples of non-isomorphic groups, D8 and Q8, with the
same character table.
Summary of properties of character tables 287
288
25
Characters of groups of order pq
By the end of the next chapter, we shall have determined the character
tables of all groups of order less than 32. A number of these groups
are so-called Frobenius groups, and in this chapter we shall describe a
class of Frobenius groups and ®nd the character tables of the groups in
this class. In particular, this will give the character tables of all groups
whose order is the product of two prime numbers.
Throughout the chapter, p will denote a prime number.
Primitive roots modulo p
Recall that the set
Z p � f0, 1, : : : , pÿ 1g,with addition and multiplication modulo p, is a ®eld; that is, Z p is an
abelian group under addition, and Z�p � Z p ÿ f0g is an abelian group
under multiplication. Clearly Z p is a cyclic group under addition,
generated by 1. It is also true, but not at all obvious, that Z�p is cyclic:
25.1 Theorem
The multiplicative group Z�p is cyclic; that is, there exists an integer n
such that
n pÿ1 � 1 mod p, and
nr 6� 1 mod p for 0 , r , pÿ 1:
An integer n of order p ÿ 1 in Z�p is called a primitive root modulo
p. We shall not provide a proof of Theorem 25.1, but for a good
account, we refer you to Theorem 45.3 of the book by J. B. Fraleigh
listed in the Bibliography.
25.2 Example
The number 2 is a primitive root modulo 3, 5, 11 and 13, but not
modulo 7; and 3 is a primitive root modulo 7.
As an immediate consequence of Theorem 25.1 we have
25.3 Proposition
If q| p ÿ 1 then there is an integer u such that u has order q modulo p
± that is, such that
uq � 1 mod p, and
ur 6� 1 mod p for 0 , r , q:
Frobenius groups of order pq, where q| p 2 1
25.4 Example
De®ne
G � 1 y
0 x
� �: x 2 Z�p, y 2 Z p
� �:
Under matrix multiplication, G is a group of order p( p ÿ 1) (see
Exercise 25.1).
Now let q| p ÿ 1, and let u be an element of order q in the
multiplicative group Z�p. De®ne
A � 1 1
0 1
� �, B � 1 0
0 u
� �,
and let F � hA, Bi, the subgroup of G generated by A and B. Then
Bÿ1 AB � 1 u
0 1
� �� Au,
and so we have the relations
Ap � Bq � I , Bÿ1 AB � Au:(25:5)
Using these relations, we see that every element of F is of the form
Ai Bj with 0 < i < p ÿ 1, 0 < j < q ÿ 1. These pq elements are dis-
Characters of groups of order pq 289
tinct, so jFj � pq. Moreover the relations (25.5) determine all products
in F, so we have the presentation
F � hA, B: Ap � Bq � I , Bÿ1 AB � Aui:
25.6 De®nition
If p is a prime and q| p ÿ 1, then we write F p,q for the group of order
pq with presentation
F p,q � ha, b: a p � bq � 1, bÿ1ab � aui,where u is an element of order q in Z�p.
It is not hard to show that, up to isomorphism, F p,q does not depend
on which integer u of order q we choose (see Exercise 25.3).
The groups F p,q belong to a wider class of groups known as
Frobenius groups. We shall not give the general de®nition of these
here, as we shall only be dealing with F p,q; the interested reader can
®nd more information in the book by D. S. Passman listed in the
Bibliography.
The next result classi®es all groups whose order is the product of
two distinct prime numbers.
25.7 Proposition
Suppose that G is a group of order pq, where p and q are prime
numbers with p . q. Then either G is abelian, or q divides p ÿ 1 and
G � F p,q.
Proof Assume that G is non-abelian. It follows from Exercise 22.3
that q divides p ÿ 1 and G has a normal subgroup H of order p.
(Alternatively, these facts follow readily from Sylow's Theorems (see
Chapter 18 of the book by J. B. Fraleigh listed in the Bibliography).)
Both H and G=H are cyclic, since they have prime order. Suppose
that H � kal and G=H � hHbi; then G is generated by a and b. Since
bq 2 H but b does not have order pq (as G is non-abelian), it follows
that b has order q.
Now H v G, so bÿ1ab � au for some integer u. Further,
a � bÿqabq � au q
and so uq � 1 mod p. Thus the order of u in the group Z�p divides q.
290 Representations and characters of groups
If the order of u were 1 then we would have bÿ1ab � a, and G would
be abelian. Therefore the order of u is q. We have now established that
a p � bq � 1, bÿ1ab � au, order of u in Z�p is q:
Hence G � F p,q. j
25.8 Example
By Proposition 25.7, every group of order 15 is abelian (indeed,
isomorphic to C3 3 C5); and the groups of order 21 are C3 3 C7 and
F7,3.
The character table of Fp,q
We have, in fact, already found the character tables of certain of the
groups F p,q: the dihedral group of order 2 p is the case where q � 2,
and in Example 21.25 we dealt with F7,3. We now construct the
character table of F p,q in general. Thus let
G � Fp,q � ha, b: ap � bq � 1, bÿ1ab � auiwhere p is prime, q| p ÿ 1 (q not necessarily prime), and u has order q
modulo p.
Let S be the subgroup of Z�p consisting of the powers of u. Thus
jSj � q. Write r � ( pÿ 1)=q, and choose coset representatives
v1, : : : , vr for S in Z�p.
25.9 Proposition
The conjugacy classes of G � F p,q are
f1g,(av i )G � favi s: s 2 Sg (1 < i < r),
(bn)G � fambn: 0 < m < pÿ 1g (1 < n < qÿ 1):
Proof The equation
bÿ javb j � avu j
shows that av is conjugate to avs for all s 2 S. Therefore the conjugacy
class of avi has size at least q; also the size of this conjugacy class is
equal to |G: CG(avi )|, and since kal < CG(avi ), this size is at most q.
Hence (avi )G has size q, and has the form stated in the proposition.
Characters of groups of order pq 291
Since CG(bn) contains kbl, and kbl has index p in G, it follows that
for n 6� 0 mod q, we have |CG(bn)| � q, and so the conjugacy class of
bn has size p. On the other hand, as G=hai is abelian, every conjugate
of bn has the form ambn for some m. Hence
(bn)G � fambn: 0 < m < pÿ 1gand the proof is complete. j
By Proposition 25.9, G has q � r conjugacy classes, so we seek
q � r irreducible characters.
First, observe that the derived subgroup G9 � kal, so G=G9 has order
q and therefore by Theorem 17.11, G has precisely q linear characters.
These are given by ÷n (0 < n < q ÿ 1), where
÷n(axb y) � e2ðiny=q (0 < x < pÿ 1, 0 < y < qÿ 1):
We shall show that G has r irreducible characters of degree q.
Let å � e2ði= p. For v 2 Z�p, denote by øv the character of kal which
is given by
øv(ax) � åvx (0 < x < pÿ 1):
We calculate the values of the induced character øv " G, using Proposi-
tion 21.23. We obtain
(øv " G)(axby) � 0 if 1 < y < qÿ 1, and
(øv " G)(ax) �Xs2S
åvsx (0 < x < pÿ 1):
Note that øv " G has degree q, and
øv " G � øvs " G if s 2 S:
For each coset representative v j (1 < j < r) of S in Z�p, let
ö j � øv j" G:
We now prove that each ö j is irreducible. By the Frobenius Recipro-
city Theorem 21.16, for all s 2 S,
hö j # hai, øv j sihai � hö j, øv j s " GiG � hö j, ö jiG:Hence
ö j # hai � hö j, ö jiGXs2S
øv j s � ÷,
292 Representations and characters of groups
where ÷ is either 0 or a character of kal. Taking degrees, it follows
that
ö j(1) > jSjhö j, ö jiG:Since ö j(1) � q � jSj, we deduce that kö j, ö jlG � 1. This proves that
ö j is irreducible, and also that
ö j # hai � hö j, ö jiGXs2S
øv j s:
By Theorem 14.23, the characters øv (v 2 Z�p) are linearly indepen-
dent, and hence ö1 # kal, . . . , ör # kal are distinct. Consequently the
irreducible characters ö1, . . . , ör are distinct.
We have now found q � r distinct irreducible characters ÷n, ö j of G
(0 < n < q ÿ 1, 1 < j < r), so we have the complete character table of
G. We summarize in the following theorem.
25.10 Theorem
Let p be a prime number, q| p ÿ 1 and r � ( pÿ 1)=q. Then the group
F p,q � ha, b: a p � bq � 1, bÿ1ab � aui� faxby: 0 < x < pÿ 1, 0 < y < qÿ 1g
has q � r irreducible characters. Of these, q have degree 1 and are
given by
÷n(axby) � e2ðiny=q (0 < n < qÿ 1)
and r have degree q and are given by
ö j(axby) � 0 if 1 < y < qÿ 1,
ö j(ax) �
Xs2S
e2ðiv j sx= p,
for 1 < j < r, where v1S, : : : , vrS are the cosets in Z�p of the
subgroup S generated by u.
We conclude by illustrating Theorem 25.10 in some examples.
25.11 Example
Let
G � F p, pÿ1 � ha, b: a p � b pÿ1 � 1, bÿ1ab � aui
Characters of groups of order pq 293
where u is a primitive root modulo p. Then G has p ÿ 1 linear
characters, and one irreducible character ö of degree p ÿ 1, with
values given by
ö(axby) � 0 if 1 < y < pÿ 2,
ö(ax) � ÿ1 if 1 < x < pÿ 1:
25.12 Example
Let a, b 2 S5 be the permutations
a � (1 2 3 4 5), b � (2 3 5 4):
Check that
a5 � b4 � 1, bÿ1ab � a2:
Hence if G � ka, bl, then G � F5,4, and so by the previous example
the character table of G is as shown.
25.13 Example
We consider the case p � 13, q � 4. Here
F13,4 � ha, b: a13 � b4 � 1, bÿ1ab � a5i:Write å � e2ði=13, and let
á � å� å5 � å8 � å12,
â � å2 � å3 � å10 � å11,
ã � å4 � å6 � å7 � å9:
By Theorem 25.10, the character table of F13,4 is as shown opposite.
In Example 21.25 we found the character table of F7,3. You may like
Character table of F5,4
gi 1 a b b2 b3
|CG(gi)| 20 5 4 4 4
÷0 1 1 1 1 1÷1 1 1 i ÿ1 ÿi÷2 1 1 ÿ1 1 ÿ1÷3 1 1 ÿi ÿ1 iö 4 ÿ1 0 0 0
294 Representations and characters of groups
to check that this agrees with the description of the character table
provided by Theorem 25.10.
Summary of Chapter 25
1. Suppose that p is prime and q divides p ÿ 1. Let u be an element
of order q in Z�p. Then
Fp,q � ha, b: ap � bq � 1, bÿ1ab � aui:
The irreducible characters of Fp,q are described in Theorem 25.10.
2. Let p and q be prime numbers with p . q. If G has order pq, then
either G is abelian or G � F p,q.
Exercises for Chapter 25
1. Let p be a prime number. Prove that
1 y
0 x
� �: x 2 Z�p, y 2 Z p
� �,
under matrix multiplication, is a group of order p( p ÿ 1).
2. Write down the character table of the non-abelian group F11,5 of
order 55.
3. Let p and q be positive integers, with p prime and q| p ÿ 1. Let u
and v be two integers which are of order q modulo p, and de®ne
Character table of F13,4
gi 1 a a2 a4 b b2 b3
|CG(gi)| 52 13 13 13 4 4 4
÷0 1 1 1 1 1 1 1÷1 1 1 1 1 i ÿ1 ÿi÷2 1 1 1 1 ÿ1 1 ÿ1÷3 1 1 1 1 ÿi ÿ1 iö1 4 á â ã 0 0 0ö2 4 â ã á 0 0 0ö3 4 ã á â 0 0 0
Characters of groups of order pq 295
G1 � ha, b: ap � bq � 1, bÿ1ab � aui,G2 � ha, b: ap � bq � 1, bÿ1ab � avi:
Prove that G1 � G2.
(This justi®es the comment which follows the de®nition of Fp,q in
25.6.)
4. Suppose that p is a prime number, with p 6� 2. Let q � ( p ÿ 1)=2
and let
G � Fp,q � ha, b: ap � bq � 1, bÿ1ab � aui,where u is an element of order q modulo p.
(a) Show that there exists an integer m such that um � ÿ1 mod p if
and only if p � 1 mod 4.
(b) Deduce that a is conjugate to aÿ1 if and only if p � 1 mod 4.
(c) Using the orthogonality relations, show that the two irreducible
characters ö1, ö2 of G of degree q have values
12(ÿ1�p(äp))
on the element a, where ä � 1 if p � 1 mod 4, and ä � ÿ1 if
p � ÿ1 mod 4.
(d) Deduce that if å � e2ði= p thenXs2Q
ås � (ÿ1�p(äp)),
where Q is the set of quadratic residues modulo p (that is,
Q � f12, 22, : : : , (( pÿ 1)=2)2g).5. Let E be the group of order 18 which is given by
E � ha, b, c: a3 � b3 � c2 � 1, ab � ba, cÿ1ac � aÿ1,
cÿ1bc � bÿ1i,as in Exercise 5.4. Note that ka, bl is a normal subgroup of E
which is isomorphic to C3 3 C3. By inducing linear characters of
this subgroup, obtain the character table of E.
6. Show that the group E of Exercise 5 has the properties that Z(E) is
cyclic, but E has no faithful irreducible representation. (Thus, E
provides a counterexample to the converse of Proposition 9.16.)
7. (a) Find a group whose irreducible character degrees are
296 Representations and characters of groups
1, 1, 1, 3, 3, 3, 3:
(b) Find a group whose irreducible character degrees are
1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3:
(c) Find a group whose irreducible character degrees are
1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 6, 6, 6, 6:
8. Let G be the group of order 54 which is given by
G � ha, b: a9 � b6 � 1, bÿ1ab � a2i:Find the character table of G.
Characters of groups of order pq 297
298
26
Characters of some p-groups
Throughout this chapter, p will be a prime number. We shall show
how to obtain the character tables of all groups of order pn for n < 4.
The method consists of examining the characters of those p-groups
which contain an abelian subgroup of index p, and before explaining
the method, we show that all groups of order pn with 1 < n < 4 do,
indeed, have an abelian subgroup of index p. We later give explicitly
the irreducible characters of all groups of order p3 and of all groups
of order 16. At the end of the chapter we point out, with references,
that we have found the character tables of all groups of order less
than 32.
Elementary properties of p-groups
A p-group is a group whose order is a power of the prime number p.
In the ®rst lemma we collect several well known properties of p-
groups. Recall that Z(G) denotes the centre of G (see De®nition 9.15).
26.1 Lemma
Let G be a group of order pn with n > 1.
(1) If {1} 6� H v G then H \ Z(G) 6� {1}. In particular, Z(G) 6� {1}.
(2) If K < Z(G) and G=K is cyclic, then G is abelian.
(3) If n < 2 then G is abelian.
Proof (1) Since H v G, H is a union of conjugacy classes of G, all
of which have size a power of p; and H \ Z(G) consists of those
conjugacy classes in H which have size 1. Therefore
jH j � jH \ Z(G)j � (a multiple of p):
As |H| is a multiple of p and |H \ Z(G)| 6� 0, we deduce that
H \ Z(G) 6� {1}.
(2) Suppose that G=K is cyclic, generated by gK. Let x1, x2 2 G.
Then
x1 � gik1, x2 � gjk2
for some integers i, j and some k1, k2 2 K. Since k1, k2 2 Z(G), it
follows that x1x2 � x2x1. Therefore G is abelian.
(3) By (1), jG=Z(G)j < pnÿ1. Hence if n < 2 then G=Z(G) is
cyclic and so G is abelian by (2). j
26.2 Lemma
Let G be a group of order pn with 1 < n < 4. Then G contains an
abelian subgroup of index p.
Proof The result is immediate if n � 1, so suppose that 2 < n < 4.
Assume that Z(G) contains a subgroup K of order pnÿ2. Then we
can ®nd a subgroup H of G such that K < H and jH j � pnÿ1. As
K < Z(H) and, by Lemma 26.1(2), H=Z(H) is not of order p, we
deduce that Z(H) � H . Therefore H is an abelian subgroup of index
p in G.
Now assume that Z(G) has no subgroup of order pnÿ2. Since
Z(G) 6� f1g by Lemma 26.1(1), the only possibility is that |G| � p4
and |Z(G)| � p. Then by Exercise 12.7, G has an element x whose
conjugacy class xG is of size p. Let H � CG(x). Then by Theorem
12.8, jH j � jGj=jxGj � p3. Moreover, Z(G) and kxl are distinct non-
identity subgroups of Z(H), and so Z(H) > p2. Hence again Z(H) � H
by Lemma 26.1(2), and H is an abelian subgroup of index p. j
For our ®nal result on the structure of p-groups, recall that the
derived subgroup of G is denoted by G9 (see De®nition 17.7).
26.3 Lemma
Let G be a non-abelian p-group which contains an abelian subgroup H
of index p. Then there exists a normal subgroup K of G such that
K < H \ G9 \ Z(G) and jKj � p:
Characters of some p-groups 299
Proof Since G is non-abelian, we have {1} 6� G9 v G, and hence
G9 \ Z(G) 6� {1} by Lemma 26.1(1). Let K be a subgroup of order p
in G9 \ Z(G). Now K < Z(G) implies that K v G and that KH is an
abelian subgroup of G (where KH � {kh: k 2 K, h 2 H}). Since G is
non-abelian and |G:H| � p, we have KH � H, and therefore K < H. j
Characters of p-groups with an abelian subgroup of index p
In view of Lemma 26.2, the next theorem provides us with all the
irreducible characters of non-abelian groups of order p3 or p4.
26.4 Theorem
Assume that G is a non-abelian p-group which contains an abelian
subgroup H of index p. Let K be a normal subgroup of G as in Lemma
26.3. Then every irreducible character of G is given by either
(1) the lift of an irreducible character of G=K, or
(2) ø " G, for some linear character ø of H which satis®es
K 6< Kerø.
Proof Let |G| � pn. By Theorem 17.3, the irreducible characters of
G=K lift to give precisely those irreducible characters of G which have
K in their kernel. The sum of the squares of the degrees of the
irreducible characters obtained in this way is jG=Kj � pnÿ1, by Theo-
rem 11.12.
We shall construct pnÿ2 ÿ pnÿ3 further irreducible characters of G,
each of degree p. Since
pnÿ1 � ( pnÿ2 ÿ pnÿ3) p2 � pn � jGj,(�)we shall then have obtained all the irreducible characters of G, again
by Theorem 11.12.
First note that if ÷ is a character of G of degree p, then either ÷ is
irreducible or ÷ is a sum of linear characters (since by Theorem 22.11,
the degree of every irreducible character of G is a power of p). In the
latter case, we have G9 < Ker ÷, as G9 is in the kernel of every linear
character, and hence K < G9 < Ker ÷. This establishes
if ÷(1) � p and K 6< Ker ÷, then ÷ is irreducible:(26:5)
We know by Theorem 9.8 that all the pnÿ1 irreducible characters of
the abelian group H are linear. Let Ö denote the set of linear
300 Representations and characters of groups
characters of H which do not have K in their kernel. Since the linear
characters of H which do have K in their kernel are precisely the lifts
of linear characters of H=K, we have
jÖj � pnÿ1 ÿ pnÿ2:
Let ø 2 Ö. By Proposition 21.23, since K < Z(G),
(ø " G)(k) � pø(k) for all k 2 K:
Thus ø " G has degree p and does not have K in its kernel. Therefore
by (26.5), ø " G is an irreducible character of G.
Suppose now that ø1 is a linear character of H such that
ø " G � ø1 " G. Then by the Frobenius Reciprocity Theorem 21.16,
1 � hø " G, ø1 " GiG � h(ø " G) # H , ø1iH :
Since (ø " G) # H has degree p, this implies that there are at most p
elements ø1 of Ö such that ø1 " G � ø " G. It follows that
fø " G: ø 2 Öggives at least jÖj= p � ( pnÿ1 ÿ pnÿ2)=p distinct irreducible characters
of G of degree p which do not have K in their kernel. As we saw
in (�), G has at most pnÿ2 ÿ pnÿ3 such characters. Therefore
fø " G: ø 2 Ög consists precisely of the pnÿ2 ÿ pnÿ3 irreducible
characters we seek, and the proof is complete. j
We now use Theorem 26.4 to give an explicit construction of the
irreducible characters of the non-abelian groups of order p3. We shall
then illustrate Theorem 26.4 further by constructing the character tables
of all the non-abelian groups of order 16.
Groups of order p3
By Theorem 9.6, the abelian groups of order p3 are
C p3 , C p2 3 Cp and Cp 3 Cp 3 Cp:
The character tables of these groups are given by Theorem 9.8.
Now let G be a non-abelian group of order p3. Write Z � Z(G). By
Lemma 26.1, Z 6� {1} and G=Z is non-cyclic. Hence G=Z � Cp 3 Cp
and Z � kzl � Cp. Choose aZ, bZ such that G=Z � haZ, bZi. Then
Characters of some p-groups 301
G=Z � farbsZ: 0 < r < pÿ 1, 0 < s < pÿ 1gand in particular, every element of G is of the form
arbszt
for some r, s, t with 0 < r, s, t < p ÿ 1.
26.6 Theorem
Let G � {arbszt: 0 < r, s, t < p ÿ 1} be a non-abelian group of order
p3, as above. Write å � e2ði= p. Then the irreducible characters of G
are
÷u,v (0 < u < pÿ 1, 0 < v < pÿ 1), and
öu (1 < u < pÿ 1),
where for all r, s, t,
÷u,v(arbszt) � å ru�sv,
öu(arbszt) �påut, if r � s � 0,
0, otherwise:
(
Proof By Theorem 9.8, the irreducible characters of G=Z are øu,v
(0 < u, v < p ÿ 1), where
øu,v(arbsZ) � å ru�sv:
The lift to G of øu,v is the linear character ÷u,v which appears in the
statement of the theorem.
Let H � ka, zl, so that H is an abelian subgroup of order p2. For
1 < u < p ÿ 1, choose a character øu of H which satis®es
øu(zt) � åut (0 < t < pÿ 1):
We shall calculate øu " G.
Let r be an integer with 1 < r < p ÿ 1. If ar is conjugate to an
element g of G, then arZ is conjugate to gZ in the abelian group
G=Z, so arZ � gZ, and therefore g � arzt for some t. Since ar =2 Z,
the conjugacy class (ar)G does not have size 1, and hence
(ar)G � farzt: 0 < t < pÿ 1g:
302 Representations and characters of groups
Then by Proposition 21.23,
(øu " G)(arzt) � øu(ar)� øu(arz)� : : :� øu(arz pÿ1)
� øu(ar)Xpÿ1
s�0
øu(zs)
� øu(ar)Xpÿ1
s�0
åus
� 0:
Also,(øu " G)(zt) � pøu(zt) � påut, and
(øu " G)(g) � 0 if g =2 H :
We have now established that if öu � øu " G, then öu takes the values
stated in the theorem.
We ®nd that
höu, öuiG � 1
p3
Xg2G
öu(g)öu(g)
� 1
p3
Xg2Z
öu(g)öu(g)
� 1
p3
Xg2Z
p2
� 1:
Therefore öu is irreducible.
Clearly the irreducible characters ÷u,v (0 < u, v < p ÿ 1) and
öu (1 < u < p ÿ 1) are all distinct, and the sum of the squares of their
degrees is
p2 . 12 � ( pÿ 1) . p2 � jGj:Hence we have found all the irreducible characters of G. j
Notice that the calculation in the proof of Theorem 26.6 is a special
case of the proof of Theorem 26.4 (with K � Z(G)).
In fact, up to isomorphism, there are precisely two non-abelian
groups of order p3. If p � 2, they are D8 and Q8. And if p is odd,
they are
Characters of some p-groups 303
H1 � ha, b: a p2 � b p � 1, bÿ1ab � a p�1i, and(26:7)
H2 � ha, b, z: a p � b p � z p � 1, az � za, bz � zb,
bÿ1ab � azi:We have Z(H1) � ka pl, Z(H2) � kzl. The elements a, b in H1 and H2
will serve for the elements a, b chosen in the statement of Theorem
26.6.
26.8 The groups of order 16
It is known that, up to isomorphism, there are precisely fourteen groups
of order 16 (see p. 134 of the book by Coxeter and Moser listed in the
Bibliography). We shall describe all these groups and their character
tables.
By Theorem 9.6, the abelian groups of order 16 are
C16, C8 3 C2, C4 3 C4, C4 3 C2 3 C2 and C2 3 C2 3 C2 3 C2,
and their character tables are given by Theorem 9.8.
For each of the nine non-abelian groups G of order 16 it is the case
that |G9 \ Z(G)| � 2 (see Exercise 26.7), so the subgroup K described
in Lemma 26.3 is given by
K � G9 \ Z(G):
Now G=K is a group of order 8. It is not C8 by Lemma 26.1(2), and
it is not Q8 by Exercise 26.8. Hence
G=K � D8, C4 3 C2 or C2 3 C2 3 C2:
We shall divide our descriptions into three parts, according to these
three possibilities for G=K. Our descriptions will be in terms of
presentations; it is possible to see, using Exercise 26.5, that all the nine
groups G1 , . . . , G9 given below do indeed have order 16.
(A) There are three non-abelian groups G of order 16 with
G=K � D8. These are
G1 � ha, b: a8 � b2 � 1, bÿ1ab � aÿ1i � D16,
G2 � ha, b: a8 � 1, b2 � a4, bÿ1ab � aÿ1i,G3 � ha, b: a8 � b2 � 1, bÿ1ab � a3i:
304 Representations and characters of groups
In each case K � ka4l. Each of the groups has seven conjugacy classes
C1, . . . , C7, and these are given in the following table.
C1 C2 C3 C4 C5 C6 C7
G1, G2 1 a4 a2, a6 a, a7 a3, a5 aib (i even) aib (i odd)G3 1 a4 a2, a6 a, a3 a5, a7 aib (i even) aib (i odd)
Using Theorem 26.4, we obtain the character tables of G1, G2 and
G3:
Class C1 C2 C3 C4 C5 C6 C7
Centralizerorder 16 16 8 8 8 4 4
1 1 1 1 1 1 11 1 1 1 1 ÿ1 ÿ11 1 1 ÿ1 ÿ1 1 ÿ11 1 1 ÿ1 ÿ1 ÿ1 12 2 ÿ2 0 0 0 02 ÿ2 0 á â 0 02 ÿ2 0 â á 0 0
where
á � p2 � ÿâ for G1, G2,
á � ip
2 � ÿâ for G3:
The ®rst ®ve characters are the lifts of the irreducible characters of
G=K � D8. The last two characters can be obtained as in Theorem
26.4(2) as induced characters ø " G, where ø is a linear character of
the abelian subgroup kal of index 2; alternatively, they can be found
by using the column orthogonality relations. Note that a is conjugate
to aÿ1 in G1 and G2, but not in G3; hence the values in the columns
C4 and C5 are all real for G1 and G2, but not for G3 (see Corollary
15.6).
(B) There are three non-abelian groups G of order 16 with G=K �C4 3 C2 (where, as before, K � G9 \ Z(G), of order 2). These are
G4 � ha, b, z: a4 � z, b2 � z2 � 1, bÿ1ab � azi,G5 � ha, b, z: a4 � 1, b2 � z, z2 � 1, bÿ1ab � azi,G6 � ha, b, z: a4 � 1, b2 � z2 � 1, bÿ1ab � az, az � za, bz � zbi:
Characters of some p-groups 305
The given presentations are somewhat cumbersome (for example, since
a4 � z in G4, z is redundant), but they are in a form which allows us
to describe the conjugacy classes C1, . . . , C10 of all three groups G4,
G5, G6 simultaneously:
C1 C2 C3 C4 C5 C6 C7 C8 C9 C10
1 z a2 a2z a, az a3, a3z b, bz a2b, a2bz ab, abz a3b, a3bz
In each case, K � kzl. The character tables of G4, G5 and G6 can again
be found using Theorem 26.4:
Class C1 C2 C3 C4 C5 C6 C7 C8 C9 C10
Centralizer order 16 16 16 16 8 8 8 8 8 8
1 1 1 1 1 1 1 1 1 11 1 ÿ1 ÿ1 i ÿi 1 ÿ1 i ÿi1 1 1 1 ÿ1 ÿ1 1 1 ÿ1 ÿ11 1 ÿ1 ÿ1 ÿi i 1 ÿ1 ÿi i1 1 1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ11 1 ÿ1 ÿ1 i ÿi ÿ1 1 ÿi i1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1 1 11 1 ÿ1 ÿ1 ÿi i ÿ1 1 i ÿi2 ÿ2 á â 0 0 0 0 0 02 ÿ2 â á 0 0 0 0 0 0
where
á � 2i � ÿâ for G4,
á � 2 � ÿâ for G5, G6:
(C) Finally, there are three non-abelian groups G of order 16 with
G=K � C2 3 C2 3 C2 (where K � G9 \ Z(G), of order 2). These are
G7 � ha, b, z: a4 � b2 � z2 � 1, bÿ1ab � aÿ1, az � za, bz � zbi� D8 3 C2,
G8 � ha, b, z: a4 � z2 � 1, a2 � b2, bÿ1ab � aÿ1, az � za, bz � zbi� Q8 3 C2,
G9 � ha, b, z: a2 � b2 � z4 � 1, bÿ1ab � az2, az � za, bz � zbi:
306 Representations and characters of groups
Each of these groups has ten conjugacy classes, which are given by
C1 C2 C3 C4 C5 C6 C7 C8 C9 C10
G7, G8 1 a2 z a2z a, a3 az, a3z b, a2b bz, a2bz ab, a3b abz, a3bzG9 1 z2 z z3 a, az2 az, az3 b, bz2 bz, bz3 ab, abz2 abz, abz3
We have
K � ha2i for G7, G8,
hz2i for G9,
(and the character tables of G7, G8 and G9, given by Theorem 26.4, are
as follows.
Class C1 C2 C3 C4 C5 C6 C7 C8 C9 C10
Centralizer order 16 16 16 16 8 8 8 8 8 8
1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ11 1 1 1 ÿ1 ÿ1 1 1 ÿ1 ÿ11 1 ÿ1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ11 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1 1 11 1 ÿ1 ÿ1 1 ÿ1 ÿ1 1 ÿ1 11 1 ÿ1 ÿ1 ÿ1 1 1 ÿ1 ÿ1 11 1 ÿ1 ÿ1 ÿ1 1 ÿ1 1 1 ÿ12 ÿ2 á â 0 0 0 0 0 02 ÿ2 â á 0 0 0 0 0 0
where
á � 2 � ÿâ for G7, G8,
á � 2i � ÿâ for G9:
26.9 The groups of order less than 32
At this point, we have in fact found the character tables of all groups
of order 31 or less. Apart from abelian groups and dihedral groups,
whose character tables are given by Theorem 9.8 and Section 18.3, the
groups, with references for their character tables, are as follows:
Characters of some p-groups 307
|G| G Reference forcharacter table
8 Q8 Exercise 17.112 A4 Section 18.2
T12 Exercise 18.316 G1, . . . , G9 Section 26.818 D6 3 C3 Theorem 19.18
E Exercise 25.520 T20 Exercise 18.3
F5,4 Theorem 25.1021 F7,3 Theorem 25.1024 D12 3 C2, A4 3 C2, T12 3 C2 Theorem 19.18
D8 3 C3, Q8 3 C3, D6 3 C4 Theorem 19.18S4 Section 18.1
SL (2, 3) Exercise 27.2T24 Exercise 18.3U24 Exercise 18.4V24 Exercise 18.5
27 H1, H2 Theorem 26.628 T28 Exercise 18.330 D6 3 C5, D10 3 C3 Theorem 19.18
Summary of Chapter 26
In this chapter, we gave the irreducible characters of various non-
abelian p-groups G, as follows.
1. Theorem 26.4: p-groups which contain an abelian subgroup of
index p.
2. Theorem 26.6: groups of order p3.
3. Section 26.8: groups of order 16.
Exercises for Chapter 26
1. Suppose that G is a group of order pn ( p prime, n > 2), with an
abelian subgroup H of index p. Show that for some integer m > 2,
G has pm linear characters and pnÿ2 ÿ pmÿ2 irreducible characters
of degree p.
308 Representations and characters of groups
2. Let H be the group of order 27 which is given by
H � ha, b, z: a3 � b3 � z3 � 1, az � za, bz � zb, bÿ1ab � azi(see (26.7)).
Find the conjugacy classes of H, and use Theorem 26.6 to write
down the character table of H.
3. Let G be the group of order 32 which is given by
G � ha, b: a16 � 1, b2 � a8, bÿ1ab � aÿ1i:Using Theorem 26.4, or otherwise, ®nd the character table of G.
4. Let A, B, C, D be the following 4 3 4 matrices:
A �
ÿ1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 ÿ1
0BBBBB@
1CCCCCA, B �
0 0 i 0
0 0 0 ÿi
i 0 0 0
0 ÿi 0 0
0BBBBB@
1CCCCCA,
C �
0 i 0 0
i 0 0 0
0 0 0 ÿi
0 0 ÿi 0
0BBBBB@
1CCCCCA, D �
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
0BBBBB@
1CCCCCA,
and let G � kA, B, C, Dl. Write Z � ÿI.
(a) Prove that all pairs of generators commute modulo hZi, and
deduce that G9 � hZi.(b) Show that for all g in G, g2 2 hZi, and deduce that G is a 2-
group of order at most 32.
(c) Prove that the given representation of G of degree 4 is
irreducible. (Hint: use Corollary 9.3.)
(d) Show that |G| � 32, and ®nd all the irreducible representations
of G.
5. Let G1, . . . , G9 be the non-abelian groups of order 16 with pre-
sentations as given in the text.
(a) Find faithful irreducible representations of degree 2 for G1, G2,
G3, G4 and G9.
(b) Why do the remaining groups G5, G6, G7 and G8 have no
faithful irreducible representations?
Characters of some p-groups 309
(c) Check that the following give faithful representations of G5
and G6:
G5: a!0 1 0
1 0 0
0 0 i
0@ 1A, b!i 0 0
0 ÿi 0
0 0 1
0@ 1A,
z!ÿ1 0 0
0 ÿ1 0
0 0 1
0@ 1A;
G6: a!i 0 0
0 ÿi 0
0 0 i
0@ 1A, b!0 1 0
1 0 0
0 0 1
0@ 1A,
z!ÿ1 0 0
0 ÿ1 0
0 0 1
0@ 1A:(d) Find faithful representations of degree 3 for G7 � D8 3 C2 and
G8 � Q8 3 C2.
(Note: This exercise can be used to con®rm that the presentations
of G1, . . . , G9 given in the text do indeed give groups of order 16.)
6. Prove that no two of the groups G1, . . . , G9 are isomorphic.
7. Let G be a non-abelian group of order p4.
(a) Prove that |Z(G)| � p or p2, and that if |Z(G)| � p2 then G
has p3 � p2 ÿ p conjugacy classes.
(b) Prove that |G9| � p or p2, and that if |G9| � p2 then G has
2 p2 ÿ 1 conjugacy classes.
(c) Deduce that |G9 \ Z(G)| � p.
8. (a) Prove that if G is any group, then G=Z(G) 6� Q8.
(Hint: assume that G=Z � haZ, bZ: a4 2 Z, a2 � b2 mod Z,
bÿ1ab � aÿ1 mod Zi. Prove that a2 commutes with b, and hence
that a2 2 Z.)
(b) Deduce from the result of Exercise 7 that if G is a group of
order 16, then G=(G9 \ Z(G)) 6� Q8.
310 Representations and characters of groups
311
27
Character table of the simple groupof order 168
Recall that a simple group is a non-trivial group G such that the only
normal subgroups of G are f1g and G itself. We discussed brie¯y in
Chapter 1 the signi®cance of simple groups in the theory of ®nite
groups. Examples of simple groups which we have met so far are
cyclic groups of prime order, A5 and A6. In fact the group A5, of
order 60, is the smallest non-abelian simple group. The next smallest
is a certain group of order 168, and in this chapter we shall describe
this group and ®nd its character table. The group belongs to a whole
family of simple groups, and we begin with a description of this
family.
Special linear groups
Let p be a prime number, and recall that Z p is the ®eld which consists
of the numbers 0, . . . , p ÿ 1, with addition and multiplication modulo
p. Denote by SL (2, p) the set of all 2 3 2 matrices M with entries in
Z p such that det M � 1. Then SL (2, p) is a group under matrix
multiplication, and is called the 2-dimensional special linear group
over Z p.
To calculate the order of the group SL (2, p), we count the
matrices
a b
c d
� �(a, b, c, d 2 Z p, ad ÿ bc � 1):
If c � 0, then there are p( p ÿ 1) choices for a, b, d which make
ad ÿ bc � 1 (since a, b are arbitrary, except that a 6� 0; and d is
determined by a). And there are p2( p ÿ 1) choices for a, b, c, d
with c 6� 0, such that ad ÿ bc � 1 (since a, d may be chosen
arbitrarily, c is any non-zero element of Z p; and then b is deter-
mined). Therefore
jSL (2, p)j � p( pÿ 1)� p2( pÿ 1)
� p( p2 ÿ 1):
If p � 2 then SL (2, p) has order 6, and it is easy to see that this
group is isomorphic to S3; so assume that p is an odd prime. By
Exercise 27.1, the centre of SL (2, p) is
Z � fI , ÿIg(where I is the 2 3 2 identity matrix). The factor group SL (2, p)=Z is
called the 2-dimensional projective special linear group, and is written
as PSL (2, p). Thus
PSL (2, p) � SL (2, p)=f�Ig:Since |SL (2, p)| � p( p2 ÿ 1), we have
jPSL(2, p)j � p( p2 ÿ 1)=2:
It is known that PSL (2, 3) � A4, PSL (2, 5) � A5, and that for p > 5,
the group PSL (2, p) is simple (see Theorem 8.19 of the book by J. J.
Rotman listed in the Bibliography).
The simple group G � PSL (2, 7) has order 168, and we shall
construct the character table of this group. After ®nding the conjugacy
classes of G, we shall ®nd the character table using only numerical
calculations, notably the orthogonality relations (Theorem 16.4) and
congruence properties (Corollary 22.26). The power of these techniques
is therefore well illustrated. In the exercises, we indicate other ways of
obtaining characters of G, using information about subgroups.
The conjugacy classes of PSL (2, 7)
27.1 Lemma
The group PSL (2, 7) has exactly six conjugacy classes. The following
table records representatives gi (1 < i < 6) for the conjugacy classes,
together with the order of gi, the order of CG(gi), and the size of the
conjugacy class containing gi.
312 Representations and characters of groups
Order of gi |CG(gi)| |gGi |
g1 � 1 0
0 1
� �Z 1 168 1
g2 � 0 1
ÿ1 0
� �Z 2 8 21
g3 � 2 ÿ2
2 2
� �Z 4 4 42
g4 � 2 0
0 4
� �Z 3 3 56
g5 � 1 1
0 1
� �Z 7 7 24
g6 � 1 ÿ1
0 1
� �Z 7 7 24
Proof For each i, we verify that gi has the stated order, and then use
direct calculation to ®nd all the elements of G which commute with
gi. Consider, for example, g4. Suppose that
a b
c d
� �Z
commutes with g4. Then
a b
c d
� �2 0
0 4
� �� � 2 0
0 4
� �a b
c d
� �,
and hence b � c � 0. Consequently
CG(g4) � 1 0
0 1
� �Z,
2 0
0 4
� �Z,
4 0
0 2
� �Z
� �:
Similarly
CG(g2) � MZ: M � 1 0
0 1
� �,
0 ÿ1
1 0
� �,
3 2
2 4
� �,
3 ÿ2
ÿ2 4
� �,
�2 3
3 ÿ2
� �,
2 4
4 ÿ2
� �,
2 2
ÿ2 2
� �,
2 ÿ2
2 2
� ��:
Also, CG(gi) � kgil for i � 3, 5, 6.
Among g1, . . . , g6, the only elements with the same order are g5
Character table of the simple group of order 168 313
and g6; so no two of these six elements are conjugate, except possibly
g5 and g6. Suppose that gÿ1 g6 g � g5 with
g � a b
c d
� �Z 2 G:
Then gg5 � g6 g, and so
a a� b
c c� d
� �� � aÿ c bÿ d
c d
� �with ad ÿ bc � 1:
It follows that c � 0, a 6� 0, d � aÿ1 and
a a� b
0 aÿ1
� �� a bÿ aÿ1
0 aÿ1
� �:
Therefore a2 � ÿ1, which is impossible for a 2 Z7. Thus g5 is not
conjugate to g6, and we have established that no two of the elements
g1, . . . , g6 are conjugate.
The size of the conjugacy class gGi is obtained by dividing 168 by
|CG(gi)| (Theorem 12.8). Since the sum of the sizes of the six
conjugacy classes gGi (1 < i < 6) is 168, these exhaust the conjugacy
classes of G. j
Notice that using Lemma 27.1, it is easy to check that G is indeed
simple, since any normal subgroup is a union of conjugacy classes (see
Proposition 12.19).
27.2 Corollary
(1) If 1 < i < 4 and ÷ is a character of G, then ÷(gi) is an integer.
(2) For some character ÷ of G, ÷(g5) is non-real.
Proof (1) By Lemma 27.1, for 1 < i < 4, gi is conjugate to (gi)k
whenever gi and (gi)k have the same order. Hence the conclusion
follows from Theorem 22.16.
(2) Notice that g6 � gÿ15 , so g5 is not conjugate to its inverse.
Therefore (2) follows from Corollary 15.6. j
The character table of G PSL (2, 7)
Since G has six conjugacy classes, it also has six irreducible charac-
ters. Let ÷1, . . . , ÷6 be the irreducible characters of G, where ÷1 is the
314 Representations and characters of groups
trivial character (so that ÷1(g) � 1 for all g 2 G). Recall that the
character table is the 6 3 6 matrix with ij-entry ÷i(gj).
We shall repeatedly exploit the column orthogonality relations,
Theorem 16.4(2), and the congruence properties given by Corollaries
22.26 and 22.27 for the elements g2, g3, g4, for which the character
values are known to be integers, by Corollary 27.2.
The entries in the column of g4 are integers, and the sum of the
squares of these integers is equal to |CG(g4)| � 3. The entries must
therefore be 1, �1, �1, 0, 0, 0 in some order. (We know that the entry
in the ®rst row is ÷1(g4) � 1.)
Similarly the entries in the column of g3 are 1, �1, �1, �1, 0, 0 in
some order, and the entries in column g2 are 1, �1, �1, �1, �2, 0 in
some order. Now for all characters ÷ of G, we have by Corollary
22.27,
÷(g2) � ÷(1) mod 2, and
÷(g3) � ÷(1) mod 2,
and so
÷(g2) � ÷(g3) mod 2:
Since we also know that X6
i�1
÷i(g3)÷i(g4) � 0,
we see that, with a suitable ordering of ÷2, . . . , ÷6, part of the
character table of G is as follows:
We shall determine the signs later. For the moment we concentrate on
the entries in the ®rst column of the character table (i.e. the degrees
÷i(1)). Let di � ÷i(1), so di is the entry on row i of column 1. By
Class representative g2 g3 g4
Centralizer order 8 4 3
÷1 1 1 1÷2 �1 �1 �1÷3 0 0 �1÷4 �1 �1 0÷5 �1 �1 0÷6 �2 0 0
Character table of the simple group of order 168 315
Corollary 22.27, Theorem 22.11 and the fact thatP6
i�1 d2i � 168, we
have
d4 � 0 mod 3,
d4 � 1 mod 2,
d4 divides jGj � 168, and
d24 < 168:
The only positive integer d4 which satis®es these conditions is d4 � 3.
In the same way, d5 � 3.
Next,
d6 � 0 mod 3,
d6 � 0 mod 2,
d6 divides 168, and
d26 < 168:
Therefore d6 is 6 or 12. But
0 �X6
i�1
÷i(g2)di � 1� d2 � 3� 3� 2d6,
so as d22 < 168, we have d6 6� 12, and hence d6 � 6.
Now
1� d22 � d2
3 � 32 � 32 � 62 � 168,
so d22 � d2
3 � 113. The only solutions to this equation with d2, d3
positive integers have d2, d3 equal to 7, 8 in some order. Since
d2 � 1 mod 2, we have d2 � 7 and d3 � 8.
We have now found the ®rst column of the character table, and have
the following portion:
Class representative g1 g2 g3 g4
Centralizer order 168 8 4 3
÷1 1 1 1 1÷2 7 �1 �1 �1÷3 8 0 0 �1÷4 3 �1 �1 0÷5 3 �1 �1 0÷6 6 �2 0 0
316 Representations and characters of groups
The equations
X6
i�1
÷i(g1)÷i(gj) � 0 for j � 2, 3, 4
now enable us to determine the signs in the columns for g2, g3, g4.
We obtain
Class representative g1 g2 g3 g4
Centralizer order 168 8 4 3
÷1 1 1 1 1÷2 7 ÿ1 ÿ1 1÷3 8 0 0 ÿ1÷4 3 ÿ1 1 0÷5 3 ÿ1 1 0÷6 6 2 0 0
Next, the equation
1 � h÷2, ÷2i �X6
i�1
÷2(gi)÷2(gi)
jCG(gi)j
� 7:7
168� 1
8� 1
4� 1
3� ÷2(g5)÷2(g5)
7� ÷2(g6)÷2(g6)
7
gives ÷2(g5) � ÷2(g6) � 0. (Note that ÷2(g5) � ÷2(1) mod 7, but we
could not use this fact as we were not sure that ÷2(g5) was an integer.)
Also, for j � 5, 6,
0 �X6
i�1
÷i(g4)÷i(gj) � 1ÿ ÷3(gj)
and so ÷3(g5) � ÷3(g6) � 1.
By Corollary 27.2, there is an irreducible character ÷ of G such that
÷(g5) is non-real. For this character ÷, the complex conjugate ÷ will be
a different character of the same degree. Hence ÷4 and ÷5 (being the
only two irreducible characters with the same degree) must be complex
conjugates of each other.
Let ÷4(g5) � ÷5(g5) � z, and let ÷6(g5) � t. Thus the column for g5
is
Character table of the simple group of order 168 317
Now
0 �X6
i�1
÷i(g2)÷i(g5) � 1ÿ zÿ z� 2t,
0 �X6
i�1
÷i(g3)÷i(g5) � 1� z� z,
7 �X6
i�1
÷i(g5)÷i(g5) � 2� 2zz� t t:
Solving these equations, we obtain
t � ÿ1, z � (ÿ1� ip
7)=2:
Since g6 � gÿ15 , we have ÷(g6) � ÷(g5) for all characters ÷ of G.
We have now completely determined the character table of
G � PSL (2, 7), as shown.
It is known that there are precisely ®ve non-abelian simple groups of
order less than 1000. We give you the character tables of all of these,
Class representative g5
Centralizer order 7
÷1 1÷2 0÷3 1÷4 z÷5 z÷6 t
Character table of PSL (2, 7)
Class representative g1 g2 g3 g4 g5 g6
Centralizer order 168 8 4 3 7 7
÷1 1 1 1 1 1 1÷2 7 ÿ1 ÿ1 1 0 0÷3 8 0 0 ÿ1 1 1÷4 3 ÿ1 1 0 á á÷5 3 ÿ1 1 0 á á÷6 6 2 0 0 ÿ1 ÿ1
where á � (ÿ1� ip
7)=2.
318 Representations and characters of groups
as follows:
G Order of G Reference for character table
A5 60 Example 20.13PSL (2, 7) 168 This chapter
A6 360 Exercise 20.2PSL (2, 8) 504 Exercise 28.3PSL (2, 11) 660 Exercise 27.6
Summary of Chapter 27
1. SL (2, p) � a b
c d
� �: a, b, c, d 2 Z p, ad ÿ bc � 1
� �.
jSL (2, p)j � p( p2 ÿ 1).
2. PSL (2, p) � SL (2, p)=f�Ig.jPSL (2, p)j � p( p2 ÿ 1)=2 ( p odd).
3. We constructed the character table of PSL (2, 7), the simple group
of order 168.
Exercises for Chapter 27
1. Prove that Z(SL (2, p)) � f�Ig.2. Find the character table of SL (2, 3).
3. Deduce directly from the character table of PSL (2, 7) that this
group is simple.
4. In this exercise we present an alternative construction of the
character table of G � PSL (2, 7), given the conjugacy classes of G,
as in Lemma 27.1.
(a) De®ne the subgroup T of G, of order 21, as follows:
T � a b
0 aÿ1
� �Z : a 2 Z�7, b 2 Z7
� �(where Z � {�I}). Calculate the values of the induced char-
acter (1T ) " G, and show that
(1T ) " G � 1G � ÷,
where ÷ is an irreducible character of G.
Character table of the simple group of order 168 319
(b) Let ë be a non-trivial linear character of T. Calculate the values
of ë " G and prove that this is an irreducible character
of G.
(c) By considering ÷S (see Proposition 19.14), obtain an irreducible
character of G of degree 6.
(d) From (a), (b), (c), we now have irreducible characters of G of
degrees 1, 7, 8 and 6. Use orthogonality relations to complete
the character table of G.
5. The character table of SL (2, 7).
Let G � SL (2, 7), the group of all 2 3 2 matrices of determinant
1, with entries in the ®eld Z7.
(a) Show that G has 11 conjugacy classes with representatives gi as
follows:
gi Order of gi |CG(gi)| | gGi |
g1 � 1 0
0 1
� �1 336 1
g2 � ÿ1 0
0 ÿ1
� �2 336 1
g3 � 0 1
ÿ1 0
� �4 8 42
g4 � 2 ÿ2
2 2
� �8 8 42
g5 � ÿ2 2
ÿ2 ÿ2
� �8 8 42
g6 � 2 0
0 4
� �3 6 56
g7 � ÿ2 0
0 ÿ4
� �6 6 56
g8 � 1 1
0 1
� �7 14 24
g9 � ÿ1 ÿ1
0 ÿ1
� �14 14 24
g10 � 1 ÿ1
0 1
� �7 14 24
g11 � ÿ1 1
0 ÿ1
� �14 14 24
320 Representations and characters of groups
(b) Use the character table of PSL (2,7) to write down the six
irreducible characters of G with kernel containing Z � {�I}.
(c) Let ÷7, ÷8, ÷9, ÷10, ÷11 be the remaining irreducible characters
of G. Show that for any j with 7 < j < 11 and any g 2 G, we
have ÷ j(g) � ÿ÷ j(ÿg).
(d) Prove that ÷ j(g3) � 0 for 7 < j < 11, and deduce that ÷ j(1) is
even.
(e) By considering the column of g6 in the character table, and
congruences modulo 3, show that the degrees of ÷7, . . . , ÷11 are
4, 4, 6, 6, 8, and ®nd ÷ j(g6) for 7 < j < 11.
(f) Let ø be one of the irreducible characters of degree 4. By
considering the values of øA on g1, g2, g3 and g6 (see
Proposition 19.14), prove that øA is equal to the irreducible
character of G of degree 6 whose kernel contains Z. Deduce the
values of the irreducible characters of degree 4 on all gi.
(g) Complete the character table of G.
6. The character table of PSL (2, 11).
Let G � PSL (2, 11). This group has eight conjugacy classes with
representatives g1, . . . , g8 having orders and centralizer orders as
follows:
Also, gÿ15 , gÿ1
6 , gÿ17 , gÿ1
8 are conjugate to g5, g6, g8, g7, respec-
tively.
Find the character table of G.
gi g1 g2 g3 g4 g5 g6 g7 g8
Order of gi 1 2 3 6 5 5 11 11|CG(gi)| 660 12 6 6 5 5 11 11
Character table of the simple group of order 168 321
322
28
Character table of GL(2, q)
We are now going to calculate the character tables of an important
in®nite series of groups, and one of the exercises will show you how
to use the results of this chapter to determine the character tables of
in®nitely many simple groups. In the last chapter and its exercises, we
found the character tables of certain groups of 2 3 2 matrices with
entries in Z7 and Z11. We shall determine the character tables of some
matrix groups with entries from an arbitrary ®nite ®eld. At ®rst sight,
this is a daunting task, since the number of irreducible characters
increases with the size of the ®eld. However, we shall see that the
conjugacy classes of our groups fall into four families, as do the
irreducible characters. Consequently, we can display the character
values in a 4 3 4 matrix.
The ®elds Fq and Fq2
We consider ®nite ®elds, and we shall tell you which properties of
these ®elds we will use; if you are unfamiliar with ®nite ®elds then
you might like to consult the book by J. B. Fraleigh listed in the
Bibliography.
Recall that a ®eld (F, �, 3) is a set F with two binary operations
� and 3 such that the following properties hold. First, (F, �) is an
abelian group, with identity element 0. Secondly, if we write
F� � Fnf0g, then (F�, 3) is an abelian group, with identity element
1. Finally, the distributive law holds; that is (a� b)c � ac� bc for all
a, b, c 2 F. For example, R, C and Z p ( p prime) are ®elds, with the
usual de®nitions of � and 3.
The basic properties of ®nite ®elds which we will use without proof
are these:
(28.1) Let p be a prime and n be a positive integer, and write
q � pn. Then there exists a ®eld Fq of order q and
every ®eld of order q is isomorphic to Fq.
For every s 2 Fq the sum of s with itself p times is
zero; in short, ps � 0.
The group (F�q , 3) is cyclic.
Notice that the binomial coef®cients (pi ) with 1 < i < pÿ 1 are all
divisible by p; it follows that (s� t) p � s p � t p for all s, t 2 Fq, and
hence (s� t) p k � s p k � t p k
for all positive integers k. We use this
remark in the proof of the next proposition.
28.2 Proposition
Let F � Fq2 and S � fs 2 F : sq � sg.(1) The set S is a sub®eld of F of order q, and hence S � Fq.
(2) If r 2 F then r � rq, r1�q 2 S.
Proof (1) Suppose that s, t 2 S. Then (s� t)q � sq � tq � s� t, so
s� t 2 S. It is now easy to check that (S, �) and (Snf0g, 3) are
abelian groups, so S is a ®eld.
(2) Since F�q2 is a group of order q2 ÿ 1, we see that rq2 � r for all
r 2 F. This implies that (r � rq)q � rq � rq2 � r � rq and (r1�q)q �r1�q, so r � rq, r1�q 2 S. j
Hereafter, we shall identify the sub®eld S of Fq2 in Proposition 28.2
with the ®eld Fq.
We introduce the following useful notation.
(28.3) Let å be a generator of the cyclic group F�q2 and let
ù � e(2ði=(q2ÿ1)). Suppose that r 2 F�q2 . We may write
r � åm for some m and we let r � ùm.
Then r! r is an irreducible character of F�q2 . More-
over, every irreducible character of F�q2 has the form
r ! r j for some integer j.
You are now in a position to appreciate the statement of the main
result of this chapter, namely Theorem 28.5.
Character table of GL(2, q) 323
The conjugacy classes of GL(2, q)
The general linear group GL(2, q) is de®ned to be the group of
invertible 2 3 2 matrices with entries in Fq. The subgroup consisting of
all matrices of determinant 1 is the special linear group SL(2, q), and
we talked about some special linear groups in the last chapter. Here,
we are going to calculate the character table of GL(2, q).
Let G � GL(2, q), and remember that the matrix
a b
c d
� �belongs to G if and only if its rows are linearly independent. The
number of such matrices is found by noting that (a, b) can be any
non-zero row vector, giving us q2 ÿ 1 choices; and once (a, b) has
been chosen, (c, d) can be any row vector which is not a multiple of
(a, b), giving us q2 ÿ q choices. Therefore,
jGj � (q2 ÿ 1)(q2 ÿ q) � q(qÿ 1)2(q� 1):
There are four families of conjugacy classes of G, of which three
are easy to describe.
First,
a b
0 c
� �can be conjugate to
a9 b90 c9
� �only if fa, cg � fa9, c9g, since conjugate matrices have the same
eigenvalues. Keep this in mind during the following discussion.
The matrices
sI � s 0
0 s
� �(s 2 F�q )
belong to the centre of G. They give us qÿ 1 conjugacy classes of
size 1.
Next, consider the matrices
us � s 1
0 s
� �(s 2 F�q ):
Let
324 Representations and characters of groups
g � a b
c d
� �2 G:
Then
gus � as a� bs
cs c� ds
� �and usg � as d � bs
cs ds
� �so g belongs to the centralizer of us if and only if c � 0 and a � d.
Thus, the matrices us (s 2 F�q ) give us qÿ 1 conjugacy classes; the
centralizer order is (qÿ 1)q, so, by Theorem 12.8, each conjugacy
class contains q2 ÿ 1 elements.
Now, let
ds, t � s 0
0 t
� �2 G (s, t 2 F�q )
and note that
0 1
1 0
� �ÿ1
ds, t0 1
1 0
� �� d t,s:
On the other hand, if s 6� t, then we have that gds, t � ds, t g if and only
if b � c � 0. Thus, the matrices ds, t (s, t 2 F�q , s 6� t) give us
(qÿ 1)(qÿ 2)=2 conjugacy classes; the centralizer order is (qÿ 1)2, so
each conjugacy class contains q(q� 1) elements.
Finally, consider
vr � 0 1
ÿr1�q r � rq
� �(r 2 Fq2nFq):
By Proposition 28.2, vr 2 G. The characteristic polynomial of vr is
det(xI ÿ vr) � x(xÿ (r � rq))� r1�q � (xÿ r)(xÿ rq),
so vr has eigenvalues r and rq. Since r =2 Fq we see that vr lies in
none of the conjugacy classes we have constructed so far.
Now,
gvr �ÿbr1�q a� b(r � rq)
ÿdr1�q c� d(r � rq)
!and
vr g �c d
ÿar1�q � c(r � rq) ÿbr1�q � d(r � rq)
!:
Character table of GL(2, q) 325
Hence gvr � vr g only if c � ÿbr1�q and d � a� b(r � rq). If these
conditions hold, then
ad ÿ bc � a2 � ab(r � rq)� b2 r1�q � (a� br)(a� brq):
Since (a, b) 6� (0, 0) and r, rq =2 Fq, we see that a� br and a� brq
are non-zero. Therefore, g 2 CG(vr) if and only if
g � a b
ÿbr1�q a� b(r � rq)
� �:
Thus, jCG(vr)j � q2 ÿ 1, and the conjugacy class containing vr has
size q2 ÿ q.
The matrix v t has eigenvalues t and tq, so it is not conjugate to vr
unless t � r or t � rq. We therefore partition Fq2nFq into subsets
fr, rqg; each subset gives us a conjugacy class representative vr and
different subsets give us representatives of different conjugacy classes.
We have now found all the conjugacy classes of G.
28.4 Proposition
There are q2 ÿ 1 conjugacy classes in GL(2, q), described as follows.
Class rep. g sI us ds, t vr
|CG(g)| (q2 ÿ 1)(q2 ÿ q) (qÿ 1)q (qÿ 1)2 q2 ÿ 1
No. of classes qÿ 1 qÿ 1 (qÿ 1)(qÿ 2)=2 (q2 ÿ q)=2
The families of conjugacy class representatives sI and us are indexed
by elements s of F�q .
The family of conjugacy class representatives ds, t is indexed by
unordered pairs fs, tg of distinct elements of F�q .
The family of conjugacy class representatives vr is indexed by
unordered pairs fr, rqg of elements of F�q2nF�q .
Proof The conjugacy classes we have found account for
(qÿ 1)� (qÿ 1)(q2ÿ 1)� (qÿ 1)(qÿ 2)q(q� 1)=2� (q2ÿ q)(q2ÿ q)=2
elements altogether. But this sum is equal to the order of GL(2, q), so
we have found all the conjugacy classes. j
326 Representations and characters of groups
The characters of GL(2,q)
We are now in a position to describe the character table of G.
28.5 Theorem
Label the conjugacy classes of GL(2, q) as in Proposition 28.4, and let
r ! r be the function from F�q2 to C described in (28.3). Then the
irreducible characters of GL(2, q) are given by ëi, øi, øi, j, ÷i as
follows.
sI us ds, t vr
ëi s2i s2i (st)i r i(1�q)
øi qs2i 0 (st)i ÿr i(1�q)
øi, j (q� 1)si� j si� j si t j � s j t i 0÷i (qÿ 1)si ÿsi 0 ÿ(r i � r iq)
Here, we have the following restrictions on the subscripts.
(a) For ëi we have 0 < i < qÿ 2. Thus, there are qÿ 1 characters
ëi, each of degree 1.
(b) For øi we have 0 < i < qÿ 2. Thus, there are qÿ 1 characters
øi, each of degree q.
(c) For øi, j we have 0 < i , j < qÿ 2. Thus, there are
(qÿ 1)(qÿ 2)=2 characters øi, j, each of degree q� 1.
(d) For ÷i, we ®rst consider the set of integers j with
0 < j < q2 ÿ 1 and (q� 1) 6 j j; if j1 and j2 belong to this set
and j1 � j2q mod (q2 ÿ 1) then we choose precisely one of j1
and j2 to belong to the indexing set for the characters ÷i.
Hence, there are (q2 ÿ q)=2 characters ÷i, each of degree qÿ 1.
Before we embark upon the task of calculating the irreducible
characters of G, we present a proposition which will be useful later.
Recall that å is our chosen generator for F�q2 and
vå � 0, 1
ÿå1�q å� åq
� �:
28.6 Proposition
Let K � hvåi. Then jKj � q2 ÿ 1. The group K contains the qÿ 1
scalar matrices sI in G, and of the remaining q2 ÿ q elements of K,
Character table of GL(2, q) 327
precisely two belong to each of the conjugacy classes represented by
vr with r 2 F�q2nF�q .
Proof The eigenvalues of vå are å and åq, so vå has order q2 ÿ 1.
The eigenvalues of viå and of viq
å are å i and å iq. If å i 6� å iq then
å i =2 Fq and viå and viq
å must be conjugate to vå i . Hence two elements
of K belong to the conjugacy class of vå i .
If å i � å iq then viå � å iI , since vi
å has eigenvalues in Fq and viå is
diagonalizable, and this case accounts for the qÿ 1 scalar matrices. j
We shall construct, in turn, the irreducible characters ëi, øi, øi, j and
÷i which appear in Theorem 28.5.
28.7 Proposition
There are qÿ 1 linear characters ëi of G, and they are given in
Theorem 28.5.
Proof The map det : g ! det g is a homomorphism from G onto F�q .
As i varies between 0 and qÿ 2 inclusive, the functions
ëi : g ! (det g)i (g 2 G)
give qÿ 1 distinct linear characters of G, whose values appear in
Theorem 28.5. j
We will see later that the linear characters ëi (0 < i < qÿ 2) which
appear in Proposition 28.7 are all the linear characters of G.
28.8 Proposition
For all integers i, j there is a character øi, j of G whose values on the
conjugacy class representatives, as described in Proposition 28.4, are
as follows.
sI us ds, t vr
øi, j (q� 1)si� j si� j si t j � s j t i 0
328 Representations and characters of groups
Proof Let
B � a b
0 c
� �2 G
� �:
Then B is a subgroup of G with jBj � (qÿ 1)2q. De®ne ëi, j : B! C
by
ëi, j :s r
0 t
� �! si t j:
Then ëi, j is a character of B. We let øi, j � ëi, j " G.
We use Proposition 21.23 to calculate øi, j(g) for each conjugacy
class representative g, as follows.
g � sI : øi, j(g) � jCG(g)jjCB(g)j ëi, j(g)
g � us : øi, j(g) � jCG(g)jjCB(g)j ëi, j(g)
g � ds, t : øi, j(g) � jCG(g)j ëi, j(g)
jCB(g)j �ëi, j(g9)
jCB(g9)j� �
, where g9 � d t,s
g � vr : øi, j(g) � 0:
Hence, the values of øi, j are as stated in the proposition. j
28.9 Proposition
For each integer i, there is an irreducible character øi of G whose
values are given in Theorem 28.5. The characters øi for 0 < i
< qÿ 2 are all different.
Proof We shall demonstrate that the character øi,i which appears in
Proposition 28.8 gives us øi,i � ëi � øi. To this end, we calculate
høi,i, øi,ii and høi,i, ëii. Remember that the complex conjugate of si is
sÿi. We have
høi,i, øi,ii � (q� 1)2
(q2 ÿ 1)(q2 ÿ q)(qÿ 1)� 1
(qÿ 1)q(qÿ 1)
� 4
(qÿ 1)2
(qÿ 1)(qÿ 2)
2
� 2:
Here, the ®rst term corresponds to the conjugacy classes of elements
Character table of GL(2, q) 329
sI , and the calculation of this ®rst term involves the following three
observations.
(1) øi,i(sI)øi,i(sI) � (q� 1)2.
(2) jCG(sI)j � (q2 ÿ 1)(q2 ÿ q):
(3) There are qÿ 1 conjugacy classes with representatives of the
forms sI .
The remaining terms in høi,i, øi,ii are calculated in a similar
fashion.
Next,
høi,i, ëii � (q� 1)
(q2 ÿ 1)(q2 ÿ q)(qÿ 1)� 1
(qÿ 1)q(qÿ 1)
� 2
(qÿ 1)2
(qÿ 1)(qÿ 2)
2
� 1:
The facts that høi,i, ëii � 1 and høi,i, øi,ii � 2 imply that øi,i �ëi � øi for some irreducible character øi. Subtract ëi from øi,i to get
the values of øi as given in Theorem 28.5.
Let s be an element of F�q of order qÿ 1. Then øi : ds,1 ! si.
Hence the characters øi for 0 < i < qÿ 2 are all different. j
28.10 Proposition
Suppose that 0 < i , j < qÿ 2. Then the character øi, j which appears
in Proposition 28.8 is irreducible.
Proof We shall show that høi, j, øi, ji � 1. Using the values of øi, j
which are given in Proposition 28.8, we obtain høi, j, øi, ji �A� B� C, where
A � (q� 1)2
(q2 ÿ 1)(q2 ÿ q)(qÿ 1), B � 1
(qÿ 1)q(qÿ 1) and
C � 1
(qÿ 1)2
1
2
Xs 6� t
(si t j � s j t i)(sÿi tÿ j � sÿ j tÿi):
The coef®cent 12
appears in C because we have just one conjugacy
class for each unordered pair fs, tg of distinct elements of F�q .
To evaluate C, note that fds, t : s, t 2 F�q g is an abelian group of
order (qÿ 1)2, and if ó : ds, t ! si t j � s j t i then ó is a sum of two
330 Representations and characters of groups
inequivalent irreducible characters of this group. Thus, hó , ó i � 2:
That is,
1
(qÿ 1)24(qÿ 1)�
Xs 6� t
(si t j � s j t i)(sÿi tÿ j � sÿ j tÿi)
!� 2:
Hence,
C � qÿ 3
qÿ 1:
And now we ®nd that A� B� C � 1. Therefore, høi, j, øi, ji � 1, and
øi, j is irreducible. j
28.11 Corollary
The characters øi, j for 0 < i , j < qÿ 2 are distinct irreducible
characters of G.
Proof Suppose that 0 < i , j < qÿ 2 and 0 < i9 , j9 < qÿ 2, and
(i, j) 6� (i9, j9). We must prove that øi, j 6� øi9, j9.
Consider the group B and its linear characters ëi, j which were used
in the proof of Proposition 28.8. We have
ëi, j � ë j,i :s b
0 t
� �! si t j � s j t i:
Since ëi, j � ë j,i 6� ëi9, j9 � ë j9,i9, there exists complex (qÿ 1)th roots of
unity s and t such that
either s 6� t and si t j � s j t i 6� si9 t j9 � s j9 t i9
or s � t and si� j 6� si9� j9.
In either case, we see that øi, j differs from øi9, j9 on a conjugacy class
of G. Therefore, øi, j 6� øi9, j9. j
28.12 Proposition
For each integer i, there exists a character öi of G which takes the
following values.
sI us ds, t vr
öi q(qÿ 1)si 0 0 r i � r iq
Character table of GL(2, q) 331
Proof Let K � hvåi, as in Proposition 28.6, and consider the linear
character ái of K which sends the generator vå of K to å i. Suppose
that g 2 K and g is conjugate in G to vr. Then g has eigenvalues r
and rq. Hence ái(g) � r i or r iq and
ái(g)� ái(gq) � r i � r iq:
Let öi � ái " G.
In order to calculate öi, ®rst recall that ái " G is zero on all
elements which are not conjugate to an element of K. Thus, by
Proposition 28.6, öi is zero on the elements of the form us and
ds, t (s 6� t).
If g � sI with s 2 F�q then g 2 K and
öi(g) � jCG(g)jjCK(g)jái(g) � q(qÿ 1)si:
Suppose that r 2 Fq2nFq. Then, by Proposition 28.6, vr is conjugate to
an element of g of K. Also,
öi(g) � jCG(g)j ái(g)
jCK(g)j �ái(gq)
jCK (g)j� �
� ái(g)� ái(gq)
� r i � r iq:
Thus, öi has the values stated in the proposition. j
To be able to work out certain inner products involving our
characters öi, we shall the use the following lemma.
28.13 Lemma
Assume that i is an integer and (q� 1) 6 j i. ThenXr2F
q2nFq
(r i � r iq)(rÿi � rÿiq) � 2(qÿ 1)2:
Proof Note that
G1 � r 0
0 rq
� �: r 2 F�q2
� �and G2 � r 0
0 rq
� �: r 2 F�q
� �are abelian groups of orders q2 ÿ 1 and qÿ 1, respectively. Now,
332 Representations and characters of groups
r 0
0 rq
� �! r i � r iq
gives a character ÷ of degree 2 for each group. For G1, the character ÷is a sum of two inequivalent irreducible characters, since (q� 1) 6 j i
implies that å i 6� å iq; and for G2, the character ÷ is twice an irreducible
character, since rq � r for r 2 F�q . Taking the inner product of the
character ÷ of G1 with itself, we get
1
q2 ÿ 1
Xr2F�
q2
(r i � r iq)(rÿi � rÿiq) � 2
and doing the same for the character ÷ of G2, we get
1
qÿ 1
Xr2F�q
(r i � r iq)(rÿi � rÿiq) � 4:
Hence Xr2F
q2nFq
(r i � r iq)(rÿi � rÿiq) � 2(q2 ÿ 1)ÿ 4(qÿ 1)
� 2(qÿ 1)2: j
28.14 Proposition
For each integer i, let ÷i be the class function on G with the following
values.
sI us ds, t vr
÷i (qÿ 1)si ÿsi 0 ÿ(r i � r iq)
If (q� 1) 6 j i then ÷i is an irreducible character of G.
Proof We can justify the manoeuvre which we now perform only by
saying that it gives the correct answer.
Recall the characters øi, j, øi and öi given in Propositions 28.8, 28.9
and 28.12. Now, ÷i is the class function on G which is given by
÷i � ø0,ÿiøi ÿ ø0,i ÿ öi:
Character table of GL(2, q) 333
The table below allows us to verify this.
sI us ds, t vr
ø0,ÿi (q� 1)sÿi sÿi sÿi � tÿi 0øi qs2i 0 (st)i ÿr i(1�q)
ø0,ÿiøi q(q� 1)si 0 si � t i 0ø0,i (q� 1)si si si � t i 0öi q(qÿ 1)si 0 0 r i � r iq
÷i (qÿ 1)si ÿsÿi 0 ÿ(r i � r iq)
Next, assume that (q� 1) 6 j i. We work out h÷i, ÷ii using Lemma
28.13.
h÷i, ÷ii � (qÿ 1)2
(q2 ÿ 1)(q2 ÿ q)(qÿ 1)� 1
q2 ÿ q(qÿ 1)� (qÿ 1)2
q2 ÿ 1� 1:
Since ÷i is a linear combination of irreducible characters of G, with
integer coef®cients, and h÷i, ÷ii � 1 and ÷i(1) . 0, it follows that ÷i is
an irreducible character of G. j
28.15 Proposition
Suppose that i and j are integers with (q� 1) 6 j i and (q� 1) 6 j j and
j 6� i, iq mod(q2 ÿ 1). Then the characters ÷i and ÷ j of G are different.
Proof Let K � hvåi, as in Proposition 28.6, and consider the linear
character ái of K which sends the generator vå of K to å i. Suppose
that g 2 K.
If g � sI where s 2 F�q then (ái � áiq)(g) � 2si.
If g is conjugate to vr where r 2 Fq2nFq then (ái � áiq)(g) �r i � r iq.
Since j 6� i, iq mod(q2 ÿ 1), the characters ái � áiq and á j � á jq of
K are different, so either si 6� s j for some s 2 F�q or
r i � r iq 6� r j � r jq for some r 2 Fq2nFq. Therefore, ÷i 6� ÷ j, as we
wished to show. j
We have now completed the proof of Theorem 28.5, since we have
shown that the class functions given in the theorem are inequivalent
irreducible characters; and the number of them is q2 ÿ 1, which is the
same as the number of conjugacy classes of G.
It is possible to use the character table of GL(2, q) to ®nd the
334 Representations and characters of groups
character table of SL(2, q), since most of the irreducible characters
remain irreducible when restricted. We do not go fully into this, since
the answers are quite complicated, and they depend upon whether q is
a power of 2 or q � 1 mod 4 or q � 3 mod 4. In Exercise 28.2, though,
you are asked to consider the easiest case, namely that where q is a
power of 2. Since SL(2, q) � PSL(2, q) when q is a power of 2, this
gives the character tables of an in®nite series of simple groups
PSL(2, q).
Among the characters of SL(2, q), those with kernel containing the
centre of SL(2, q) provide the characters of the groups PSL(2, q) ±
compare Chapter 27 ± and so the character table of PSL(2, q) is rather
easier to ®nd than that of SL(2, q). A challenging exercise is to
determine the character table of PSL(2, q) when q � 1 mod 4 or
q � 3 mod 4 from the character table of GL(2, q).
Although the character table of GL(2, q) was ®rst given in 1907, it
was not until the 1950's that the character table of GL(3, q) was found.
Then, in 1955, J. A. Green determined the character table of GL(n, q)
for all positive integers n.
Summary of Chapter 28
The character table of GL(2, q) has the following properties.
(a) There are qÿ 1 conjugacy classes with representatives of the form
sI � s 0
0 s
� �, and there are qÿ 1 irreducible characters of
degree 1.
(b) There are qÿ 1 conjugacy classes with representatives of the form
us � s 1
0 s
� �, and there are qÿ 1 irreducible characters of degree
q.
(c) There are (qÿ 1)(qÿ 2)=2 conjugacy classes with representatives
of the form ds, t � s 0
0 t
� �(s 6� t), and there are (qÿ 1)(qÿ 2)=2
irreducible characters of degree q� 1.
(d) There are (q2 ÿ q)=2 conjugacy classes with representatives of the
Character table of GL(2, q) 335
form vr � 0 1
ÿr1�q r � rq
� �, and there are (q2 ÿ q)=2 irreducible
characters of degree qÿ 1.
Exercises for Chapter 28
1. Use Theorem 28.5 to write down explicitly the character table of
GL(2, 3).
2. Suppose that q is a power of 2. Let Z � fsI : s 2 F�q g. Prove that
GL(2, q) � Z 3 SL(2, q):
Deduce the character table of SL(2, q) from that of GL(2, q). Prove
that if q 6� 2 then SL(2, q) is simple.
3. Use your solution to Exercise 28.2 to write down explicitly the
character table of PSL(2, 8).
336 Representations and characters of groups
337
29
Permutations and characters
We have already seen in Chapter 13 that if G is a permutation group,
i.e. a subgroup of Sn for some n, then G has a permutation character
ð de®ned by ð(g) � jfix(g)j for g 2 G, a fact which proved useful in
many of our subsequent character table calculations. In this chapter we
take the theory of permutation groups and characters somewhat further,
and develop some useful results, particularly about irreducible charac-
ters of symmetric groups (see Theorem 29.12 below).
Group actions
We begin with a more general notion than that of a permutation group.
If Ù is a set, denote by Sym(Ù) the group of all permutations of Ù.
In particular, if Ù � f1, 2, . . . , ng then Sym(Ù) � Sn.
De®nition
Let G be a group and Ù a set. An action of G on Ù is a
homomorphism ö: G! Sym(Ù). We also say that G acts on Ù(via ö).
29.1 Examples
(1) If G < Sn then the identity map is an action of G on f1, . . . , ng.(2) Let G � Sn and let Ù be the set consisting of all pairs fi, jg of
elements of f1, . . . , ng. De®ne ö: G! Sym(Ù) by setting
fi, jg(gö) � fig, jggfor all g 2 Sn and 1 < i , j < n. (So for example, (1 2)ö sends
f1, 3g ! f2, 3g.) Check that ö is an action of Sn; it is called the
action of Sn on pairs.
(3) Let G � GL(2, q), the group of invertible 2 3 2 matrices over the
®nite ®eld Fq, as de®ned in Chapter 28. Let V be the 2-dimensional
vector space over Fq consisting of all row vectors (a, b) with
a, b 2 Fq, and let Ù be the set of all 1-dimensional subspaces hvi of
V. De®ne ö: G! Sym(Ù) by setting
hvi(gö) � hvgifor all hvi 2 Ù and g 2 G. For example, if
g � 1 1
0 1
� �then gö sends
h(a, b)i ! h(a, a� b)i:Then ö is an action of G on Ù.
(4) Let G be a group with a subgroup H of index n, and let Ù be the
set of all right cosets Hx of H in G (so jÙj � n). De®ne ö:
G! Sym(Ù) by
(Hx)(gö) � Hxg
for all x, g 2 G. Then by Exercise 9 of Chapter 23, ö is an action of
G, and Kerö � T x2G xÿ1 Hx < H .
To simplify notation, if ö: G! Sym(Ù) is an action, for ù 2 Ù and
g 2 G we usually just write ùg instead of ù(gö). With this notation,
the fact that ö is a homomorphism simply says that ù(gh) � (ùg)h
for all ù 2 Ù and g, h 2 G.
Adopting this notation, de®ne a relation � on Ù as follows: for
á, â 2 Ù, we have á � â if and only if there exists g 2 G such that
ág � â. It is easy to see that � is an equivalence relation on Ù. The
equivalence classes are called the orbits of G on Ù. Thus Ù is the
disjoint union of the orbits of G. Write
orb(G, Ù)
for the number of orbits of G on Ù. The group G is said to be
transitive on Ù if orb(G, Ù) � 1; in other words, G is transitive if,
given any á, â 2 Ù, there exists g 2 G such that ág � â.
29.2 Examples
(1) Let G � C4, generated by x, say, and let ö: G! S8 be the action
338 Representations and characters of groups
de®ned by xö � (1 2 3 4)(5 6)(7 8) (and of course xkö �((1 2 3 4)(5 6)(7 8))k for any k). Then G has three orbits on
Ù � f1, . . . , 8g, namely f1, 2, 3, 4g, f5, 6g and f7, 8g.(2) The group G is transitive on the set Ù in each of Examples 29.1(2,
3, 4). This is clear in Example (2); to verify it for Example (3) you
need to convince yourself that for any two non-zero row vectors
v, w 2 V there is an invertible 2 3 2 matrix A 2 GL(2, q) such that
vA � w; and in Example (4), simply observe that, given two right
cosets Hx, Hy 2 Ù, the element g � xÿ1 y 2 G has the property that
(Hx)g � Hy.
Let G be a group acting on a set Ù. For ù 2 Ù, write ùG for the
orbit of G which contains ù, so ùG � fùg : g 2 Gg; and de®ne
Gù � fg 2 G : ùg � ùg:We call Gù the stabilizer of ù in G.
29.3 Proposition
The stabilizer Gù is a subgroup of G. Moreover, the size of the orbit
ùG is equal to the index of Gù in G; that is,
jùGj � jG : Gùj:
Proof If g, h 2 Gù then ù(gh) � (ùg)h � ùh � ù, hence gh 2 Gù.
Also gÿ1 2 Gù, and Gù contains the identity, so Gù is a subgroup.
Now let Ä be the set of right cosets Gùx of Gù in G. Observe that
for x, y 2 G,
Gùx � Gù y, xyÿ1 2 Gù , ùxyÿ1 � ù, ùx � ùy:
Hence we can de®ne an injective function ã : Ä! ùG by
ã(Gùx) � ùx
for all x 2 G. Clearly ã is also surjective, and hence jÄj � jùGj, as
required. j
Permutation characters
Let G be a group acting on a ®nite set Ù. Denote by CÙ the vector
space over C for which Ù is a basis. In other words, CÙ consists of
all expressions of the form
Permutations and characters 339
Xù2Ù
ëùù (ëù 2 C)
with the obvious addition and scalar multiplication. As in Chapter 13,
we can make CÙ into a CG-module, called the permutation module,
by de®ning Xëùù
!g �
Xëù(ùg)
for all g 2 G. We see just as in Chapter 13 (p. 129) that if ð is the
character of this permutation module, then for g 2 G,
ð(g) � jfixÙ(g)j,where fixÙ(g) � fù 2 Ù : ùg � ùg. We call ð the permutation char-
acter of G on Ù.
The next result, though elementary, is rather famous, and provides a
basic link between the permutation character and the action of G. It is
often referred to as `̀ Burnside's Lemma'', but is in fact due to Cauchy
and Frobenius.
29.4 Proposition
Let G be a group acting on a ®nite set Ù, and let ð be the
permutation character. Then
hð, 1Gi � 1
jGjXg2G
jfixÙ(g)j � orb(G, Ù):
Proof First note that
hð, 1Gi � 1
jGjXg2G
ð(g) � 1
jGjXg2G
jfixÙ(g)j:
Let Ä1, . . . , Ä t be the orbits of G on Ù, and for each i, pick ùi 2 Äi.
By Proposition 29.3, for 1 < i < t we have
jÄij � jùGi j � jG : Gùi
j:Hence jÄij jGùi
j � jGj. Now de®ne Ö � f(ù, g) : ù 2 Ù, g 2 G,
ùg � ùg. We calculate jÖj in two different ways. First, for each g,
the number of ù 2 Ù such that ùg � ù is equal to jfixÙ(g)j, hence
340 Representations and characters of groups
jÖj �Xg2G
jfixÙ(g)j:
Secondly, for each ù, the number of g 2 G such that ùg � ù is equal
to jGùj, hence
jÖj �Xù2ÙjGùj �
Xt
i�1
jÄij jGùij �
Xt
1
jGj � tjGj:
ThereforeP
g2G jfixÙ(g)j � tjGj, and the conclusion follows. j
29.5 Corollary
G is transitive on Ù if and only if hð, 1Gi � 1.
Now let G be a group, and suppose that G acts on two sets Ù1 and
Ù2, with corresponding permutation characters ð1 and ð2 respectively.
Then we can de®ne an action of G on the Cartesian product Ù1 3 Ù2
by setting
(ù1, ù2)g � (ù1 g, ù2 g)
for all ùi 2 Ùi, g 2 G. It is clear that fixÙ13Ù2(g) �
fixÙ1(g) 3 fixÙ2
(g) for any g 2 G. Hence if ð is the permutation
character of G on Ù1 3 Ù2, then
ð(g) � ð1(g)ð2(g)
for all g 2 G.
29.6 Proposition
Let G act on Ù1 and Ù2, with permutation characters ð1 and ð2
respectively. Then
hð1, ð2i � orb(G, Ù1 3 Ù2):
Proof We have
hð1, ð2i � 1
jGjXg2G
jfixÙ1(g)kfixÙ2
(g)j � 1
jGjXg2G
jfixÙ13Ù2(g)j,
which is equal to orb(G, Ù1 3 Ù2) by Proposition 29.4. j
In the rest of the chapter we apply Proposition 29.6 in a number of
situations, the ®rst being the case where Ù1 � Ù2.
Permutations and characters 341
Suppose G acts on Ù. Then G also acts on Ù 3 Ù in the way
de®ned above, namely (ù1, ù2)g � (ù1 g, ù2 g) for all ù1, ù2 2Ù, g 2 G.
29.7 De®nition
The number of orbits of G on Ù 3 Ù is called the rank of G on Ù,
written r(G, Ù). Thus
r(G, Ù) � orb(G, Ù 3 Ù):
The next result is immediate from Proposition 29.6.
29.8 Proposition
Let G act on Ù, with permutation character ð. Then
r(G, Ù) � hð, ði:
Now suppose G is transitive on Ù and jÙj. 1. Then
Ä � f(ù, ù) : ù 2 Ùgis an orbit of G on Ù 3 Ù, and hence certainly r(G, Ù) > 2. The case
where equality holds is of particular interest.
29.9 De®nition
Let G be transitive on Ù. Then G is said to be 2-transitive on Ù if
r(G, Ù) � 2.
In other words, G is 2-transitive if, for any ordered pairs (á1, á2)
and (â1, â2) in Ù 3 Ù, with á1 6� á2, â1 6� â2, there exists g 2 G such
that á1 g � â1 and á2 g � â2.
29.10 Corollary
If G is 2-transitive on Ù, with permutation character ð, then
ð � 1G � ÷,
where ÷ is an irreducible character of G.
Proof We have hð, 1Gi � 1 by Corollary 29.5, and hð, ði � 2 by
Proposition 29.8. The result follows, using Theorem 14.17. j
342 Representations and characters of groups
29.11 Examples
(1) The symmetric group Sn is 2-transitive on f1, . . . , ng. Also An is
2-transitive, provided n > 4. Hence these groups have an irreducible
character ÷ given by
÷(g) � jfix(g)j ÿ 1:
We have seen this irreducible character in a number of previous
examples (see 18.1, 19.16, 19.17).
(2) Consider the action of G � GL(2, q) given in Example 29.1(3).
Here Ù is the set of all 1-dimensional subspaces of the 2-dimensional
vector space V. We claim that G is 2-transitive on Ù. To see this, let
(hv1i, hv2i) and (hw1i, hw2i) be two pairs of distinct 1-spaces in Ù.
Then v1, v2 and w1, w2 are both bases of V. The linear transformation
from V to V which sends v1 ! w1, v2 ! w2 is therefore invertible,
giving an element of GL(2, q) which sends hv1i ! hw1i, hv2i ! hw2i.Hence G is 2-transitive on Ù, as claimed. Since jÙj � q� 1 (see
Exercise 1 at the end of the chapter), the irreducible character ÷ of G
given by Corollary 29.10 is the character ø0 of degree q in Theorem
28.5.
(3) Consider the action of Sn on pairs de®ned in Example 29.1(2), with
n > 4. This action is not 2-transitive, since, for example, there is no
element of Sn which sends (f1, 2g, f3, 4g) to (f1, 2g, f2, 3g). In fact
it is easy to see that the orbits of G � Sn on Ù 3 Ù are Ä, Ä1 and
Ä2, where Ä is as above, and
Ä1 � f(fi, jg, fk, lg) : jfi, jg \ fk, lgj � 1g,Ä2 � f(fi, jg, fk, lg) : jfi, jg \ fk, lgj � 0g:
Thus hð, ði � r(G, Ù) � 3, and so ð � 1G � ÷� ø, where ÷ and øare irreducible characters of Sn.
Some irreducible characters of Sn
By Theorem 12.15 we know that the conjugacy classes of Sn are in
bijective correspondence with the set of all possible cycle-shapes of
permutations. Each cycle-shape (including 1-cycles) is a sequence
ë � (ë1, . . . , ës) of positive integers ëi such that ë1 > ë2 > . . . > ës
and ë1 � . . . � ës � n, and we call such a sequence a partition of n.
By Theorem 15.3, the irreducible characters of Sn are also in
bijective correspondence with the partitions ë of n. A key aim is
therefore to construct, for each partition ë, an irreducible character ÷ë
Permutations and characters 343
of Sn, in a natural way. We shall apply the material of this chapter to
carry out this aim in the case where ë � (nÿ k, k), a 2-part partition
(see Theorem 29.13 below). The ideas can be developed to carry out
the aim in general, but we do not do this; if you want to see this, and
much more, on the character theory of Sn, we refer you to the book by
G. James listed in the Bibliography.
Let G � Sn and I � f1, 2, . . . , ng. For an integer k < n=2, de®ne
Ik to be the set consisting of all subsets of I of size k. Just as in
Example 29.1(2) we can de®ne an action of G on Ik as follows: for
any subset A � fi1, . . . , ikg 2 Ik and any g 2 G, let
Ag � fi1 g, . . . , ikgg:Let ðk be the permutation character of G in its action on Ik . Observe
that
ðk(1) � jIk j � n
k
� �:
29.12 Proposition
If l < k < n=2, then hðk , ð li � l � 1.
Proof By Proposition 29.6, hðk , ð li � orb(G, Ik 3 Il). The orbits of
G � Sn on Ik 3 Il are easily seen to be J0, J1, . . . Jl, where for
0 < s < l,
Js � f(A, B) 2 Ik 3 Il : jA \ Bj � sg:Hence orb(G, Ik 3 Il) � l � 1, giving the conclusion. j
29.13 Theorem
Let m � n=2 if n is even, and m � (nÿ 1)=2 if n is odd. Then Sn has
distinct irreducible characters ÷(n) � 1G, ÷(nÿ1,1), ÷(nÿ2,2), . . . , ÷(nÿm,m)
such that for all k < m,
ðk � ÷(n) � ÷(nÿ1,1) � . . . � ÷(nÿk,k):
In particular, ÷(nÿk,k) � ðk ÿ ðkÿ1.
Proof We prove the existence of irreducible characters ÷(n),
÷(nÿ1,1), . . . , ÷(nÿk,k) such that ðk � ÷(n) � ÷(nÿ1,1) � . . . � ÷(nÿk,k), by
induction on k. This holds for k � 1 by Corollary 29.10.
Now assume the statement holds for all values less than k. Then
344 Representations and characters of groups
there exist irreducible characters ÷(n), ÷(nÿ1,1), . . . , ÷(nÿk�1,kÿ1) such
that
ði � ÷(n) � ÷(nÿ1,1) � . . . � ÷(nÿi,i)
for all i , k. Now by Proposition 29.12,
hðk , 1Gi � 1, hðk , ð1i � 2, . . . , hðk , ðkÿ1i � k, hðk , ðki � k � 1:
It follows that ðk � ðkÿ1 � ÷ for some irreducible character ÷. Writing
÷ � ÷(nÿk,k), we have ðk � ÷(n) � ÷(nÿ1,1) � . . . � ÷(nÿk,k), as re-
quired. j
29.14 Examples
(1) The formula ÷(nÿk,k) � ðk ÿ ðkÿ1 makes it easy to calculate the
values of the characters ÷(nÿk,k). For example, the degree is
÷(nÿk,k)(1) � ðk(1)ÿ ðkÿ1(1) � n
k
� �ÿ n
k ÿ 1
� �:
As another example, suppose n � 7 and let us calculate the value of
the irreducible character ÷(5,2) on a 3-cycle:
÷(5,2)(123) � ð2(123)ÿ ð1(123) � jfix I2(123)j ÿ jfix I1
(123)j � 6ÿ 4 � 2:
(2) In the character table of S6 given in Example 19.17, the irreducible
characters ÷1, ÷3, ÷7, ÷9 are equal to ÷(6), ÷(5,1), ÷(4,2), ÷(3,3), respectively.
Summary of Chapter 29
1. An action of G on Ù is a homomorphism G! Sym(Ù). The orbits
are the equivalence classes in Ù of the relation de®ned by
á � â, ág � â for some g 2 G. The size of the orbit ùG contain-
ing ù is jùGj � jG : Gùj.2. If G acts on Ù then CÙ is the permutation module, and the
corresponding character of G is ð, where ð(g) � jfixÙ(g)j. The
number of orbits is equal to hð, 1Gi.3. The rank r(G, Ù) is the number of orbits of G on Ù 3 Ù, and
r(G, Ù) � hð, ði. If G is 2-transitive then r(G, Ù) � 2 and ð �1G � ÷ with ÷ irreducible.
4. The irreducible characters of Sn are in bijective correspondence with
partitions of n. The irreducible characters ÷(nÿk,k) corresponding to
2-part partitions have values given by Theorem 29.13.
Permutations and characters 345
Exercises for Chapter 29
1. Let G be a ®nite group, and de®ne a function ö : G 3 G !Sym(G) by x((g, h)ö) � gÿ1xh for all x, g, h 2 G.
(a) Show that ö is an action of G 3 G on G, which is transitive.
(b) Find the stabilizer (G 3 G)1 of the identity 1 2 G, and ®nd the
kernel of ö.
(c) Show that the rank of this action r(G 3 G, G) is equal to the
number of conjugacy classes of G, and the permutation char-
acter ð is
ð �X÷
÷ 3 ÷,
where the sum is over all irreducible characters ÷ of G, and
÷ 3 ÷ is the irreducible character of G 3 G given by Theorem
19.18.
2. Show that if Ù is the set of all 1-dimensional subspaces of a 2-
dimensional vector space over Fq (as in Example 29.1(3)), then
jÙj � q� 1.
3. Let G � GL(2, q) and let V � F2q as in Example 29.1(2). Let
V� � V ÿ f0g, and de®ne an action ö : G! Sym(V�) by
v(gö) � vg for v 2 V�, g 2 G. Let ð be the permutation character
of G in this action.
Decompose ð as a sum of irreducible characters of GL(2, q) (the
latter are given by Theorem 28.5).
(Hint: one way to do this is to write down the values of ð on the
conjugacy classes of G, and take inner products with the irreducible
characters of G given in 28.5.)
4. Let G be a ®nite group, and let H1, H2 be subgroups of G. For
i � 1, 2 de®ne Ùi to be the set of right cosets of Hi in G, so that
G acts on Ùi as in Example 29.1(4); let ði be the permutation
character of G in the action on Ùi. Suppose that ð1 � ð2.
Prove that if G is abelian, then H1 � H2:
Give an example to show that this need not be the case in
general.
5. Let G be a ®nite group acting transitively on a set Ù of size greater
than 1. Prove that G contains an element g such that jfixÙ(g)j � 0.
(Such an element is called a ®xed-point-free element of G.)
346 Representations and characters of groups
6. Let n be a positive integer, and let Ù be the set of all ordered pairs
(i, j) with i, j 2 f1, . . . , ng and i 6� j. Let Sn act on Ù in the
obvious way (namely, (i, j)g � (ig, jg) for g 2 Sn), and let the
permutation character of Sn in this action be ð(nÿ2,1,1).
By considering inner products as in the proof of Theorem 29.13,
prove that
ð(nÿ2,1,1) � 1� 2÷(nÿ1,1) � ÷(nÿ2,2) � ÷,
where ÷ is an irreducible character.
Writing ÷ � ÷(nÿ2,1,1), calculate the degree of ÷(nÿ2,1,1), and calcu-
late its value on the elements (12) and (123) of Sn.
In the character table of S6 given in Example 19.17, which
irreducible character is equal to ÷(4,1,1)?
Permutations and characters 347
348
30
Applications to group theory
There are several ways of using the character theory of a group to
determine information about the structure of the group. The examples
which we have come across so far ± ®nding the centre of the group,
seeing whether or not the group is simple, and so on ± require little
calculation. In this chapter we present some rather deeper applications.
The ®rst involves doing arithmetic with character values to determine
certain numbers, known as the class algebra constants. These constants
carry information about the multiplication in G, and they can be used
to investigate the subgroup structure of G, as we shall demonstrate.
The second application takes this much further: the Brauer±Fowler
Theorem 23.19 motivates the study of simple groups containing an
involution with centralizer isomorphic to a given group C. Using a
little group theory and a lot of character theory we shall carry out such
a study in the case where C � D8, the dihedral group of order 8.
Class algebra constants
Let G be a ®nite group and let C1, . . . , Cl be the distinct conjugacy
classes of G. Recall from Proposition 12.22 that the class sums
C1, . . . , Cl form a basis for the centre of the group algebra CG
(where Ci �P
g2Cig).
30.1 Proposition
There exist non-negative integers aijk such that for 1 < i < l and
1 < j < l,
CiC j �Xl
k�1
aijk Ck :
Proof For g 2 Ck the coef®cient of g in the product CiC j is equal to
the number of pairs (a, b) with a 2 Ci, b 2 Cj and ab � g. This
number is a non-negative integer, and is independent of the chosen
element g of Ck . The result follows. j
Another way of looking at Proposition 30.1 is to note that CiC j
belongs to Z(CG), so it must be a linear combination of C1, . . . , Cl.
30.2 De®nition
The integers aijk in the formula
CiC j �Xl
k�1
aijk Ck
are the class algebra constants of G.
From their very de®nition, the numbers aijk carry information about
the multiplication in G:
(30:3) For all g 2 Ck and all i, j we have
aijk � the number of pairs (a, b) with a 2 Ci, b 2 Cj and ab � g:
Also, the constants aijk determine the product of any two elements in
the centre Z(CG) of the group algebra, since C1, . . . , Cl is a basis of
Z(CG). As the centre of the group algebra plays an important role in
representation theory, you might suspect that the class algebra constants
are determined by the character table of G. Our next theorem shows
that this is indeed the case.
30.4 Theorem
Let gi 2 Ci for 1 < i < l. Then for all i, j, k, we have
aijk � jGjjCG(gi)j jCG(gj)j
X÷
÷(gi)÷(gj)÷(gk)
÷(1)
where the sum is over all the irreducible characters ÷ of G.
Applications to group theory 349
Proof Let ÷ be an irreducible character of G, and let U be a CG-
module with character ÷. Then by Lemma 22.7, for all u 2 U we have
uCi � jGj÷(gi)
jCG(gi)j÷(1)u:
Therefore
uCiC j � jGj2jCG(gi)j jCG(gj)j
÷(gi)÷(gj)
(÷(1))2u
and
Xl
m�1
aijmuCm �Xl
m�1
aijm
jGj÷(gm)
jCG(gm)j÷(1)u:
Since CiC j �P
maijmCm, we deduce that
Xl
m�1
aijm
÷(gm)
jCG(gm)j �jGj
jCG(gi)j jCG(gj)j÷(gi)÷(gj)
÷(1):(30:5)
Pick k with 1 < k < l. Multiply both sides of equation (30.5) by ÷(gk)
and sum over all irreducible characters ÷ of G, to obtain
Xl
m�1
aijm
X÷
÷(gm)÷(gk)
jCG(gm)j �jGj
jCG(gi)j jCG(gj)jX÷
÷(gi)÷(gj)÷(gk)
÷(1):
By the column orthogonality relations, Theorem 16.4(2), this yields
aijk � jGjjCG(gi)j jCG(gj)j
X÷
÷(gi)÷(gj)÷(gk)
÷(1):
j
Examples
30.6 Example
In this example we shall use the class algebra constants to prove some
facts about the elements and subgroups of the symmetric group S4;
these results can readily be proved directly, but they serve as a useful
illustration of the method.
Let G � S4. By Section 18.1, the character table of G is as shown:
350 Representations and characters of groups
(1) We use Theorem 30.4 to calculate the class algebra constant
a555:
a555 � 24
4:4
1
1�ÿ1
1� 0
2�ÿ1
3� 1
3
� �� 0:
Hence, by (30.3), S4 does not possess elements a, b of order 4 such
that the product ab also has order 4. We deduce from this that S4 does
not have a subgroup which is isomorphic to the quaternion group Q8:
for Q8 does have two elements of order 4 with product of order 4.
(2) By Theorem 30.4,
a245 � 24
4:81� 1� 1
3� 1
3
� �� 2:
Hence S4 has elements a, b of order 2 such that ab has order 4.
Writing x � ab, we have
x4 � 1, aÿ1xa � ba � (ab)ÿ1 � xÿ1,
so ka, bl � D8. We deduce the fact (which we already know from
Exercise 18.1) that S4 has a subgroup which is isomorphic to D8.
(3) Finally,
a235 � 24
4:3(1� 1) � 4,
so S4 has elements a of order 2 and b of order 3 with ab of order 4.
In fact, it can be shown that S4 has a presentation as follows:
S4 � ha, b: a2 � b3 � (ab)4 � 1i:In other words, S4 is generated by a and b, and all products of
elements of S4 are determined by the given relations. We supply a
Character table of S4
Class Ci C1 C2 C3 C4 C5
gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4
÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0÷4 3 1 0 ÿ1 ÿ1÷5 3 ÿ1 0 ÿ1 1
Applications to group theory 351
proof in the solution to Exercise 30.6 ± in the meantime, you may
wish to puzzle out the relevance of the ®gure above.
30.7 Example
We use Theorem 30.4 to ®nd a subgroup H of the simple group
PSL (2, 7) with H isomorphic to S4. That such a subgroup exists is not
obvious, and it is quite tricky to construct directly.
We found in Chapter 27 that the character table of G � PSL (2, 7) is
as follows.
Character table of PSL (2, 7)
Class rep. gi g1 g2 g3 g4 g5 g6
Order of gi 1 2 4 3 7 7|CG(gi)| 168 8 4 3 7 7
÷1 1 1 1 1 1 1÷2 7 ÿ1 ÿ1 1 0 0÷3 8 0 0 ÿ1 1 1÷4 3 ÿ1 1 0 á á÷5 3 ÿ1 1 0 á á÷6 6 2 0 0 ÿ1 ÿ1
where á � (ÿ1� ip
7)=2.
352 Representations and characters of groups
We calculate the class algebra constant a243. By Theorem 30.4,
a243 � 168
8:31� 1
7� 0� 0� 0� 0
� �� 8:
Hence, by (30.3), G contains elements x and y such that x has order
2, y has order 3 and xy has order 4. Let H be the subgroup kx, yl of
G. From Example 30.6, we know that
S4 � ha, b: a2 � b3 � (ab)4 � 1i:Hence there is a homomorphism ö from S4 onto H (ö sends a to x
and b to y). By Theorem 1.10, S4=Kerö � H . Now Kerö, being a
normal subgroup of S4, is {1}, V4, A4 or S4 (see Example 12.20), so
H is isomorphic to S4, S3, C2 or {1}. Since H has an element of order
4, namely xy, we conclude that
H � S4:
Thus we have shown that PSL (2, 7) has a subgroup which is iso-
morphic to S4.
The Brauer programme
The Brauer±Fowler Theorem 23.19 states that there are only ®nitely
many non-isomorphic ®nite simple groups containing an involution with
a given centralizer. This fact led Brauer to initiate a programme to
®nd, given a ®nite group C, all ®nite simple groups G possessing an
involution t such that CG(t) � C. This programme formed an important
part of the effort of many mathematicians to classify all the ®nite
simple groups, an effort which was ®nally completed in the early
1980s (see the book by D. Gorenstein listed in the Bibliography).
The next result carries out part of Brauer's programme in the case
where C � D8, a dihedral group of order 8. It determines the possible
orders of simple groups G having an involution t such that
CG(t) � D8. We have chosen to present this result because it provides
a wonderful illustration of the use of character theory in the service of
group theory.
30.8 Theorem
Let G be a ®nite non-abelian simple group which has an involution t
such that CG(t) � D8. Then G has order 168 or 360.
Applications to group theory 353
Observe that PSL(2, 7) is a simple group of order 168 having an
involution with centralizer D8 (see Lemma 27.1); and A6 is a simple
group of order 360 with this property (see Exercise 7 at the end of the
chapter). Using some rather more sophisticated group theory than that
covered in this book, one can show that PSL(2, 7) and A6 are the only
simple groups of order 168 or 360.
Before embarking upon the proof of Theorem 30.8, we require a
couple of preliminary results. The ®rst is Sylow's Theorem, a basic
result in ®nite group theory. We shall not prove this, but refer you to
Theorems 18.3 and 18.4 of the book by J. Fraleigh listed in the
Bibliography.
30.9 Sylow's Theorem
Let p be a prime number, and let G be a ®nite group of order pab,
where a, b are positive integers and p 6 j b. Then
(1) G contains a subgroup of order pa; such a subgroup is call a
Sylow p-subgroup of G;
(2) all Sylow p-subgroups are conjugate in G (i.e. if P, Q are Sylow p-
subgroups, then there exists g 2 G such that Q � gÿ1 Pg);
(3) if R is a subgroup of G with jRj � pc for some c, then there is a
Sylow p-subgroup of G containing R.
30.10 Lemma
Let G be a ®nite non-abelian simple group, and let P be a Sylow 2-
subgroup of G. Suppose Q is a subgroup of P with jP : Qj � 2. If u is
an involution in G, then u is conjugate to an element of Q.
Proof Suppose u is not conjugate to an element of Q. Let Ù be the
set of right cosets Qx of Q in G, and de®ne an action of G on Ù by
(Qx)g � Qxg for x, g 2 G (see Example 29.2(4)). Observe that
jÙj � 2jG : Pj � 2m, where m is odd since P is a Sylow 2-subgroup.
Now consider fixÙ(u) � fù 2 Ù : ùu � ùg If Qx 2 fixÙ(u), then
Qxu � Qx and hence xuxÿ1 2 Q, contrary to assumption. Hence
fixÙ(u) � Æ. This means that in its action on Ù, the involution u is a
product of m disjoint 2-cycles, hence is an odd permutation. Hence the
subgroup
fg 2 G : g acts as an even permutation on Ùg
354 Representations and characters of groups
is a normal subgroup of index 2 in G. This is impossible since G is
non-abelian and simple. This contradiction completes the proof. j
We also need to introduce the idea of a generalized character of a
group H . This is simply a class function of the form
ø �X
x
n÷÷
where the sum is over all the irreducible characters of H , and each
n÷ 2 Z. If n÷ > 0 for all ÷ then of course ø is a character, but this
need not be the case for a generalized character. In particular, the
degree ø(1) can be 0 or negative for a generalized character ø. Notice
also that the orthogonality relations give the usual inner products
hø, ÷i � n÷, hø, øi �X
n2÷
for a generalized character ø as above.
The generalized character ø can be expressed as a difference áÿ â,
where á and â are characters of H : take
á �Xn÷>0
n÷÷, â � ÿXn÷,0
n÷÷:
Finally, if H is a subgroup of a group G, we de®ne the induced
generalized character ø " G by
ø " G � (á " G)ÿ (â " G)
where ø � áÿ â as above. It is clear from this de®nition that the
formulae for the values of ø " G given in Proposition 21.19 and
Corollary 21.20 hold for generalized characters ø.
Proof of Theorem 30.8
Let G be a ®nite non-abelian simple group with an involution t such
that CG(t) � D � D8. Certainly t commutes with itself, so t 2 D; and
as t commutes with all elements of D, we have t 2 Z(D), the centre
of D. The centre of D8 is a cyclic group of order 2 (see (12.12)), and
hence Z(D) � hti.By Theorem 30.9(3), there is a Sylow 2-subgroup P of G such that
D < P. Then Z(P) < CG(t) � D, so Z(P) < Z(D) � hti. By Lemma
26.1(1) we have Z(P) 6� 1, and hence Z(P) � hti. Therefore P <
Applications to group theory 355
CG(t) � D, and so P � D. In other words, D is a Sylow 2-subgroup
of G.
Write D � ha, bi where a4 � b2 � 1 and bÿ1ab � aÿ1. Then t � a2.
Let C � hai be the cyclic subgroup of index 2 in D. By Lemma
30.10, every involution of G is conjugate to an involution in C. As
t � a2 is the only such involution, we conclude that tG is the unique
conjugacy class of involutions in G.
Next, let g 2 G and suppose that gÿ1cg 2 C for some non-identity
element c 2 C. Since t � c or c2, we must have gÿ1 tg � t, hence
g 2 CG(t) � D and so gÿ1Cg � C.
We summarise what we have proved so far:
(30.11) D is a Sylow 2-subgroup of G; tG is the unique
conjugacy class of involutions in G; for any g 2 G we
have C \ gÿ1Cg � f1g or C; and if C \ gÿ1Cg � C
then g 2 D.
This is all the group theory we will need for the proof. The rest is
character theory.
Let ë be the linear character of C such that ë(a) � i, and de®ne
è � (1C " D)ÿ (ë " D),
a generalized character of D. Then è takes the value 2 on a, aÿ1, the
value 4 on t, and 0 elsewhere. Referring to the character table of D8
in Example 16.3(3), we have è � ÷1 � ÷2 ÿ ÷5. (In particular,
è(1) � 0.) Hence hè, èi � 3.
We next establish
hè " G, è " Gi � 3:(30:12)
To see this, observe ®rst that by Frobenius reciprocity, hè " G,
è " Gi � h(è " G) # D, èi Now for 1 6� c 2 C, Proposition 21.19 gives
(è " G)(c) � 1
8
Xy2G
_è(yÿ1cy):
By (30.11), if yÿ1cy 2 C then y 2 D, whence yÿ1cy � c�1 and
è(yÿ1cy) � è(c). And if yÿ1cy 2 Dÿ C then è(yÿ1cy) � 0. It follows
that (è " G)(c) � è(c). Since è vanishes on Dÿ C, we therefore have
h(è " G) # D, èi � hè, èi � 3, giving (30.12).
Now hè " G, 1Gi � h1C ÿ ë, 1Ci � 1. Also (è " G)(1) � 0 (see Corol-
lary 21.20), and so it follows from (30.12) that
356 Representations and characters of groups
è " G � 1G � áÿ â,
where á, â are irreducible characters of G. Since we have shown that
(è " G)(t) � è(t) � 4, we have now proved the following.
(30.13) We have è " G � 1G � áÿ â, where á, â are irreduci-
ble, 1� á(1)ÿ â(1) � 0 and 1� á(t)ÿ â(t) � 4.
Note that by Corollary 13.10, á(t) and â(t) are integers.
We now introduce another class function of G into the picture. For
g 2 G, de®ne ã(g) to be number of ordered pairs (x, y) 2 tG 3 tG
such that g � xy. If we write tG � Ci and g lies in the conjugacy
class Ck of G, then ã(g) � aiik in the notation of (30.3). Hence
Theorem 30.4 yields the following.
(30.14) We have
ã � jGjjDj2X÷
÷(t)2
÷(1)÷,
where the sum is over all irreducible characters ÷ of G.
We shall calculate the inner product of ã and è " G in two ways.
First, from (30.13) and (30.14) we have
hè " G, ãi � jGj64
1� á(t)2
á(1)ÿ â(t)2
â(1)
!:(30:15)
On the other hand, by Frobenius Reciprocity, hè " G, ãi �h1C ÿ ë, ã # Ci. Consider ã(c) for 1 6� c 2 C. If c � xy with x, y 2 tG,
then xÿ1cx � yx � cÿ1, and hence x 2 D by (30.11); similarly y 2 D.
Now calculation in D8 shows that ã(c) � 4. Therefore
h1C ÿ ë, ã # Ci � 1
jCj :4:((1ÿ i)� 2� (1� i)) � 4:
Hence from (30.15) we deduce
jGj 1� á(t)2
á(1)ÿ â(t)2
â(1)
!� 28:(30:16)
This equation gives us enough number-theoretic information about
jGj to ®nish the proof fairly quickly. Write d � á(1) and e � á(t) 2 Z.
By (30.13) we have
Applications to group theory 357
â(1) � d � 1, â(t) � eÿ 3:
From the column orthogonality relations 16.4(2), we have
8 � jCG(t)j > 1� á(t)2 � â(t)2 � 1� e2 � (eÿ 3)2,
from which it follows that e � 1 or 2.
Suppose now that e � 1. Then (30.16) gives
jGj 1� 1
dÿ 4
d � 1
� �� 28,
whence
jGj � 28 d(d � 1)
(d ÿ 1)2:
Now the highest common factor hcf(d ÿ 1, d � 1) is 1 or 2, and
hcf(d ÿ 1, d) � 1. Hence (d ÿ 1)2 must divide 210, and so d ÿ 1 � 2r
with r < 5. Moreover, a Sylow 2-subgroup of G has order 8. It follows
that r � 3 and d � 9, giving jGj � 360, one of the possibilities in the
conclusion of Theorem 30.8.
Finally, suppose that e � 2. Then (30.16) yields
jGj � 28 d(d � 1)
(d � 2)2:
Reasoning as above, we deduce that d � 2 � 23, giving d � 6 and
jGj � 168. This completes the proof of Theorem 30.8. j
Summary of Chapter 30
1. The class algebra constants aijk are given by
CiC j �X
k
aijk Ck :
They can be calculated from the character table, by using the
formula
aijk � jGjjCG(gi)j jCG(gj)j
X÷
÷(gi)÷(gj)÷(gk)
÷(1):
2. Given groups G and H, the class algebra constants of G can
sometimes be used to determine whether or not G has a subgroup
which isomorphic to H.
358 Representations and characters of groups
3. Using Sylow's Theorem, together with lots of ingenious character
theory, it can be shown that any simple group possessing an
involution with centralizer isomorphic to D8 must have order 168 or
360.
Exercises for Chapter 30
1. Use the character table of PSL (2, 7), given at the end of Chapter
27, to prove that PSL (2, 7) contains elements a and b such that a
has order 2, b has order 3 and ab has order 7.
2. Does PSL (2, 7) contain a subgroup isomorphic to D14?
(Hint: D14 � ka, b: a2 � b2 � 1, (ab)7 � 1l.)
For the next three exercises, you may assume that A5 is a simple
group, and that A5 has the following presentation:
A5 � ha, b: a2 � b3 � (ab)5 � 1i:3. The character table of PSL (2, 11) is given in the solution to
Exercise 27.6. Does PSL (2, 11) contain a subgroup which is iso-
morphic to A5?
4. Prove that A5 is characterized by its character table ± that is, if G
is a group with the same character table as A5 (see Example 20.13),
then G � A5.
5. Suppose that G is a group, and that G has the character table
shown.
(a) Show that G is a simple group of order 360.
(b) Use the Frobenius±Schur Count of Involutions to obtain an
g1 g2 g3 g4 g5 g6 g7
÷1 1 1 1 1 1 1 1÷2 5 1 ÿ1 2 ÿ1 0 0÷3 5 1 ÿ1 ÿ1 2 0 0÷4 8 0 0 ÿ1 ÿ1 á â÷5 8 0 0 ÿ1 ÿ1 â á÷6 9 1 1 0 0 ÿ1 ÿ1÷7 10 ÿ2 0 1 1 0 0
where á � (1�p5)=2, â � (1ÿp5)=2.
Applications to group theory 359
upper bound for the number of involutions in G, and deduce
that g2 has order 2 and g3 has order 4.
(c) Prove that G has a subgroup H which is isomorphic to A5.
(d) Using Exercise 23.9, show that G � A6.
6. Use the ®gure which appears in Example 30.6(3) to show that every
group G which is generated by two elements a and b which satisfy
a2 � b3 � (ab)4 � 1
has order at most 24.
7. Prove that PSL(2, 7) and A6 are simple groups of order 168, 360
respectively, both of which contain an involution with centralizer
isomorphic to D8.
8. Find a simple group G having an involution t such that
CG(t) � D16.
(Hint: look for a suitable simple group PSL(2, p).)
360 Representations and characters of groups
361
31
Burnside's Theorem
One of the most famous applications of representation theory is
Burnside's Theorem, which states that if p and q are prime numbers
and a and b are positive integers, then no group of order paqb is
simple. In the ®rst edition of his book Theory of groups of ®nite order
(1897), Burnside presented group-theoretic arguments which proved the
theorem for many special choices of the integers a, b, but it was only
after studying Frobenius's new theory of group representations that he
was able to prove the theorem in general. Indeed, many later attempts
to ®nd a proof which does not use representation theory were unsuc-
cessful, until H. Bender found one in 1972.
A preliminary lemma
We prepare for the proof of Burnside's Theorem with a lemma (31.2)
which is concerned with character values. In order to establish this
lemma we require some basic facts about algebraic integers and
algebraic numbers, which we now describe. We omit proofs of these ±
for a good account, see for instance the book by Pollard and Diamond
listed in the Bibliography.
An algebraic number is a complex number which is a root of some
non-zero polynomial over Q. We call a polynomial in x monic if the
coef®cient of the highest power of x in it is 1.
Let á be an algebraic number; and let p(x) be a monic polynomial
over Q of smallest possible degree having á as a root. Then p(x) is
unique and irreducible; it is called the minimal polynomial of á. The
roots of p(x) are called the conjugates of á.
For example, if ù is an nth root of unity then the minimal poly-
nomial of ù divides xn ÿ 1, and so every conjugate of ù is also an
nth root of unity.
If á is an algebraic integer, then á is a root of a monic polynomial
with integer coef®cients (see Chapter 22), and it turns out that the
minimal polynomial of á also has integer coef®cients.
We shall require the following fact about conjugates:
(31.1) Let á and â be algebraic numbers. Then every conju-
gate of á � â is of the form á9 � â9, where á9 is a
conjugate of á and â9 is a conjugate of â. Moreover, if
r 2 Q then every conjugate of rá is of the form rá9,
where á9 is a conjugate of á.
For an elementary proof of this, see Pollard and Diamond, Chapter
V, Section 3. Alternatively, (31.1) can be proved easily using some
Galois theory.
31.2 Lemma
Let ÷ be a character of a ®nite group G, and let g 2 G. Then
j÷(g)=÷(1)j < 1, and if
0 , j÷(g)=÷(1)j, 1
then ÷(g)=÷(1) is not an algebraic integer.
Proof Let ÷(1) � d. By Proposition 13.9 we have ÷(g) � ù1 �. . . � ùd , where each ùi is a root of unity, so
÷(g)=÷(1) � (ù1 � : : : � ùd)=d:
Since |÷(g)| � |ù1 � . . . � ùd | < |ù1| � . . . � |ùd | � d, it follows that
j÷(g)=÷(1)j < 1.
Now suppose that ÷(g)=÷(1) is an algebraic integer and
j÷(g)=÷(1)j, 1. We prove that ÷(g) � 0.
Write ã � ÷(g)=÷(1), and let p(x) be the minimal polynomial of ã,
so that
p(x) � xn � anÿ1x nÿ1 � : : : � a1x� a0
where ai 2 Z for all i. By (31.1), each conjugate of ã is of the form
(ù91 � : : :� ù9d)=d
where ù91, . . . , ù9d are roots of unity. Hence each conjugate of ã has
362 Representations and characters of groups
modulus at most 1. It follows that if ë is the product of all the
conjugates of ã (including ã), then
jëj, 1:
But the conjugates of ã are, by de®nition, the roots of the polynomial
p(x), and the product of all these roots is equal to �a0. Thus
ë � �a0:
Since a0 2 Z and |ë| , 1, it follows that a0 � 0. As p(x) is irreducible,
this implies that
p(x) � x,
which in turn forces ã � 0. Thus ÷(g) � 0, and the proof is complete.
j
Burnside's paqb Theorem
We deduce the main result, Theorem 31.4, from another interesting
theorem of Burnside.
31.3 Theorem
Let p be a prime number and let r be an integer with r > 1. Suppose
that G is a ®nite group with a conjugacy class of size pr. Then G is
not simple.
Proof Let g 2 G with |gG | � pr. Since pr . 1, G is not abelian and
g 6� 1. As usual, denote the irreducible characters of G by ÷1, . . . , ÷k ,
and take ÷1 to be the trivial character.
The column orthogonality relations, Theorem 16.4(2), applied to the
columns corresponding to 1 and g in the character table of G, give
1�Xk
i�2
÷i(g)÷i(1) � 0:
Therefore Xk
i�2
÷i(g) .÷i(1)
p� ÿ 1
p:
Now ÿ1=p is not an algebraic integer, by Proposition 22.5. Therefore,
for some i > 2, ÷i(g)÷i(1)= p is not an algebraic integer (see Theorem
22.3). Since ÷i(g) is an algebraic integer (Corollary 22.4), it follows
Burnside's Theorem 363
that ÷i(1)= p is not an algebraic integer; in other words, p does not
divide ÷i(1). Thus
÷i(g) 6� 0 and p 6 j ÷i(1):
As |gG | � pr, this means that ÷i(1) and |gG | are coprime integers, and
so there are integers a and b such that
ajG:CG(g)j � b÷i(1) � 1:
Hence
ajGj÷i(g)
jCG(g)j÷i(1)� b÷i(g) � ÷i(g)
÷i(1):
By Corollaries 22.10 and 22.4, the left-hand side of this equation is an
algebraic integer; and since ÷i(g) 6� 0, it is non-zero. Now Lemma 31.2
implies that
j÷i(g)=÷i(1)j � 1:
Let r be a representation of G with character ÷i. By Theorem
13.11(1), there exists ë 2 C such that
gr � ëI :
Let K � Ker r, so that K is a normal subgroup of G. Since ÷i is not
the trivial character, K 6� G. If K 6� {1} then G is not simple, as
required; so assume that K � {1}, that is, r is a faithful representation.
Since gr is a scalar multiple of the identity, gr commutes with hrfor all h 2 G. As r is faithful, it follows that g commutes with all
h 2 G; in other words,
g 2 Z(G):
Therefore Z(G) 6� {1}. As Z(G) is a normal subgroup of G and
Z(G) 6� G, we conclude that G is not simple. j
We now come to the main result of the chapter, Burnside's Theorem.
31.4 Burnside's paqb Theorem
Let p and q be prime numbers, and let a and b be non-negative
integers with a � b > 2. If G is a group of order paqb, then G is not
simple.
Proof First suppose that either a � 0 or b � 0. Then the order of G is
a power of a prime, so by Lemma 26.1(1) we have Z(G) 6� {1}.
364 Representations and characters of groups
Choose g 2 Z(G) of prime order. Then kgl v G and kgl is not equal to
{1} or G. Hence G is not simple.
Now assume that a . 0 and b . 0. By Sylow's Theorem 30.9, G has
a subgroup Q of order qb. We have Z(Q) 6� {1} by Lemma 26.1(1).
Let g 2 Z(Q) with g 6� 1. Then Q < CG(g), so
jgGj � jG:CG(g)j � pr
for some r. If pr � 1 then g 2 Z(G), so Z(G) 6� {1} and G is not
simple as before. And if pr . 1 then G is not simple, by Theorem
31.3. j
In fact Burnside's paqb Theorem leads to a somewhat more informa-
tive result about groups of order paqb:
(31.5) Every group of order paqb is soluble.
Here, by a soluble group we mean a group G which has subgroups
G0, G1, . . . , Gr with
1 � G0 , G1 , : : : , Gr � G
such that for 1 < i < r, Giÿ1 v Gi and the factor group Gi=Giÿ1 is
cyclic of prime order.
We sketch a proof of (31.5), using induction on a � b. The result is
clear if a � b < 1, so assume that a � b > 2 and let G be a group of
order paqb. By Burnside's Theorem 31.4, G has a normal subgroup H
such that H is not {1} or G. Both H and the factor group G=H have
order equal to a product of powers of p and q, and these orders are
less than paqb. Hence by induction, H and G=H are both soluble.
Therefore there are subgroups
1 � G0 v G1 v : : : v Gs � H ,
1 � Gs=H v Gs�1=H v : : : v Gr=H � G=H
with all factor groups Gi=Giÿ1 of prime order. Then the series
1 � G0 v G1 v : : : v Gr � G
shows that G is soluble.
Summary of Chapter 31
1. If G has a conjugacy class of size pr ( p prime, r > 1), then G is
not simple.
Burnside's Theorem 365
2. If |G| � paqb ( p, q primes, a� b > 2), then G is not simple.
Exercises for Chapter 31
1. Show that a non-abelian simple group cannot have an abelian
subgroup of prime power index.
2. Prove that if G is a non-abelian simple group of order less than 80,
then |G| � 60.
(Hint: use Exercise 13.8.)
366 Representations and characters of groups
367
32
An application of representation theoryto molecular vibration
Representation theory is used extensively in many of the physical
sciences. Such applications come about because every physical system
has a symmetry group G, and certain vector spaces associated with the
system turn out to be RG-modules. For example, the vibration of a
molecule is governed by various differential equations, and the symme-
try group of the molecule acts on the space of solutions of these
equations. It is on this application ± the theory of molecular vibrations
± that we concentrate in this ®nal chapter. In order to keep our
treatment elementary, we stay within the framework of classical mech-
anics throughout. (Quantum mechanical effects can be incorporated
subsequently, but we shall not go into this ± for an account, consult
the book by D. S. Schonland listed in the Bibliography.)
Symmetry groups
Let V be R2 or R3, and for v, w 2 V, let d(v, w) denote the distance
between v and w ± in other words, if v � (x1, x2, . . .) and
w � (y1, y2, . . .), then
d(v, w) �r X
(xi ÿ yi)2
!:
An isometry of V is an invertible endomorphism W of V such that
d(vW, wW) � d(v, w) for all v, w 2 V :
The set of all isometries of V forms a group under composition, called
the orthogonal group of V, and denoted by O(V).
Any rotation of R3 about an axis through the origin is an example
of an isometry; so is any re¯ection in a plane through the origin. The
endomorphism ÿ1R3 which sends every vector v to ÿv is another
example of an element in the orthogonal group O(R3). It turns out that
the composition of two rotations is again a rotation, and that for every
isometry g in O(R3), either g or ÿg is a rotation (see Exercise 32.1).
The orthogonal group O(R3) therefore contains a subgroup of index 2
which consists of the rotations. The same is true of the group O(R2).
If Ä is a subset of V , where V � R2 or R3, then we de®ne G(Ä) to
be the set of isometries which leave Ä invariant ± that is,
G(Ä) � fg 2 O(V ): Äg � Äg(where Äg � {vg: v 2 Ä}). Then G(Ä) is a subgroup of O(V ), and is
called the symmetry group of Ä. The subgroup of G(Ä) consisting of
the rotations in G(Ä) is called the rotation group of Ä. The index of
the rotation group of Ä in the symmetry group G(Ä) is 1 or 2.
32.1 Example
Let V � R2, and let Ä be a regular n-sided polygon, with n > 3,
centred at the origin. The symmetry group of Ä is easily seen to be
the dihedral group D2n, which was de®ned as a group of n rotations
and n re¯ections preserving Ä (see Example 1.1(3)).
Now let V � R3, and again let Ä be a regular n-sided polygon
(n > 3) centred at the origin. This time, G(Ä) � D2n 3 C2; the extra
elements arise because there is an isometry which ®xes all points of Ä,
namely the re¯ection in the plane of Ä.
32.2 Example
Let Ä be a regular tetrahedron in R3 centred at the origin:
368 Representations and characters of groups
Label the corners of the tetrahedron 1, 2, 3, 4. We claim that each
permutation of the numbers 1, 2, 3, 4 corresponds to an isometry of Ä.
For example, the 2-cycle (1 2) corresponds to a re¯ection in the plane
which contains the origin and the edge 34; similarly each 2-cycle
corresponds to a re¯ection. Since S4 is generated by the 2-cycles, each
of the 24 permutations of 1, 2, 3, 4 corresponds to an isometry, as
claimed.
No non-identity endomorphism of R3 ®xes all the corners of Ä,
since Ä contains three linearly independent vectors. Therefore we have
found all the isometries, and G(Ä) � S4.
Notice that the rotation group of Ä is isomorphic to A4; for
example, (1 2)(3 4) corresponds to a rotation through ð about the axis
through the mid-points of the edges 12 and 34, and (1 2 3) corresponds
to a rotation through 2ð=3 about the axis through the origin and the
corner 4.
Finally, observe that the group G(Ä) is unchanged if we take Ä to
consist of just the four corners of the tetrahedron.
32.3 Example
In this example we describe the symmetry groups of the molecules
H2O (water), CH3Cl (methyl chloride) and CH4 (methane). The symme-
try group of a molecule is de®ned to be the group of isometries which
not only preserve the position of the molecule in space, but also send
each atom to an atom of the same kind.
The shapes of the three molecules are as follows.
An application of representation theory to molecular vibration 369
We always assume that the centroid of our molecule lies at the origin
in R3.
The CH4 molecule has four hydrogen atoms at the corners of a
regular tetrahedron, and a carbon atom at the centre of the tetrahedron.
So the symmetry group of the molecule CH4 is equal to the symmetry
group of the tetrahedron, as given in Example 32.2. This group is
isomorphic to S4, permuting the four hydrogen atoms among them-
selves and ®xing the carbon atom.
As for the CH3Cl molecule, this possesses a rotation symmetry a of
order 3 about the vertical axis, and three re¯ection symmetries in the
planes containing the C, Cl and one of the H atoms. If b is one of
these re¯ections, then the symmetry group is
f1, a, a2, b, ab, a2bgand is isomorphic to S3, permuting the three H atoms and ®xing the C
and Cl atoms.
Finally, the H2O molecule possesses two re¯ection symmetries, one
in the plane of the molecule, and one in a plane perpendicular to this
one passing through the O atom; and it has a rotation symmetry of
order 2. Hence the symmetry group is isomorphic to C2 3 C2.
Vibration of a physical system
We prepare for a description of the general problem with an example.
32.4 Example
Suppose we have a spring stretched between two points P and Q on a
smooth horizontal table, with equal masses m attached at the points of
trisection of the spring:
The masses are displaced slightly, and released. What can we say about
the subsequent motion of the system?
To investigate this problem, we let x1 and x2 be the displacements of
the two masses at time t. We measure x1 from left to right, and x2
from right to left, as indicated in the ®gure above. Let k be the
370 Representations and characters of groups
stiffness of the spring ± in other words, if the extension in the spring
is x, then the restoring force is kx.
The spring pulls the left-hand mass towards P with force kx1 and
towards Q with force ÿk(x1 � x2). By dealing with the right-hand mass
similarly, we obtain the following equations of motion of the system:
m�x1 � ÿkx1 ÿ k(x1 � x2) � ÿ2kx1 ÿ kx2,
m�x2 � ÿkx2 ÿ k(x1 � x2) � ÿkx1 ÿ 2kx2,
where �xi denotes the second derivative of xi with respect to t.
These are second order linear differential equations in two unknowns
x1 and x2, so the general solution involves four arbitrary constants. We
shall ®nd the general solution, using a method which can be applied in
a much wider context.
Write x � (x1, x2), �x � (�x1, �x2) and q � k=m. Then the equations of
motion are equivalent to the matrix equation
�x � xA, where A � ÿ2q ÿq
ÿq ÿ2q
� �:(32:5)
Notice that A is symmetric. Hence the eigenvalues of A are real, and
A has two linearly independent eigenvectors. It is this property which
we wish to emphasize and exploit in the present example.
Before we explicitly ®nd the eigenvectors of A, let us explain why
they allow us to solve the equation of motion (32.5).
Suppose that u is an eigenvector of A, with eigenvalue ÿù2. For an
arbitrary constant â, let
x � sin (ùt � â) u:
Then
�x � ÿù2 sin (ùt � â) u
� sin (ùt � â) uA (since uA � ÿù2u)
� xA:
Thus x is a solution of the equation of motion. If u1 and u2 are
linearly independent eigenvectors of A, with eigenvalues ÿù21 and
ÿù22, respectively, then
á1 sin (ù1 t � â1) u1 � á2 sin (ù2 t � â2) u2
An application of representation theory to molecular vibration 371
is a solution of the equation of motion which involves four arbitrary
constants á1, á2, â1, â2, so it is the general solution.
We now adopt this line of attack in the problem to hand. For the
matrix given in (32.5), the eigenvalues are ÿ3q and ÿq, with corre-
sponding eigenvectors (1, 1) and (1, ÿ1). Therefore the general solution
of the equation of motion (32.5) is
á1 sin (p
(3q) t � â1) (1, 1)� á2 sin (p
q . t � â2) (1, ÿ1):
The solutions which involve just one eigenvector of A are called the
normal modes of vibration. They are as follows.
sin (p
(3q) t � â1) (1, 1)Mode 1:
Here, x1 � x2 � sin (p
(3q) t � â1) and the vibration is
sin (p
q . t � â2) (1, ÿ1)Mode 2:
Here, x1 � ÿx2 � sin (p
q . t � â2) and the vibration is
The general molecular vibration problem
Suppose we have a molecule consisting of n atoms which vibrate under
internal forces. At the equilibrium position of each atom, we assign
three coordinate axes, which we use to measure the displacement of
the atom. Thus, the state of the molecule at a given time is described
by a vector in the 3n-dimensional vector space R3n. We assume that
the internal forces are linear functions of the displacements. It follows
that when we apply Newton's Second Law of Motion, we obtain
equations which may be expressed in the form
�x � xA:(32:6)
(Compare (32.5).) Here x is the row vector in R3n which measures the
displacements of all the atoms, and A is a 3n 3 3n matrix with real
entries which are determined by the internal forces.
Assume, for the moment, that at each atom the three coordinate axes
372 Representations and characters of groups
which we have chosen are at right angles to each other. It can be shown,
from physical considerations, that in this special case the matrix A is
symmetric. In particular, A has real eigenvalues, and A has 3n linearly
independent eigenvectors. Now, the effect of changing coordinate axes is
merely to replace A by a matrix which is conjugate to A. Therefore we
have the following proposition, for the general case, where our chosen
coordinate axes are not necessarily at right angles to each other.
32.7 Proposition
All the eigenvalues of A are real, and A has 3n linearly independent
eigenvectors.
To solve the equation of motion (32.6), we look for normal modes
of the system, which we de®ne next.
32.8 De®nition
A normal mode of vibration for our molecule is a vector in R3n of one
of the following forms:
sin (ùt � â) u (â constant)(1)
where ÿù2 is a non-zero eigenvalue of A and u is a corresponding
eigenvector;
(t � â) u (â constant)(2)
where u is an eigenvector of A corresponding to the eigenvalue 0.
32.9 Proposition
Each normal mode of vibration is a solution of the equation of motion
(32.6), and for this solution all the atoms vibrate with the same
frequency. The general solution of the equation of motion is a linear
combination of the normal modes of vibration.
Proof If uA � ÿù2 A and x � sin (ùt � â) u, then
�x � ÿù2 sin (ùt � â) u � sin (ùt � â) uA � xA:
If uA � 0 and x � (t � â)u, then
�x � 0 � (t � â)uA � xA:
This proves that the normal modes of vibration are solutions of the
An application of representation theory to molecular vibration 373
equation of motion (32.6). By construction, all the atoms vibrate with
the same frequency (namely, ù or 0) in a normal mode.
Note that A can have no strictly positive eigenvalue; for if ë were
such an eigenvalue, with eigenvector u, then x � epë tu would be a
solution to the equation of motion, which is nonsense. Therefore there
exist 3n linearly independent normal modes, by Proposition 32.7. Since
each normal mode involves an arbitrary constant, the general linear
combination of normal modes involves 6n arbitrary constants, so it is
the general solution to the equation of motion (32.6) (as (32.6) consists
of second order differential equations in 3n unknowns). j
Proposition 32.9 reduces the problem of solving the equations of
motion to that of ®nding all the eigenvalues and eigenvectors of the
3n 3 3n matrix A. However, this can be a huge and unwieldy task if
it is attempted directly ± even writing down the matrix A for a given
molecule can be a painful operation!
The symmetry group of the molecule and its representation theory
can often be used to simplify greatly the calculation of the eigenvectors
of A, and we shall describe a method for doing this.
Use of the symmetry group
We continue the discussion of the previous section. Let G be the
symmetry group of the molecule in question. Since G permutes the
atoms among themselves, each element of G acts as an endomorphism
of the space R3n of displacement vectors. Thus, R3n is an RG-module.
32.10 Example
Let g be the rotation of order 2 of the H2O molecule:
Assign coordinate axes at the initial positions of each atom as shown,
and for 1 < i < 9, let vi denote a unit vector along coordinate axis i.
Then g ®xes v1, negates v2 and v3, interchanges v4 and v7, and
374 Representations and characters of groups
interchanges v5 and v6 with the negatives of v8 and v9. Therefore g
acts on R9 as follows:
(x1, x2, x3, x4, x5, x6, x7, x8, x9)g
� (x1, ÿx2, ÿx3, x7, ÿx8, ÿx9, x4, ÿx5, ÿx6):
We return to the general set-up. The equations of motion are �x � xA,
and we are trying to ®nd the eigenvectors of A. The eigenspace for the
eigenvalue ë of A is, by de®nition,
fx 2 R3n: xA � ëxg:We can now present the crucial proposition which allows us to
exploit the symmetry group G of our molecule. In effect, it tells us
that the function x! xA (x 2 R3n) is an RG-homomorphism from R3n
to itself.
32.11 Proposition
For all g 2 G and x 2 R3n,
(xg)A � (xA)g,
and the eigenspaces of A are RG-submodules of R3n.
Proof Let ÿù2 be a non-zero eigenvalue of A, and let v be a
corresponding eigenvector. Then v speci®es the directions and relative
magnitudes of the displacements of the atoms from the equilibrium
position when the molecule is vibrating in a normal mode of frequency
ù. For all g in G, vg must also specify the directions and relative
magnitudes of displacements in a normal mode of frequency ù, since
the relative con®guration of the atoms is unaltered by applying g.
Therefore, vg is an eigenvector of A, with eigenvalue ÿù2. This shows
that the eigenspace for ÿù2 is an RG-submodule of R3n. A similar
argument applies to the eigenspace for the eigenvalue 0.
Choose a basis of R3n which consists of eigenvectors of A (see
Proposition 32.7), and let g 2 G. For all vectors v in the basis,
vA � ëv for some ë 2 R, and
(vg)A � ë(vg) � (ëv)g � (vA)g:
Hence (xg)A � (xA)g for all x 2 R3n. j
The idea now is to use representation theory to express the RG-
module R3n as a direct sum of irreducible RG-submodules, and hence
An application of representation theory to molecular vibration 375
to determine the eigenspaces of A, and the normal modes of the
molecule.
We can use character theory to see which irreducible RG-modules
are contained in R3n; and if ÷ is the character of an irreducible RG-
module which occurs, then the elementXg2G
÷(gÿ1)g
sends R3n onto the sum of those irreducible RG-submodules of R3n
which have character ÷ (see (14.27)). (This procedure sometimes needs
to be modi®ed, since the character of the RG-module R3n might
contain an irreducible character which cannot be realized over R ± but
in practice, problems like this are uncommon.)
32.12 De®nition
Suppose that ÷ is the character of an irreducible RG-module. Let V÷
denote the sum of those irreducible RG-submodules of R3n which have
character ÷. We call V÷ a homogeneous component of R3n.
The problem of ®nding the eigenspaces of A is considerably simpli-
®ed as a consequence of our next proposition.
32.13 Proposition
Each homogeneous component V÷ of R3n is A-invariant ± that is,
xA 2 V÷ for all x 2 V÷:
Proof By Maschke's Theorem we may write R3n � V÷ � W for some
RG-module W, and no RG-submodule of W has character ÷. The
function
å: v� w! w (v 2 V÷, w 2 W )
is an RG-homomorphism. Therefore, by Proposition 32.11, the function
x! (xA)å (x 2 V÷)
is an RG-homomorphism from V÷ into W. By Proposition 11.3, this
function is zero, so xA 2 V÷ for all x 2 V÷. (Although Proposition 11.3
is stated in terms of CG-modules, its proof works equally well for
RG-modules ± compare Exercise 23.8.) j
376 Representations and characters of groups
32.14 Corollary
If V÷ is an irreducible RG-module, then all the non-zero vectors in V÷
are eigenvectors of A.
Proof (Compare the proof of Schur's Lemma.) Since V÷ is A-invariant,
we may choose v 2 V÷ such that v is an eigenvector of A, with
eigenvalue ë, say. Then the intersection of V÷ with the eigenspace for ëis a non-zero RG-submodule of V÷, so it must equal V÷. j
We now summarize the steps in the procedure for ®nding the normal
modes of vibration of a given molecule.
32.15 Summary
(1) Assign three coordinate axes at each of the n atoms of the
molecule, to obtain R3n.
(2) Calculate the symmetry group G of the molecule. Then R3n is an
RG-module.
(3) Calculate the character ÷ of the RG-module R3n and express ÷ as a
linear combination of the irreducible characters of G.
(4) Express R3n as a direct sum of homogeneous components. This can
be done by applying the elementP
g2G÷i(gÿ1)g to R3n for each
irreducible character ÷i of G which appears in ÷, or by some other
method.
(5) Consider, in turn, each homogeneous component V÷iof R3n, and
®nd the eigenvectors of A in V÷i(see Proposition 32.13). This involves
no extra work if V÷iis irreducible (see Corollary 32.14). If V÷i
is
reducible, then see Remark 32.19 below, or Exercise 32.7, to make
further progress.
(6) If v is an eigenvector of A, with eigenvalue ÿù2, then
sin (ùt � â) v (or (t � â)v if ù � 0)
is a normal mode, where â is an arbitrary constant. It is usually
necessary to know the equations of motion in order to determine the
frequency ù.
This programme can often be successfully completed, as we shall
illustrate in the examples which make up the rest of this chapter.
An application of representation theory to molecular vibration 377
32.16 Example
We ®rst return to Example 32.4, with the spring and two vibrating
masses:
The symmetry group of this system is G � hg: g2 � 1i, where g is the
re¯ection in the mid-point of PQ. The displacement vectors (x1, x2)
form an RG-module R2.
Since (x1, x2)g � (x2, x1), the RG-submodules of R2 are sp (u1) and
sp (u2), where
u1 � (1, 1), u2 � (1, ÿ1):
It follows that the normal modes of the system are given by
sin (ù1 t � â1)(1, 1), sin (ù2 t � â2)(1, ÿ1),
where â1, â2 are constants and ù1, ù2 are the frequencies. This agrees
with the conclusion of Example 32.4.
Notice that we have determined the normal modes of vibration (but
not their frequencies) using the symmetry group alone.
32.17 Example
Consider a hypothetical triatomic molecule, where the three identical
atoms are at the corners of an equilateral triangle. For simplicity, we
consider only vibrations of the molecule in the plane, so we assign two
displacement coordinates to each atom. We choose to take our axes
along the edges of the triangle, as shown, as this eases the calcula-
tions:
Thus the position of the molecule is given by a vector (x1, . . . , x6) in
R6, where xi is the displacement along axis i (1 < i < 6).
The symmetry group (in two dimensions) of the molecule is the
dihedral group D6, generated by a rotation a of order 3 and a re¯ection
378 Representations and characters of groups
b (see Example 32.1). It is easy to work out the action of each element
of D6 on R6. For example, if b is the re¯ection which ®xes the top
atom, then
(x1, x2, x3, x4, x5, x6)b � (x2, x1, x6, x5, x4, x3):
We want to express the RD6-module R6 as a direct sum of
irreducible RD6-modules. To do this, we ®rst calculate the character ÷of the module. Since the rotation a does not ®x any of the atoms,
÷(a) � 0. And from the action of b given above, we see that ÷(b) � 0.
Thus the values of ÷ are
By Section 18.3, the character table of D6 is
Hence ÷ � ÷1 � ÷2 � 2÷3.
Thus, we seek to express R6 as a direct sum of RD6-submodules
with characters ÷1, ÷2, ÷3 and ÷3.
As a matter of notation, if v1, v2, v3 are 2-dimensional displacement
vectors for the three atoms, then we represent the displacement vector
(v1, v2, v3) 2 R6 pictorially by the diagram
We ®rst calculate the normal modes of the form (t � â)v, corre-
sponding to the eigenspace of A for eigenvalue 0. These include the
rotation and translation modes, which occur for every molecule.
1 a b
÷ 6 0 0
1 a b
÷1 1 1 1÷2 1 1 ÿ1÷3 2 ÿ1 0
An application of representation theory to molecular vibration 379
Rotation mode In this mode, the molecule rotates with constant angular
velocity about the centre. The mode is given by (t � â)v, where
v � (1, ÿ1, 1, ÿ1, 1, ÿ1); pictorially,
We call sp (v) the rotation submodule of R6. If ÷R is the character of
sp (v), then
÷R(1) � 1, ÷R(a) � 1, ÷R(b) � ÿ1,
and so ÷R � ÷2. Indeed, sp (v) � R6å2, where
å2 �Xg2D6
÷2(gÿ1)g � 1� a� a2 ÿ bÿ abÿ a2b
(compare (14.27)).
Translation modes These are modes in which all atoms move in the
same constant direction with the same constant speed. The modes are
of the form (t � â)v, where v is a vector in the span of v1, v2 and v3,
these vectors being given pictorially by
(thus v1 � (ÿ1, 1, 0, ÿ1, 1, 0), v2 � (1, 0, ÿ1, 1, 0, ÿ1), v3 � (0, ÿ1,
1, 0, ÿ1, 1)).
Since v1 � v2 � v3 � 0, the subspace sp (v1, v2, v3) has dimension 2.
It is clearly an RD6-submodule of R6, and is called the translation
submodule; it does not contain the rotation submodule. Thus the
character of the translation submodule is part of ÷ ÿ ÷2 � ÷1 � 2÷3, so
the character must be ÷3.
380 Representations and characters of groups
Vibratory modes The remaining normal modes correspond to eigen-
spaces of the matrix A with non-zero eigenvalues, and are called
vibratory modes. The sum of these eigenspaces forms an RD6-submo-
dule R6vib of R6 (by Proposition 32.11), with character ÷vib, where
÷vib � ÷ÿ (÷2 � ÷3) � ÷1 � ÷3:
In particular, R6vib has dimension 3. Since no mode in R6
vib can have
any translation component, if w 2 R6vib then the total component of w
in each direction is zero; moreover, R6vib does not contain the rotation
submodule, so that total moment of each vector in R6vib about the
centre is zero. These constraints imply three independent linear equa-
tions in the coordinates of the vectors in R6vib, and since R6
vib has
dimension 3, every vector in R6 which satis®es these equations lies in
R6vib. Therefore a basis of R6
vib is u1, u2, u3, where
Clearly sp (u1 � u2 � u3) is an RD6-submodule of R6vib with charac-
ter ÷1. The vibratory mode given by u1 � u2 � u3 is sometimes called
the expansion±contraction mode (you will see the reason for this name
in the picture in (32.18(3)) below).
Finally, since D6 permutes the vectors u1, u2, u3 among themselves,
it is easy to see that sp (u1 ÿ u2, u1 ÿ u3) is an RD6-submodule of
R6vib. Its character is ÷3 and it gives us the last eigenspace for the
matrix A.
Our calculation of the normal modes is now complete, and we
summarize our ®ndings below.
An application of representation theory to molecular vibration 381
(32.18) (1) Rotation mode:
(2) Translation modes: linear combinations of
(3) Vibratory mode: expansion±contraction mode
(4) Vibratory modes: linear combinations of
(We have chosen 2u2 ÿ u1 ÿ u3, 2u1 ÿ u2 ÿ u3 as the basis for the
vibratory modes in (4) merely because these modes look simpler than
u1 ÿ u2, u1 ÿ u3 pictorially.)
We emphasize that we have found the normal modes of vibration
without explicit knowledge of the equations of motion. In order to
382 Representations and characters of groups
check our results, we now calculate the equations of motion.
Let m be the mass of each atom, and assume that the magnitude of
the force between two atoms is k times the decrease in distance
between them.
For a general displacement (x1, x2, x3, x4, x5, x6), denote the new
positions of the atoms by P9, Q9, R9. From the diagram, the difference
in length between QR and Q9R9 is
(x4 � x5)� 12(x3 � x6):
(We always assume that x1, : : : , x6 are small compared with the
distance between the atoms, so that we may ignore second order
terms.)
Similarly,
PRÿ P9R9 � (x1 � x6)� 12(x2 � x5),
PQÿ P9Q9 � (x2 � x3)� 12(x1 � x4):
Hence the force on the molecule at P in the direction of the ®rst
coordinate axis is
ÿk(PRÿ P9R9) � ÿk(x1 � x6)ÿ 12k(x2 � x5):
Therefore,m
k�x1 � ÿ(x1 � x6)ÿ 1
2(x2 � x5):
In the same way,
m
k�x2 � ÿ(x2 � x3)ÿ 1
2(x1 � x4),
An application of representation theory to molecular vibration 383
and we obtain similar equations for �x3, . . . , �x6. The matrix A for
which �x � xA is therefore given by
A � ÿk
m
1 1=2 1=2 0 0 1
1=2 1 1 0 0 1=2
0 1 1 1=2 1=2 0
0 1=2 1=2 1 1 0
1=2 0 0 1 1 1=2
1 0 0 1=2 1=2 1
0BBBBBB@
1CCCCCCA:
You should check that the vectors which we gave in (32.18) are indeed
eigenvectors of A.
32.19 Remark
In Example 32.17, the character ÷ of the RG-module R6 was given by
÷ � ÷1 � ÷2 � 2÷3:
All the non-zero vectors in the homogeneous components for ÷1 and ÷2
gave normal modes, since these homogenous components were irreduci-
ble (see Corollary 32.14). The homogeneous component V÷3for ÷3 was
reducible, but we were able to write it as a sum of two subspaces of
eigenvectors (those appearing in (32.18)(2) and (4)) because V÷3\ R6
vib
was an A-invariant RG-submodule of V÷3different from {0} and V÷3
.
This illustrates a method which sometimes helps to deal with reducible
homogeneous components. In our next example, the situation is more
complicated.
32.20 Example
We analyse the normal modes for the methane molecule CH4.
We determined the symmetry group G of this molecule in Example
32.2, where we found that G is isomorphic to S4. Label the corners of
384 Representations and characters of groups
the tetrahedron 1, 2, 3, 4, as shown below, and identify G with S4;
thus, for example, the rotations about the vertical axis through 1 are
written as1, (2 3 4), (2 4 3):
In order fully to exploit the symmetry of the methane molecule, at
each hydrogen atom we choose displacement axes along the edges of
the tetrahedron. Let v12, v13, v14 be unit vectors at corner 1 in the
directions of the edges 12, 13, 14, respectively; similarly, let v21, v23,
v24 be unit vectors at corner 2 in the directions of the edges 21, 23,
24, and so on, giving twelve vectors vij, in all.
We now introduce a new idea, by taking four unit vectors w1, w2,
w3 and w4 at the carbon atom, with wi pointing towards corner i
(1 < i < 4). Since w1 � w2 � w3 � w4 � 0, these four vectors span a
3-dimensional space.
Let V be the vector space over R with basis
v12, v13, v14, v21, v23, v24, v31, v32, v34, v41, v42, v43,
and let W be the vector space over R spanned by w1, w2, w3, w4. Then
V � R12, W � R3 and V and W are RG-modules. Our main task is to
®nd RG-submodules of R15 � V � W.
The action of G on V is easy to describe; for g in G, we have
vijg � vig,jg for all i, j:
Thus, G permutes our twelve basis vectors of V, and we can quickly
calculate the character ÷ of V:
1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
÷ 12 2 0 0 0
An application of representation theory to molecular vibration 385
For example, (1 2) ®xes the basis vectors v34 and v43 only; all the basis
vectors are moved by (1 2 3); and so on.
The group G acts on W as follows; for g in G, we have
wig � wig (1 < i < 4):
After recalling that w1 � . . . � w4 � 0, it is easy to calculate the
character ö of W ± it is
1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
ö 3 1 0 ÿ1 ÿ1
By Section 18.1, the character table of S4 is as shown at the top of
p. 387. We ®nd that
÷ � ÷1 � ÷3 � 2÷4 � ÷5,
ö � ÷4:
By applying the elementsXg2G
÷i(gÿ1)g (i � 1, 3, 5, 4)
to R15, we can ®nd RG-submodules with characters ÷1, ÷3, ÷5 and 3÷4
(see (14.27)).
The RG-submodule W1 with character ÷1 is spanned byXi, j
vij
and this gives the expansion±contraction normal mode:
We next describe the RG-submodule W5 with character ÷5. Let
p1 � (v23 ÿ v32)� (v34 ÿ v43)� (v42 ÿ v24),
p2 � (v31 ÿ v13)� (v14 ÿ v41)� (v43 ÿ v34),
386 Representations and characters of groups
p3 � (v12 ÿ v21)� (v41 ÿ v14)� (v24 ÿ v42),
p4 � (v21 ÿ v12)� (v13 ÿ v31)� (v32 ÿ v23):
The vector pi gives a rotation about the axis through the corner i and
the centroid of the tetrahedron.
It should be clear from the pictures that for all i with 1 < i < 4 and
Character table of S4
1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0÷4 3 1 0 ÿ1 ÿ1÷5 3 ÿ1 0 ÿ1 1
An application of representation theory to molecular vibration 387
all g in G, we have pig � � pj for some j. Therefore, if we let
W5 � sp ( p1, p2, p3, p4), then W5 is an RG-submodule of V. Now
p1 � p2 � p3 � p4 � 0, so dim W5 � 3. Check that the character of W5
is ÷5. The RG-module W5 is the rotation submodule. (Compare, for
example, the picture for p4 with the picture for the rotation vector v in
Example 32.17.)
Now we construct the RG-submodule W3 of V with character ÷3. Let
q1 � (v12 � v21)� (v34 � v43)ÿ (v13 � v31)ÿ (v24 � v42),
q2 � (v13 � v31)� (v24 � v42)ÿ (v14 � v41)ÿ (v23 � v32),
q3 � (v14 � v41)� (v23 � v32)ÿ (v12 � v21)ÿ (v34 � v43):
(Each qi is associated with an `opposite pair of edges'.)
For all i with 1 < i < 4 and g in G, we have qi g � �q j for some j.
Let W3 � sp (q1, q2, q3). Then W3 is an RG-submodule of V. Since
q1 � q2 � q3 � 0, the dimension of W3 is 2; its character is ÷3.
388 Representations and characters of groups
In the RG-submodules W1, W5 and W3 which we have found so far,
all the non-zero vectors are eigenvectors of A, by Corollary 32.14.
We now come to the homogeneous component (V � W)÷4of R15.
De®ne the vectors r1, r2, r3, r4 by
r1 � (v12 � v21)� (v13 � v31)� (v14 � v41)
ÿ (v23 � v32)ÿ (v24 � v42)ÿ (v34 � v43),
r2 � (v12 � v21)� (v23 � v32)� (v24 � v42)
ÿ (v13 � v31)ÿ (v14 � v41)ÿ (v34 � v43),
r3 � (v13 � v31)� (v23 � v32)� (v34 � v43)
ÿ (v12 � v21)ÿ (v14 � v41)ÿ (v24 � v42),
r4 � (v14 � v41)� (v24 � v42)� (v34 � v43)
ÿ (v13 � v31)ÿ (v12 � v21)ÿ (v23 � v32):
(The vector ri is associated with corner i.)
An application of representation theory to molecular vibration 389
For each g in G and i with 1 < i < 4, we have ri g � rig. Thus G
permutes the vectors r1, r2, r3, r4 among themselves. Note that
r1 � r2 � r3 � r4 � 0, so r1, r2, r3, r4 span a 3-dimensional RG-
submodule W4 of V. The character of W4 is ÷4 (see Proposition 13.24).
Next, de®ne the vectors s1, s2, s3, s4 by
s1 � (v12 � v13 � v14)ÿ (v21 � v31 � v41),
s2 � (v21 � v23 � v24)ÿ (v12 � v32 � v42),
s3 � (v31 � v32 � v34)ÿ (v13 � v23 � v43),
s4 � (v41 � v42 � v43)ÿ (v14 � v24 � v34):
We have
sig � sig (g 2 G, 1 < i < 4),
s1 � s2 � s3 � s4 � 0,
390 Representations and characters of groups
and s1, s2, s3, s4 span a 3-dimensional RG-submodule W94 of V with
character ÷4.
Now recall that w1, w2, w3, w4 span W; we have
wig � wig (g 2 G, 1 < i < 4),
w1 � w2 � w3 � w4 � 0,
and the character of W is ÷4. The sum of W4, W 94 and W is direct, so
(V � W )÷4� W4 � W 94 � W :
We now break off temporarily from studying the methane molecule,
in order to deal with the easier case of a molecule with identical atoms
at the corners of the tetrahedron, and no central atom. In this case, the
space W does not enter our calculations, and we can decompose
V÷4� W4 � W 94 in the following way.
(32.21) (1) The vectors r1 ÿ 2s1, r2 ÿ 2s2, r3 ÿ 2s3, r4 ÿ 2s4
span the 3-dimensional space of translation modes.
(2) The vectors r1, r2, r3, r4 span the subspace
V÷4\ R12
vib of V÷4, and so they give the ®nal 3-dimen-
sional space of eigenvectors (see Remark 32.19). The
normal mode (sinùt)r1 is sometimes called an `um-
brella mode'. To see why, look at the picture of the
vector r1!
We return now to the methane molecule.
The task in front of us is to ®nd the eigenvectors of A which lie in
(V � W )÷4� W4 � W 94 � W :
The solution of this problem depends, in fact, upon the constants which
appear in the equations of motion, so we cannot complete the work
using only representation theory. Since dim (V � W )÷4� 9, it is very
dif®cult to calculate the eigenvectors of A in (V � W )÷4directly. We
shall therefore press on and explain how to reduce the dif®culty to that
of calculating the eigenvectors of a 3 3 3 matrix.
Let H be the subgroup of S4 consisting of those permutations which
®x the number 1, and let
U1 � fv 2 (V � W )÷4: vh � v for all h 2 Hg:
Since (vh)A � (vA)h for all v 2 V÷4and all h 2 H, it follows that U1
is A-invariant.
An application of representation theory to molecular vibration 391
We ®nd that
h3÷4 # H , 1 HiH � 3,
and so dim U1 � 3. But for all h 2 H, r1 h � r1, s1 h � s1 and
w1 h � w1. Therefore
U1 � sp (r1, s1, w1):
Once the equations of motion, and hence the matrix A, have been
calculated, it is possible to calculate the 3 3 3 matrix B of the action
of A on r1, s1, w1 (see Exercise 32.5). The eigenvectors of B then give
three eigenvectors of A.
Better still,
r1(1 2) � r2, s1(1 2) � s2, w1(1 2) � w2,
and since A commutes with the action of G, the space U2, de®ned by
U2 � sp (r2, s2, w2)
is A-invariant, and the matrix of A acting on r2, s2, w2 is again B. A
similar remark applies to U3, where
U3 � sp (r3, s3, w3):
Therefore, the process of calculating the eigenvectors of the 3 3 3
matrix B gives nine eigenvectors of A which form a basis of
(V � W )÷4.
One eigenvector of A acting on r1, s1, w1 is easy to ®nd, namely the
translation vector
r1 ÿ 2s1 � 3 cos Ww1,
where W is the angle between an edge of the tetrahedron and the line
joining a corner on the edge to the centroid. Thus we obtain the
translation submodule sp (r1 ÿ 2s1 � 3 cos Ww1, r2 ÿ 2s2 � 3 cos Ww2,
r3 ÿ 2s3 � 3 cos Ww3).
By means of representation theory, we have therefore reduced the
initial problem of ®nding the eigenvectors of a 15 3 15 matrix A to
that of calculating two of the eigenvectors of a 3 3 3 matrix. It is hard
to imagine a more spectacular application of representation theory with
which to conclude our text.
392 Representations and characters of groups
Summary of Chapter 32
1. The symmetry group G of a molecule with n atoms consists of
those distance-preserving endomorphisms of R3 which send each
atom to an atom of the same kind.
2. The equations of motion of the molecule have the form
�x � xA,
where x 2 R3n and the 3n 3 3n matrix A has 3n linearly indepen-
dent eigenvectors.
3. If u is an eigenvector of A, with eigenvalue ÿù2, then x �sin (ùt � â)u (or x � (t � â)u if ù � 0) is a solution of the equa-
tions of motion, and is called a normal mode. All solutions are
linear combinations of normal modes.
4. The space R3n of displacement vectors is an RG-module. The action
of any g 2 G on R3n commutes with A.
5. To determine the eigenvectors of A (and hence all solutions of the
equations of motion), it is suf®cient to ®nd the eigenvectors of A
restricted to each homogeneous component V÷iof the RG-module
R3n. If V÷ iis irreducible, then all non-zero vectors in V÷i
are
eigenvectors of A.
Exercises for Chapter 32
1. Suppose that b 2 O(R3), and let e1 � (1, 0, 0), e2 � (0, 1, 0),
e3 � (0, 0, 1).
(a) Show that the matrix B of b with respect to the basis e1, e2, e3
of R3 satis®es BBt � I (where Bt denotes the transpose of B).
Deduce that det B � �1.
(b) Let C � (det B)B. Prove successively that
(i) C has a real eigenvalue,
(ii) C has a positive eigenvalue,
(iii) 1 is an eigenvalue of C.
(c) Deduce from (a) and (b) that b is a rotation if det B � 1, and
ÿb is a rotation otherwise.
(d) Show that if b is a rotation through an angle ö about some
axis, then tr B � 1 � 2 cosö.
2. Suppose that G is the symmetry group of some molecule in R3.
An application of representation theory to molecular vibration 393
Show that the sum of the character ÷T of the translation submodule
and the character ÷R of the rotation submodule has value at g 2 G
given by
(÷T � ÷R)(g) �2(1� 2 cosö), if g is a rotation
through angle öabout some axis,
0, if g is not a rotation:
8>><>>:3. Consider the triatomic molecule studied in Example 32.17. Take the
axes for the displacement coordinates as shown below:
Calculate the equations of motion �x � xA with respect to these axes,
and verify that A is symmetric. (See the paragraph before Proposi-
tion 32.7.)
4. Consider the space spanned by the vectors r1, r2, r3, r4 given in
Example 32.20. Find a basis for this space which is simpler than
the one which we have used. What property of r1, r2, r3, r4
prompted us to use these vectors?
5. The purpose of this exercise is to derive the equations of motion of
the methane molecule, and so ®nd explicitly the 3 3 3 matrix B
which appears at the end of Example 32.20.
Label the corners of the tetrahedron 1, 2, 3, 4 and let 0 denote
the centroid of the tetrahedron. Work with 15 unit displacement
vectors
v12, v13, : : : , v43, w1, w2, w3
as described in Example 32.20, and let the position vector of the
molecule be Xi6� j
xijvij �X3
i�1
yiwi:
(a) Prove that cos (/ 012) � p(2=3) and cos (/ 102) � ÿ1=3.
394 Representations and characters of groups
(b) Show that the decrease in the length of the edge 12 from its
original length is
x12 � x21 � 12(x13 � x14 � x23 � x24),
with similar expressions for the edges 13, 14, 23, 24, 34.
Also, show that the length of the edge 01 has decreased by
p(2=3)(x12 � x13 � x14)� y1 ÿ 1
3(y2 � y3),
with similar expressions for the edges 02, 03.
Finally, show that the length of the edge 04 has decreased byp
(2=3)(x41 � x42 � x43)ÿ 13(y1 � y2 � y3):
(c) Let m1 denote the mass of a hydrogen atom, and m2 the mass
of a carbon atom. Assume that the magnitude of the force
between hydrogen atoms is k1 times the decrease in distance
between them, and the magnitude of the force between a
hydrogen atom and a carbon atom is k2 times the decrease in
distance between them.
Prove that
m1�x12 �ÿ k1[x12 � x21 � 12(x13 � x14 � x23 � x24)]
ÿ 13k2[x12 � x13 � x14 �p(3=2)(y1 ÿ 1
3y2 ÿ 1
3y3)],
with similar expressions for �x13, �x14, �x21, �x23, �x24, �x31, �x32, �x34.
Also, show
m1�x41 �ÿ k1[x14 � x41 � 12(x42 � x43 � x12 � x13)]
ÿ 13k2[x41 � x42 � x43 ÿ 1
3
p(3=2)(y1 � y2 � y3)],
with similar expressions for �x42 and �x43.
Finally, show
m2 �y1 � ÿk2[p
(2=3)(x12 � x13 � x14 ÿ x41
ÿ x42 ÿ x43)� 43y1],
with similar expressions for �y2 and �y3.
(d) The equations in part (c) determine the 15 3 15 matrix A in the
equations of motion �x � xA. Verify that the vectorsXi, j
vij, p1, p2, p3, q1, q2,
An application of representation theory to molecular vibration 395
which appear in Example 32.20, are eigenvectors of A.
(e) Find the entries bij in the 3 3 3 matrix B which are given by
r1 A � b11 r1 � b12s1 � b13w1,
s1 A � b21 r1 � b22s1 � b23w1,
w1 A � b31 r1 � b32s1 � b33w1,
where the vectors r1, s1, w1 are as in Example 32.20.
(f) Verify that
(1, ÿ2,p
6)
is an eigenvector of B.
6. Consider a hypothetical molecule in which there are four identical
atoms at the corners of a square. Assume that the only internal
forces are along the sides of the square.
(a) Find the normal modes of the molecule.
(b) Calculate the equations of motion, �x � xA, and check that the
vectors you found in part (a) are, indeed, eigenvectors of A.
7. In this exercise, we derive a method for simplifying the problem of
®nding the eigenvectors of A when the homogeneous component V÷i
is reducible. (See 32.15(5).) We assume that ÷i is the character of
an irreducible RG-module which remains irreducible as a CG-
module.
Suppose that V÷ i� U1 � . . . � Um, a sum of m isomorphic irredu-
cible RG-modules. We reduce our problem to that of ®nding the
eigenvectors of an m 3 m matrix.
For 1 < i < m, let Wi be an RG-isomorphism from U1 to Ui.
(a) Prove that for all non-zero u in U1,
sp (uW1, : : : , uWm)
is an A-invariant vector space of dimension m.
(Hint: compose the function w! wA with a projection, and use
Exercise 23.8.)
(b) Let Au denote the matrix of the endomorphism w! wA of
sp (uW1, . . . , uWm) with respect to the basis uW1, . . . , uWm.
Prove that if u and v are non-zero elements of U1, then
Au � Av.
(c) Assume that the eigenvectors of the m 3 m matrix Au are
known. Show how to ®nd the eigenvectors of A in V÷i.
396 Representations and characters of groups
397
Solutions to exercises
Chapter 1
1. Note that all subgroups of G are normal, since G is abelian; and G 6� {1}since G is simple. Let g be a non-identity element of G. Then kgl is anormal subgroup of G, so kgl � G. If G were in®nite, then hg2i would be anormal subgroup different from G and {1}; hence G is ®nite. Let p be aprime number which divides |G|. Then hg pi is a normal subgroup of Gwhich is not equal to G. Therefore g p � 1, and so G is cyclic of primeorder.
2. Since G is simple and Ker W v G, either Ker W � f1g or Ker W � G. IfKer W � f1g then W is an isomorphism; and if Ker W � G then H � f1g.
3. First, G \ An � fg 2 G: g is even}, so G \ An v G. Since G \ An 6� G, wemay choose h 2 G with h =2 An. For all odd g in G, we have g �(ghÿ1)h 2 (G \ An)h. Therefore G \ An and (G \ An)h are the only rightcosets of G \ An in G, and G=(G \ An) � C2.
4. (a) Using the method of Example 1.4, it is routine to verify that ö and øare homomorphisms. Kerö � {1, a2} and Kerø � {1, c2}.
(b) Since b2ë � I but (bë)2 � Y 2 � ÿI , it follows that ë is not ahomomorphism. Check using the method of Example 1.4 again that ì isa homomorphism. Also Ker ì � {1} and Im ì � L, so ì is anisomorphism.
5. Let
D4m � ha, b: a2m � b2 � 1, bÿ1ab � aÿ1i, and
D2m � hc, d: cm � d2 � 1, dÿ1cd � cÿ1i,where m is odd. The elements of D2m 3 C2 are
(cid j, (ÿ1)k)
for 0 < i < m ÿ 1, 0 < j < 1, 0 < k < 1. Let x � (c(m�1)=2, ÿ1) and y �(d, 1). Check that
x2m � y2 � 1, yÿ1xy � xÿ1:
By Example 1.4, the function W: D4m ! D2m 3 C2 de®ned by
W: aib j ! xi y j (0 < i < 2mÿ 1, 0 < j < 1)
is a homomorphism. Since Im W � kx, yl, it contains
x2 � (c, 1) and xm � (1, ÿ1)
and hence Im W � D2m 3 C2. As |D4m| � |D2m 3 C2|, we conclude that W isan isomorphism.
6. (a) Let G � kal and suppose that 1 6� H < G. First observe that there existsi . 0 such that ai 2 H. Choose k as small as possible such that k . 0and ak 2 H . If 1 6� a j 2 H then j � qk � r for some integers q, r with0 < r , k. Hence ar � a jaÿqk � a j(ak)ÿq 2 H . Since r , k, we haver � 0. Therefore a j � akq and so H � kakl; thus H is cyclic.
(b) Assume that G � hai and jGj � dn. If g 2 G and gn � 1, then g � a j
for some integer j and dnj jn, so dj j; hence g 2 kadl. It follows that
fg 2 G: gn � 1g � hadi,which is a cyclic group of order n.
(c) If x and y are elements of order n in the ®nite cyclic group G, then x,y 2 H, where H � {g 2 G: gn � 1}. Now kxl and kyl have order n; alsoH has order n, by part (b). We deduce that
hxi � H � hyi:Thus x 2 hyi, and so x is a power of y.
7. Let G be the set of non-zero complex numbers. If g, h 2 G then gh 6� 0,so gh 2 G. If g, h, k 2 G then (gh)k � g(hk); also 1 2 G and 1g � g1 � gfor all g 2 G. Finally, if g 2 G then gÿ1 � 1=g 2 G, and gÿ1 g � ggÿ1 � 1.Thus G is a group under multiplication.
If H is a subgroup of G of order n, then hn � 1 for all h 2 H (since theorder of h divides n, by Lagrange's Theorem). Therefore
H < fg 2 G: gn � 1g � he2ði=ni:Since jH j � n � jhe2ði=nij, it follows that H � ke2ði=nl.
8. Partition G into subsets fg, gÿ1g (g 2 G). Each such subset has size 1 or2, and the identity element is in a subset of size 1. Hence, if jGj is eventhen there exists g 2 G such that g 6� 1 and the subset fg, gÿ1g has size 1;so g � gÿ1 and g has order 2.
9. De®ne matrices A, B as follows:
A � eið=4 0
0 eÿið=4
� �, B � 0 1
ÿ1 0
� �:
Check that A8 � I, B2 � A4 and Bÿ1AB � Aÿ1. These relations show thatevery element of the group kA, Bl has the form A j Bk with 0 < j < 7,0 < k < 1. Moreover,
A j � ei jð=4 0
0 eÿi jð=4
� �, A j B � 0 ei jð=4
ÿeÿi jð=4 0
� �:
Since these matrices, with 0 < j < 7, are all distinct, kA, Bl has order 16.
398 Representations and characters of groups
10. Suppose jG: H j � 2 and let g 2 G. If g 2 H then gÿ1 Hg � H. And ifg =2 H then H, Hg are the two right cosets of H in G, while H, gH arethe two left cosets. Therefore Hg � gH, and so gÿ1 Hg � H again. HenceH v G.
Chapter 2
1. Let u, w 2 W and ë 2 F. Since W is a linear transformation, we have
(uWÿ1 � wWÿ1)W � (uWÿ1)W� (wWÿ1)W � u� w,
(ë(wWÿ1))W � ë(wWÿ1)W � ëw:
Hence (u � w)Wÿ1 � uWÿ1 � wWÿ1 and (ëw)Wÿ1 � ë(wWÿ1), so Wÿ1 is alinear transformation.
2. (1)) (2): If W is invertible then W is injective, so Ker W � {0}.(2)) (3): If Ker W � {0} then dim (Im W) � dim V (by (2.12)), so Im W � V(by (2.7)).(3)) (1): Assume that Im W � V, so W is surjective. By (2.12), Ker W � {0}.If u, v 2 V and uW � vW then (u ÿ v)W � 0, so u ÿ v 2 Ker W � {0}, and sou � v. Thus W is injective. As W is surjective and injective, W is invertible.
3. First suppose that V � U � W. Then V � U � W. Let v 2 U \ W. Thenv � v � 0 � 0 � v and this gives us two expressions for v as the sum of anelement in U and an element in W. Since such expressions are unique,v � 0. Thus U \ W � {0}.
Now suppose that V � U � W and U \ W � {0}. If u1 � w1 � u2 � w2
with u1, u2 2 U and w1, w2 2 W, then u1 ÿ u2 � w2 ÿ w1 2 U \ W � {0};hence u1 � u2 and w1 � w2. This shows that V � U � W.
4. Assume ®rst that V � U � W. If v 2 V then v � u � w for some u 2 U andw 2 W; since u is a linear combination of u1, . . . , ur and w is a linearcombination of w1, . . . , ws, it follows that v is a linear combination ofu1, . . . , ur, w1, . . . , ws. Therefore u1, . . . , ur, w1, . . . , ws span V. Supposethat
ë1u1 � : : :� ërur � ì1w1 � : : :� ìsws � 0
with all ëi, ì j in F. Since V � U � W, the expression 0 � 0 � 0 is theunique expression for 0 as the sum of vectors in U and W, and so
ë1u1 � : : :� ër ur � ì1w1 � : : :� ìsws � 0:
As u1, . . . , ur are linearly independent, this forces ëi � 0 for all i;similarly ìi � 0 for all i. Therefore u1, . . . , ur, w1, . . . , ws are linearlyindependent; hence they form a basis of V.
Conversely, suppose that u1, . . . , ur, w1, . . . , ws is a basis of V. Ifv 2 U \ W then v � ë1u1 � . . . � ërur � ì1w1 � . . . � ìsws for some ëi,ì j 2 F; this gives ëi � ì j � 0 for all i, j, since u1, . . . , ur, w1, . . . , ws arelinearly independent. Thus v � 0 and so U \ W � {0}. It is easy to see thatV � U � W, so by Exercise 3, V � U � W.
5. (a) Assume ®rst that V � U1 � U2 � U3. Let u 2 U1 \ (U2 � U3). Thenu � u1 � u2 � u3 for some ui 2 Ui (1 < i < 3). Since u1 � 0 � 0 �0 � u2 � u3 and the sum U1 � U2 � U3 is direct, we have u1 �
Chapter 2 399
u2 � u3 � 0. Therefore U1 \ (U2 � U3) � {0}. Similarly,U2 \ (U1 � U3) � U3 \ (U1 � U2) � f0g.Now suppose that U1 \ (U2 � U3) � U2 \ (U1 � U3) �U3 \ (U1 � U2) � {0}. Assume that ui, u9i 2 Ui (1 < i < 3) andu1 � u2 � u3 � u91 � u92 � u93. Thenu1 ÿ u91 � (u92 ÿ u2) � (u93 ÿ u3) 2 U1 \ (U2 � U3) � {0}, so u1 � u91.Similarly, u2 � u92 and u3 � u93. Therefore V � U1 � U2 � U3.
(b) Let V � R2, and U1 � sp ((1, 0)), U2 � sp ((0, 1)), U3 � sp ((1, 1)).
6. By Exercise 4, if V � U � W then dim V � dim U � dim W. More generally,if V � U1 � . . . � Ur then V � U1 � (U2 � . . . � Ur) (see (2.10)); byinduction on r, dim (U2 � . . . � Ur) � dim U2 � . . . � dim Ur, sodim V � dim U1 � . . . � dim Ur.
7. Let V � R2. De®ne W, ö: V! V by
W: (x, y)! (x, 0) and ö: (x, y)! (y, 0):
Then Im W � sp ((1, 0)), Ker W � sp ((0, 1)), so V � Im W � Ker W; and Imö �Kerö � sp ((1, 0)), so V 6� Imö � Kerö.
8. Suppose ®rst that W is a projection. Then V � Im W � Ker W by Proposition2.32. Take a basis u1, . . . , ur for Im W and a basis w1, . . . , ws for Ker W.Then u1, . . . , ur, w1, . . . , ws is a basis, say B, of V, by Exercise 4. SinceuiW � ui for all i and wjW � 0 for all j, the matrix [W]B is diagonal, thediagonal entries being r 1's followed by s 0's.
Conversely, if [W]B has the given form, then clearly W2 � W, so W is aprojection.
9. Let v 2 V. Then
v � 12(v� vW)� 1
2(vÿ vW):
Observe that 12(v � vW)W � 1
2(vW � v), so 1
2(v � vW) 2 U. Similarly,
12(v ÿ vW) 2 W. Thus V � U � W. If v 2 U \ W then v � vW � ÿv, so v � 0.
Therefore V � U � W, by Exercise 3. The construction of the basis B issimilar to that in Solution 8.
Chapter 3
1. First, suppose that r is a representation of G. Then
I � 1r � (am)r � (ar)m � Am:
Conversely, assume that Am � I. Then (ai)r � Ai for all integers i(including i . m ÿ 1 and i , 0). Therefore for all integers i, j,
(aia j)r � (ai� j)r � Ai� j � Ai A j � (air)(a jr),
and so r is a representation.
2. Check that A3 � B3 � C3 � I . Hence by Exercise 1, each r j is arepresentation. The representations r2 and r3 are faithful, but r1 is not.
400 Representations and characters of groups
3. De®ne r by
(aib j)r � (ÿ1) j (0 < i < nÿ 1, 0 < j < 1):
It is easy to check that r is a representation of G.
4. (1) For all g 2 G, Iÿ1(gr)I � gr; hence r is equivalentto r.(2) If r is equivalent to ó then there is an invertible matrix T such thatgó � Tÿ1(gr)T for all g 2 G; then gr � (Tÿ1)ÿ1(gó)Tÿ1, so ó is equivalentto r.(3) If r is equivalent to ó and ó is equivalent to ô, then there are invertiblematrices S and T such that gó � Sÿ1(gr)S and gô � Tÿ1(gó)T for allg 2 G; then gô � (ST)ÿ1(gr)(ST), so r is equivalent to ô.
5. Check that in each of the cases (1) S � A, T � B, (2) S � A3, T � ÿB, (3)S � ÿA, T � B, (4) S � C, T � D, we have
S6 � T 2 � I , Tÿ1ST � Sÿ1:
It follows that each rk is a representation (see Example 1.4).The matrices Ar Bs (0 < r < 5, 0 < s < 1) are all different, so r1 is
faithful. Similarly r4 is faithful. But r2 and r3 are not faithful, sincea2r2 � I and a3r3 � I.
The representations r1 and r4 are equivalent: to see this, let
T � 1 1
ÿi i
� �:
Then Tÿ1(gr4)T � gr1 for all g 2 G. (To ®nd T, ®rst work out theeigenvectors of C.)
If j 6� 2, then a2r j 6� I; hence r2 is not equivalent to any of the others.And if j 6� 3, then a3r j 6� I; hence r3 is not equivalent to any of the others.
6. De®ne the matrices A and B by
A �0 1 0
ÿ1 0 0
0 0 1
0@ 1A, B �1 0 0
0 ÿ1 0
0 0 1
0@ 1A:Then the function r: ar bs ! Ar Bs (0 < r < 3, 0 < s < 1) is a faithfulrepresentation of D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. Compare Example3.2(1).
7. By Theorem 1.10, G=Ker r � Imr. But Imr < GL (1, F) and GL (1, F) isabelian. Therefore G=Ker r is abelian.
8. No: let G be any non-abelian group and let r be the trivial representation.
Chapter 3 401
Chapter 4
1.
g 1 (1 2) (1 3)
[g]B 1
1 0 0
0 1 0
0 0 1
0@ 1A 0 1 0
1 0 0
0 0 1
0@ 1A 0 0 1
0 1 0
1 0 0
0@ 1A[g]B 2
1 0 0
0 1 0
0 0 1
0@ 1A 1 0 0
0 ÿ1 0
0 ÿ1 1
0@ 1A 1 0 0
0 1 ÿ1
0 0 ÿ1
0@ 1A
g (2 3) (1 2 3) (1 3 2)
[g]B 1
1 0 0
0 0 1
0 1 0
0@ 1A 0 1 0
0 0 1
1 0 0
0@ 1A 0 0 1
1 0 0
0 1 0
0@ 1A[g]B 2
1 0 0
0 0 1
0 1 0
0@ 1A 1 0 0
0 ÿ1 1
0 ÿ1 0
0@ 1A 1 0 0
0 0 ÿ1
0 1 ÿ1
0@ 1AAll the matrices [g]B 2
have the form
1 0 0
0 j j
0 j j
0@ 1A:2. Let g 2 Sn. For all u, v in V and ë in F, we have
vg 2 V , v1 � v, (ëv)g � ë(vg), (u� v)g � ug � vg:
It remains to check (2) of De®nition 4.2. Let v 2 V and g, h 2 Sn. Assume®rst that gh 2 An. Then v(gh) � v; and (vg)h � v, since either vg � v � vh(if g, h 2 An) or vg � ÿv � vh (if g, h =2 An). Next, assume that gh =2 An.Then v(gh) � ÿv; and (vg)h � ÿv, since one of g, h is in An and theother is not. We have now checked all the conditions in De®nition 4.2, so Vis an FG-module.
3. Let A �0 1 0 0
ÿ1 0 0 0
0 0 0 ÿ1
0 0 1 0
0BB@1CCA and B �
0 0 1 0
0 0 0 1
ÿ1 0 0 0
0 ÿ1 0 0
0BB@1CCA.
Check that
A4 � I , B2 � A2, Bÿ1 AB � Aÿ1:
Hence r: aib j ! Ai Bj (0 < i < 3, 0 < j < 1) is a representation of Q8 overR. Let V � R4. By Theorem 4.4(1), V becomes an RQ8-module if we de®nevg � v(gr) for all v 2 V, g 2 Q8. If we put
402 Representations and characters of groups
v1 � (1, 0, 0, 0), v2 � (0, 1, 0, 0), v3 � (0, 0, 1, 0), v4 � (0, 0, 0, 1),
then for all i, via and vib are as required in the question.
4. You may ®nd it helpful ®rst to check that if
M �
0 1
1 0
1
. ..
1
0BBBBB@
1CCCCCAthen MA is obtained from A by swapping the ®rst two rows, and AM isobtained from A by swapping the ®rst two columns. To solve the exercise,let g be the permutation in Sn which has the property that for all i,
row i of B � row ig of A:
Let P be the n 3 n matrix ( pij) de®ned by
pij �1, if j � ig,
0, if j 6� ig:
8<:Then P is a permutation matrix, and the ij-entry of PA isXn
k�1
pikakj � aig, j:
Hence PA � B.If C is a matrix obtained from A by permuting the columns, then
C � AQ for some permutation matrix Q; the proof is similar to that for therows.
Chapter 5
1. It is easy to verify that V is an FG-module. Let U be a non-zero FG-submodule of V, and let (á, â) 2 U with (á, â) 6� (0, 0). Then(á, â) � (á, â)a � (á � â, á � â) 2 U, and (á, â) ÿ (á, â)a �(á ÿ â, â ÿ á) 2 U. Since at least one of á � â and á ÿ â is non-zero, wededuce that (1, 1) or (1, ÿ1) belongs to U. Hence the FG-submodules of Vare {0}, sp ((1, 1)), sp ((1, ÿ1)) and V.
2. Suppose that r has degree n and r is reducible. Then r is equivalent to arepresentation ô of the form
ô: g !X g 0
Y g Z g
0@ 1A (g 2 G)
where Xg is a k 3 k matrix and 0 , k , n. Then ó is equivalent to ô, sinceó is equivalent to r. Therefore ó is reducible.
3. Let G � D12 and let r1, . . . , r4 be the representations of G de®ned inExercise 3.5. First consider the FG-module V � F2, where vg � v(gr1) for
0
0
Chapter 5 403
v 2 V, g 2 G. Suppose that U is a non-zero FG-submodule of V. Then U isan FH-module, where H is the subgroup {1, b}. By the solution to Exercise1, either (1, 1) or (1, ÿ1) lies in U. Since (1, 1) and (1, 1)a are linearlyindependent, and also (1, ÿ1) and (1, ÿ1)a are linearly independent, itfollows that dim U > 2. Consequently U � V and so V is irreducible. (SeeExample 5.5(2) for an alternative argument.) Therefore r1 is irreducible;also r4 is irreducible, since r1 and r4 are equivalent.
Now let V � F2 with vg � v(gr2) for v 2 V, g 2 G. Then (1, 1)a �ÿ(1, 1) � (1, 1)b. Hence sp ((1, 1)) is an FG-submodule of V and r2 isreducible.
Finally, r3 is irreducible, by an argument similar to that for r1.
4. (a) It is easy to check the given relations. Using the relations, we may writeevery element of G in the form
aib jck (0 < i < 2, 0 < j < 2, 0 < k < 1):
Thus jGj < 18. However, it is clear that |ka, bl| � 9 and ka, bl 6� G.Hence, by Lagrange's Theorem, jGj is a multiple of 9 and jGj. 9.Therefore jGj � 18.
(b) Let
A � å 0
0 åÿ1
� �, B � ç 0
0 çÿ1
� �, C � 0 1
1 0
� �:
Check that A3 � B3 � C2 � I , AB � BA, Cÿ1 AC � Aÿ1 andCÿ1 BC � Bÿ1. Hence r is a representation (compare Example 1.4).
(c) For every element g of ka, bl, there exists a cube root î of unity suchthat
gr � î 0
0 îÿ1
� �:
But there are only three distinct cube roots of unity, so there existdistinct g1, g2 2 ka, bl with g1r � g2r. Therefore r is never faithful.
(d) Let V � C2 be the CG-module obtained by de®ning vg � v(gr) for allv 2 C2, g 2 G. If U is a non-zero CG-submodule of V, then U is a CH-submodule, where H is the subgroup {1, c}. Hence either (1, 1) or(1, ÿ1) lies in U, by the solution to Exercise 1; accordingly, let u be(1, 1) or (1, ÿ1) (so that u 2 U). Now u and ua are linearly independentunless å � 1, and u and ub are linearly independent unless ç � 1. Hence,if either å 6� 1 or ç 6� 1 then dim U � 2 and so r is irreducible. On theother hand, if å � ç � 1 then sp ((1, 1)) is a CG-submodule of V, so r isreducible.
5. Let V � {0} and let 0g � 0 for all g 2 G. Then V is neither reducible norirreducible.
Chapter 6
1: (a) xy � ÿ2:1ÿ a3 � ab� 3a2b� 2a3b,
yx � ÿ2:1ÿ a3 � b� 2a2b� 3a3b,
x2 � 4:1� a2 � 4a3:
404 Representations and characters of groups
(b) az � ab � a3b � a2ba � ba � za, and bz � 1 � a2 � zb. Henceaib jz � zaib j for all i, j and so gz � zg for all g 2 G. If r 2 CG thenr � P g ë g g with ë g 2 C, so rz �Pë g gz �Pë gzg � zr.
2. Let C2 3 C2 � ka, b: a2 � b2 � 1, ab � bal. Relative to the basis 1, a, b,ab of F(C2 3 C2), the regular representation r is given by
1r �1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
0BBB@1CCCA, ar �
0 1 0 0
1 0 0 0
0 0 0 1
0 0 1 0
0BBB@1CCCA,
br �0 0 1 0
0 0 0 1
1 0 0 0
0 1 0 0
0BBB@1CCCA, (ab)r �
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
0BBB@1CCCA:
3. No: let G � ka: a2 � 1l, and take r � 1 � a, s � 1 ÿ a.
4. (a) As g runs through G, so do gh and hg. Hence ch � hc � c.(b) c2 � c
Ph2G h �Ph2G ch � jGjc.
(c) All the entries in [W]B are 1 (compare the solution to Exercise 2). Thereason is that for all i, j, there exists a unique h in G such thatgih � gj.
5. Note ®rst that if u is an element of a vector space, and u � u � u, thenu � 0. Now 0r � (0 � 0)r � 0r � 0r, and v0 � v(0 � 0) � v0 � v0; hence0r � v0 � 0.
Let V be the trivial FG-module and let 0 6� v 2 V and 1 6� g 2 G. If r �1 ÿ g, then vr � 0 and neither v nor r is 0.
6. Let v1 � 1 � ù2a � ùa2 and v2 � b � ù2ab � ùa2b. Check that v1a � ùv1,v2a � ù2v2, v1b � v2 and v2b � v1. Hence W is a CG-submodule of CG.Use the argument of either Example 5.5(2) or the solution to Exercise 5.3to prove that W is irreducible.
Chapter 7
1. For all u1, u2 2 U, ë 2 F and g 2 G, we have
(u1 � u2)Wö � (u1W� u2W)ö � u1(Wö)� u2(Wö),
(ëu1)Wö � (ë(u1W))ö � ë(u1(Wö)),
(u1 g)Wö � ((u1W)g)ö � ((u1W)ö)g � (u1(Wö))g:
2. Let a � (1 2 3 4 5) and let v1, . . . , v5 be the natural basis for thepermutation module for G over F. Then
W: ë1v1 � : : :� ë5v5 ! ë11� ë2a� ë3a2 � ë4a3 � ë5a4
is the required FG-isomorphism. (Note that viW � ai, so (via)W � vi�1W �ai�1 � (viW)a; hence W is an FG-homomorphism.)
3. It is easy to show that V0 is an FG-submodule of V. Let x 2 G. Then
Chapter 7 405
Xg2G
vxg �Xg2G
vg �Xg2G
vgx:
Hence (vx)W � vW � (vW)x; noting that VW � V0, we see that W is an FG-homomorphism from V to V0.
If v 2 V0 then (v=jGj)W � v; hence W is surjective.
4. Suppose that ö: V! W is an FG-isomorphism. Let g 2 G. For all v 2 V0,(vö)g � (vg)ö � vö, and so V0ö � W0. For all w 2 W0, (wöÿ1)g �(wg)öÿ1 � wöÿ1, so W0öÿ1 � V0. Hence the function ö, restricted to V0,is an FG-isomorphism from V0 to W0.
5. No: let v1, . . . , v4 be the natural basis of the permutation module V for Gover F. In the notation of Exercise 3,
V0 � sp (v1 � v2, v3 � v4) and (FG)0 � spXg2G
g
!:
Since V0 and (FG)0 have different dimensions, it follows from Exercise 4that V and FG are not isomorphic FG-modules.
6. (a) W is easily seen to be a linear transformation. Also
(á1� âx)xW � (â1� áx)W � (âÿ á)(1ÿ x) � (áÿ â)(1ÿ x)x
� (á1� âx)Wx:
Hence W is an FG-homomorphism.(b) (á ÿ â)(1 ÿ x)W � ((á ÿ â) ÿ (â ÿ á))(1 ÿ x) � 2(á ÿ â)(1 ÿ x). Hence
W2 � 2W.(c) Let B be the basis 1 ÿ x, 1 � x.
Chapter 8
1. V � sp (ÿùv1 � v2) � sp (ÿù2v1 � v2), where ù � e2ði=3. (Find eigenvectorsfor x.)
2. Let G � {1, a, b, ab} � C2 3 C2 (so a2 � b2 � 1, ab � ba). Then
RG � sp (1� a� b� ab)� sp (1� aÿ bÿ ab)� sp (1ÿ a� bÿ ab)
� sp (1ÿ aÿ b� ab):
3. Let G be any group, and let V be a 2-dimensional vector space over C withbasis v1, v2. De®ne vg � v for all v 2 V, g 2 G; this makes V into a CG-module. If we let
W: ëv1 � ìv2 ! ëv2 (ë, ì 2 C)
then W is a CG-homomorphism from V to V, and Ker W � Im W � sp (v2).
4. Suppose r is reducible. Then by Maschke's Theorem, r is equivalent to arepresentation ó of the form
gó � ë g 0
0 ì g
� �(ë g, ì g 2 C):
Then (gó)(hó) � (hó)(gó) for all g, h 2 G, since all diagonal matrices
406 Representations and characters of groups
commute with each other; hence also (gr)(hr) � (hr)(gr) for all g, h 2 G.This is a contradiction. Therefore r is irreducible.
5. U � sp ((1, 0)) is the only 1-dimensional CG-submodule of V, so there is noCG-submodule W of V with V � U � W.
6. (1) It is straightforward to verify for [ , ] the axioms of a complex innerproduct. For example, if u 6� 0 then (ux, ux) . 0 for all x 2 G, so[u, u] . 0. Also
[ug, vg] �Xx2G
(ugx, vgx) �Xx2G
(ux, vx) � [u, v]:
(2) It is easy to prove that U? is a subspace of V. Let v 2 U? and g 2 G.Then for all u 2 U,
[u, vg] � [ugÿ1, vggÿ1] by part (1)
� [ugÿ1, v] � 0 since ugÿ1 2 U :
Therefore vg 2 U?, and so U? is a CG-submodule of V.
(3) Let W � U?. Then V � U � W, and W is a CG-submodule of V bypart (2).
7. We know that the regular CG-module CG is faithful (Proposition 6.6). LetCG � U1 � . . . � Ur, where U1, . . . , Ur are irreducible CG-submodules ofCG. Then there exist i 2 {1, . . . , r} and g 2 G such that ug 6� u for someu 2 Ui (otherwise vg � v for all v 2 CG). De®ne
K � fx 2 G: vx � v for all v 2 Uig:Check that K is a normal subgroup of G; also K 6� G since g =2 K. Since Gis simple, we must therefore have K � {1}. This means that Ui is a faithfulirreducible CG-module.
Chapter 9
1. Let C2 � ka: a2 � 1l. Irreducible representations r1, r2:
1r1 � ar1 � (1); 1r2 � (1), ar2 � (ÿ1):
Let C3 � kb: b3 � 1l and let ù � e2ði=3. Irreducible representations r1, r2,r3:
1r1 � br1 � b2r1 � (1);
bir2 � (ùi);
bir3 � (ù2i):
Let C2 3 C2 � {(1, 1), (x, 1), (1, y), (x, y)}, where x2 � y2 � 1. Irreduciblerepresentations r1, r2, r3, r4:
Chapter 9 407
gr1 � (1) for all g 2 C2 3 C2;
(xi, y j)r2 � (ÿ1) j;
(xi, y j)r3 � (ÿ1)i;
(xi, y j)r4 � (ÿ1)i� j:
2. Let C4 3 C4 � k(x, 1), (1, y): x4 � y4 � 1l.(a) r: (xi, yj)! (ÿ1)i.(b) If g1 � (x2, 1) and g2 � (1, y2) then g1, g2 and g1 g2 all have order 2.
Since (g1 g2)ó � (g1ó)(g2ó) for all representations ó, we cannot have(g1 g2)ó � g1ó � g2ó � (ÿ1).
3. For 1 < j < r, let gj generate Cn j, and let å j � e2ði=n j . Then
r: (gi11 , : : : , gir
r )!åi1
1
. ..
åi rr
0BB@1CCA
is a faithful representation of Cn13 . . . 3 Cnr
of degree r.Yes: if r � 2, n1 � 2, n2 � 3, then
ó : (gi11 , gi2
2 )! (åi11 å
i22 )
is a faithful representation of degree 1 , r.
4. Check that A4 � B2 � I, Bÿ1AB � Aÿ1 when A � ar and B � br. Hence rgives a representation; similarly for ó.
If M(gr) � (gr)M for g � a and for g � b, then M � ëI for some ë 2 C.Hence r is irreducible (Corollary 9.3).
Notice that the matrix5 ÿ6
4 ÿ5
� �commutes with gó for all g 2 G; hence ó is reducible (Corollary 9.3).
5. Let z �Pg2G g. Then xz � z � zx for all x 2 G. Hence z 2 Z(CG), and theresult follows from Proposition 9.14.
6. (a) Clearly a commutes with a � aÿ1. Also bÿ1(a � aÿ1)b � aÿ1 � a, so bcommutes with a � aÿ1.
(b) Check that w(a � aÿ1) � ÿw for all w 2 W.
7. (a) Let Cn � kx: xn � 1l. Then r: xj ! (e2ði j=n) is a faithful irreduciblerepresentation of Cn.
(b) r: a ! 0 1
ÿ1 0
� �, b ! 1 0
0 ÿ1
� �gives a faithful irreducible representation of D8 (see Example 5.5(2)).
(c) The centre of C2 3 D8 is isomorphic to C2 3 C2, so is not cyclic.Therefore Proposition 9.16 shows that C2 3 D8 has no faithfulirreducible representation.
(d) Let C3 � kx: x3 � 1l and let ù � e2ði=3. Check that
r: (x, a)! 0 ùÿù 0
� �, (x, b)! ù 0
0 ÿù� �
0
0
408 Representations and characters of groups
gives a representation of C3 3 D8. It is irreducible (see for exampleExercise 8.4) and faithful.
Chapter 10
1. Let V � sp (P
g2G g). Then V is a trivial CG-submodule of CG. Nowsuppose that U is an arbitrary trivial CG-submodule of CG, so U � sp (u)for some u. Then ug � u for all g 2 G, so |G|u � u(
Pg2G g) �
(P
g2G g)u 2 V. Thus U � V, and so CG has exactly one trivial CG-submodule, namely V.
2. Let G � kx: x4 � 1l. Then
CG � sp (1� x� x2 � x3)� sp (1� ixÿ x2 ÿ ix3)
� sp (1ÿ x� x2 ÿ x3)� sp (1ÿ ixÿ x2 � ix3):
3. Let
u1 � 1� a� a2 � a3 ÿ bÿ abÿ a2bÿ a3b,
u2 � 1ÿ a� a2 ÿ a3 � bÿ ab� a2bÿ a3b,
u3 � 1ÿ a� a2 ÿ a3 ÿ b� abÿ a2b� a3b:
4. We decompose CG as a direct sum of irreducible CG-submodules. Let
v0 � 1� a� a2 � a3, v1 � 1� iaÿ a2 ÿ ia3,
v2 � 1ÿ a� a2 ÿ a3, v3 � 1ÿ iaÿ a2 � ia3
(compare the solution to Exercise 2). For 0 < j < 3, let wj � bv j. Then, asin Example 10.8(2), the subspaces sp (v0, w0), sp (v1, w3), sp (v2, w2) andsp (v3, w1) are CG-submodules of CG. We have
sp (v0, w0) � U0 � U1, sp (v2, w2) � U2 � U3,
where Ui � sp (ui) (0 < i < 3) and u1, u2, u3 are as in the solution toExercise 3, while u0 �
Pg2G g.
Let U4 � sp (v1, w3), U5 � sp (v3, w1). As in Example 5.5(2) (or seeExercise 8.4), U4 and U5 are irreducible CG-modules. Moreover U4 � U5,since there is a CG-isomorphism sending v1 ! w1, w3 ! v3.
Theorem 10.5 now shows that there are exactly ®ve non-isomorphicirreducible CG-modules, namely U0, U1, U2, U3 and U4. Therefore everyirreducible representation of D8 over C is equivalent to precisely one of thefollowing:
r0: a! (1), b! (1)
r1: a! (1), b! (ÿ1)
r2: a! (ÿ1), b! (1)
r3: a! (ÿ1), b! (ÿ1)
r4: a! i 0
0 ÿi
!, b! 0 1
1 0
!:
Chapter 10 409
5. Let W: U1 ! U2 be a CG-isomorphism. For ë 2 C, de®ne the functionöë: U1 ! V by
öë: u! u� ëuW (u 2 U1):
Then öë is easily seen to be a CG-homomorphism; moreover,
u 2 Keröë , u� ëuW � 0, u � 0,
since the sum U1 � U2 is direct. Thus U1 � Imöë. It is easy to check thatif ë 6� ì then Imöë 6� Imöì. Therefore we have constructed in®nitely manyCG-submodules Imöë of the required form.
6. V is irreducible, either by the method of Example 5.5(2) or by Exercise 8.4.Let u1 � 1 ÿ ia ÿ a2 � ia3, u2 � b ÿ iab ÿ a2b � ia3b. Then sp (u1, u2) is
a CG-submodule of CG which is isomorphic to V. A CG-isomorphism isgiven by v1 ! u1, v2 ! u2.
Chapter 11
1. Since G is non-abelian, not all the dimensions are 1 (see Proposition 9.18).Hence, by Theorem 11.12, the dimensions are 1, 1, 2.
2. Compare Example 11.13 to see that the possible answers are 112, 182, 1422
and 133 (where 112 means twelve 1s, 182 means eight 1s and one 2, and soon). It will be shown later (Exercises 15.4, 17.3) that 182 cannot occur.
By Exercise 5.3, D12 has at least two inequivalent irreduciblerepresentations of degree 2. Hence the answer for D12 is 1422.
3. For each g 2 G, de®ne ö g: CG! CG by rö g � gr (r 2 CG). Then{ö g: g 2 G} gives a basis of HomCG(CG, CG) (compare the proof ofProposition 11.8).
4. Let v1, . . . , vn be the natural basis of V. Then sp (v1 � . . . � vn) is theunique trivial CG-submodule of V (compare Exercise 10.1). Hence byCorollary 11.6, dim (HomCG(V, U)) � 1.
5. Let v1, w2 be the basis of U3 described in Example 10.8(2). De®ne W1 andW2 by rW1 � v1 r, rW2 � w2 r (r 2 CG). Then W1, W2 is a basis ofHomCG (CG, U3), by the proof of Proposition 11.8. Also, de®ne ö1, ö2 byuö1 � u, uö2 � bu (u 2 U3). Then ö1, ö2 is a basis ofHomCG (U3, CG).
6. Let V � X1 � . . . � Xr and W � Y1 � . . . � Ys, where each Xa and each Yb
is an irreducible CG-module. Then by (11.5)(3) and Proposition 11.2,dim (HomCG (V, W)) is equal to the number of ordered pairs (a, b) such thatX a � Yb. This, in turn, equalsXk
i�1
jf(a, b): Xa � Yb � Vigj:
Now the number of integers a with X a � Vi is dim (HomCG (V , Vi)) � di,by Corollary 11.6, and similarly the number of integers b with Yb � Vi isei. Therefore, dim (HomCG (V, W)) �Pk
i�1diei.
410 Representations and characters of groups
Chapter 12
1. Assume that g, h 2 CG(x). Then gx � xg and hx � xh, so hÿ1x � xhÿ1 andghÿ1x � gxhÿ1 � xghÿ1; thus ghÿ1 2 CG(x). Also 1x � x1, so 1 2 CG(x).Therefore CG(x) is a subgroup of G.
If z 2 Z(G) then zg � gz for all g 2 G, so zx � xz and z 2 CG(x).
2. Note that x 2 CG(g), x 2 CG(gz). Now the required result follows fromTheorem 12.8.
3. (a) (1 2)G � {(i j): 1 < i , j < n} and this set has size (n2 ). The centralizer
CG((1 2)) consists of all elements x and (1 2)x, where x is a permutation®xing 1 and 2. Thus |CG((1 2))| � 2´(n ÿ 2)!, in agreement with Theorem12.8 (since (n
2 ) � n!=(2:(nÿ 2)!)).(b) (1 2 3)G consists of all 3-cycles (i j k). There are (n
3 ) choices for thenumbers i, j, k (unordered). Each choice gives exactly two 3-cycles,namely (i j k) and (i k j).
(1 2)(3 4)G consists of all permutations of the form (i j)(k l ) with i, j,k, l distinct. There are (n
4 ) choices for the numbers i, j, k, l (unordered),and three permutations for each choice, namely (i j)(k l ), (i k)( j l ) and(i l )( j k).
(c) Every element of (1 2 3)(4 5 6)G has the form (1 i j)(k l m), with i, j, k, l,m distinct. There are ®ve choices for i; then four choices for j; then wecan make two different 3-cycles (k l m) and (k m l ) from the remainingnumbers. This gives 5 . 4 . 2 � 40 elements in all.
The elements of (1 2)(3 4)(5 6)G have the form (1 i)( j k)(l m). Thereare ®ve choices for i; then we can make three permutations ( j k)(l m),( j l )(k m) and ( j m)(k l ) of cycle shape (2, 2) from the remainingnumbers. Hence |(1 2)(3 4)(5 6)G | � 5 . 3 � 15.
The sizes of the conjugacy classes of S6 are given in the followingtable:
Cycle-shape (1) (2) (3) (22) (4) (3, 2) (5) (23) (32) (4, 2) (6)Class size 1 15 40 45 90 120 144 15 40 90 120
4. An element x of cycle-shape (5) has CS6(x) � kxl (note that |xS6 | � 144 and
use Theorem 12.8). Hence by Proposition 12.17, x A6 6� xS6 . For elements gof other cycle-shapes, gA6 � gS6 .
5. By Example 12.18(2), the conjugacy classes of A5 have sizes 1, 12, 12, 15,20. If H is a normal subgroup of A5 then jH j divides 60, and 1 2 H, andH is a union of conjugacy classes of A5. Hence jH j � 1 or 60; therefore A5
is simple.
6. We have Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l. The conjugacy classesof Q8 are
f1g, fa2g, fa, a3g, fb, a2bg, fab, a3bg,and a basis of Z(CQ8) is
1, a2, a� a3, b� a2b, ab� a3b:
7. The class equation gives
Chapter 12 411
jGj � jZ(G)j �X
xi=2Z(G)
jxGi j:
(a) For xi =2 Z(G), |xGi | divides pn and |xG
i | 6� 1 by Theorem 12.8 and (12.9).Therefore p divides |xG
i |. Hence p divides |Z(G)|, so Z(G) 6� f1g.(b) If no conjugacy class of G has size p, then p2 divides |xG
i | for allxi =2 Z(G). If, in addition, |G| > p3, then by the class equation, p2 divides|Z(G)|. This is a contradiction.
Chapter 13
1. The characters ÷i of ri (i � 1, 2) are as follows:
Also Ker r1 � {1, a3} and Ker r2 � {1, a2, a4}.
2. Let C4 � kx: x4 � 1l. The irreducible characters ÷1, . . . , ÷4 of C4 are asfollows:
We have ÷reg � ÷1 � ÷2 � ÷3 � ÷4.
3. Since ÷(g) � |®x (g)|, we have ÷((1 2)) � 5 and ÷((1 6)(2 3 5)) � 2.
4. If ÷ is a non-zero character which is a homomorphism, then÷(1) � ÷(12) � (÷(1))2, so ÷(1) � 1.
5. Let r be a representation with character ÷. Then zr � ëI for some ë 2 C,by Proposition 9.14. Thus, for all g in G, (zg)r � (zr)(gr) � ë(gr), andhence ÷(zg) � ë÷(g). Moreover, I � 1r � zmr � (zr)m � ëmI, so ëm � 1.
6. Let r be a representation with character ÷. If g 2 Z(G) then gr � ëI forsome ë 2 C, by Proposition 9.14. Conversely, if gr � ëI for some ë 2 C,then (gr)(hr) � (hr)(gr) for all h 2 G, and hence g 2 Z(G) since r isfaithful.
We have now proved that gr � ëI for some ë 2 C if and only ifg 2 Z(G). The required result now follows from Theorem 13.11(1).
7. (a) For all g, h 2G, det ((gh)r) � det ((gr)(hr)) � det (gr) det (hr). Henceg! (det (gr)) is a representation of G over C of degree 1, and so ä isa linear character of G.
(b) G=Ker ä � Imä by Theorem 1.10, and Im ä is a subgroup of the
1 a3 a, a5 a2, a4 b, a2b, a4b ab, a3b, a5b
÷1 2 2 ÿ1 ÿ1 0 0÷2 2 0 0 2 0 ÿ2
1 x x2 x3
÷1 1 1 1 1÷2 1 i ÿ1 ÿi÷3 1 ÿ1 1 ÿ1÷4 1 ÿi ÿ1 i
412 Representations and characters of groups
multiplicative group C� of non-zero complex numbers, which isabelian. Therefore G=Ker ä is abelian.
(c) Im ä is a ®nite subgroup of C�, hence is cyclic, by Exercise 1.7. Alsoÿ1 2 Im ä, so Im ä has even order. Hence Im ä contains a subgroup Hof index 2. It is easy to check that {g 2 G: ä(g) 2 H} is a normalsubgroup of G of index 2.
8. Let r be the regular representation of G, and de®ne the character ä as inExercise 7. By Exercise 1.8, G has an element x of order 2. Order thenatural basis g1, . . . , g2k of CG so as to obtain a basis B in which gand gx are adjacent for all g 2 G. Then
[x]B �
0 1
1 0
0 1
1 0
. ..
0BBBBB@
1CCCCCA:There are k blocks (0
110), and since k is odd, det ([x]B ) � (ÿ1)k � ÿ1. Thus
ä(x) � ÿ1. The required result now follows from Exercise 7.
9. Let V be a CG-module with character ÷. We may choose a basis B of Vso that [g]B is diagonal with all diagonal entries �1 (see (9.10)). Saythere are r entries 1 and s entries ÿ1. De®ne ä as in Exercise 7. If s isodd then ä(g) � ÿ1, and G has a normal subgroup of index 2 by Exercise7. And if s is even then ÿs � s mod 4, so÷(g) � r ÿ s � r � s � ÷(1) mod 4.
10. As x 6� 1, we have ÷reg(x) 6� ÷reg(1) (see Proposition 13.20), so ÷i(x) 6� ÷i(1)for some irreducible character ÷i of G, by Theorem 13.19.
Chapter 14
1. Using Proposition 14.5(2), we obtain
h÷, ÷i � 3 . 3
24� (ÿ1)(ÿ1)
4� 0� 3 . 3
8� (ÿ1)(ÿ1)
4� 2,
h÷, øi � 3 . 3
24� (ÿ1) . 1
4� 0� 3 . (ÿ1)
8� (ÿ1)(ÿ1)
4� 0,
hø, øi � 3 . 3
24� 1 . 1
4� 0� (ÿ1)(ÿ1)
8� (ÿ1)(ÿ1)
4� 1:
Hence ø is irreducible by Theorem 14.20 (but ÷ is not).
2. Let ÷i be the character of ri (i � 1, 2, 3). The values of these charactersare as follows:
0
0
Conjugacy class 1 a2 a, a3 b, a2b ab, a3b
÷1 2 ÿ2 0 0 0÷2 2 ÿ2 0 0 0÷3 2 2 0 0 ÿ2
Chapter 14 413
By Theorem 14.21, r1 and r2 are equivalent, but r3 is not equivalent to r1
or r2.
3. The representations r and ó have the same character, by Proposition 13.2;hence r and ó are equivalent, by Theorem 14.21, and this gives therequired matrix T.
4. Let ÷1 be the trivial character of G. Then
h÷, ÷1i � 1
jGjXg2G
÷(g):
Since ÷(1) . 0 and by hypothesis ÷(g) > 0 for all g 2 G, we havek÷, ÷1l 6� 0. As ÷ 6� ÷1, Theorem 14.17 shows that ÷ is reducible.
5. We have
h÷reg, ÷i � 1
jGjXg2G
÷reg(g)÷(g):
But ÷reg(g) is |G| if g � 1 and is 0 if g 6� 1. Hence k÷reg, ÷l � ÷(1).
6. This follows at once from Exercise 11.4 and Theorem 14.24.
7. Recall that hø, øi �Pki�1d2
i . Hence if kø, øl � a where a � 1, 2 or 3, thenexactly a of the integers di are 1 and the rest are 0. If kø, øl � 4, theneither exactly four of the di are 1, or exactly one of the di is 2; the rest are0.
8. No: let G � C2 and ÷ � ÷reg, the regular character of C2.
Chapter 15
h÷, ÷1i � 16(19 . 1� 3 . (ÿ1) . 1� 2 . (ÿ2) . 1) � 2,1:
h÷, ÷2i � 16(19 . 1� 3 . (ÿ1)(ÿ1)� 2 . (ÿ2) . 1) � 3,
h÷, ÷3i � 16(19 . 2� 0� 2 . (ÿ2)(ÿ1)) � 7:
Hence ÷ � 2÷1 � 3÷2 � 7÷3. Since all the coef®cients here are non-negativeintegers, it follows that ÷ is a character of S3.
2. By the method used in the solution to Exercise 1, we obtain
ø1 � 16÷1 � 1
6÷2 � 1
3÷3,
ø2 � 12÷1 ÿ 1
2÷2,
ø3 � 13÷1 � 1
3÷2 ÿ 1
3÷3:
3. We ®nd that ø � ÿ÷2 � ÷3 � ÷5 � 2÷6. Since the coef®cient of ÷2 is anegative integer, ø is not a character of G.
4. (a) For all groups G, if x 2 G then the subgroup generated by x and theelements of Z(G) is abelian (since the elements of Z(G) commute withpowers of x). Hence, if G � Z(G) [ Z(G)x then G � Z(G). Therefore thecentre of a group G never has index 2 in G.Every abelian group of order 12 has 12 conjugacy classes. If |G| � 12
414 Representations and characters of groups
and G is non-abelian, then |Z(G)| divides 12 and |Z(G)| 6� 6 or 12, so|Z(G)| < 4. Therefore, at most 4 conjugacy classes of G have size 1 (see(12.9)); the remaining conjugacy classes have size at least 2, so therecannot be as many as 9 conjugacy classes in total.
(b) Since the number of irreducible representations is equal to the numberof conjugacy classes, it follows from the solution to Exercise 11.2 andpart (a) that G has 4, 6 or 12 conjugacy classes. If G is abelian (e.g.G � C4 3 C3) then G has 12 conjugacy classes; if G � D12 then G has6 conjugacy classes (see (12.12)); and if G � A4 then G has 4 conjugacyclasses (see Example 12.18(1)).
Chapter 16
1. Let C2 3 C2 � {(1, 1), (x, 1), (1, y), (x, y): x2 � y2 � 1}. The charactertable of C2 3 C2 is (cf. Exercise 9.1)
2. The last row of the character table is (cf. Example 16.5(2))
3. The complete character table of G is
The two unknown degrees ÷3(1), ÷4(1) are 1, 2 sinceP4
i�1(÷i(1))2 � 10.
Because g4 has order 2, Corollary 13.10, together with the relationP4i�1÷i(1)÷i(g4) � 0, gives the values on g4. Then
P4i�1÷i(g2)÷i(g4) � 0
gives ÷3(g2) � 1; similarly ÷3(g3) � 1. Finally,P4
i�1÷i(1)÷i(g2) � 0 gives
÷4(g2) � (ÿ1ÿp5)=2; similarly ÷4(g3) � (ÿ1�p5)=2.
(1, 1) (x, 1) (1, y) (x, y)
÷1 1 1 1 1÷2 1 1 ÿ1 ÿ1÷3 1 ÿ1 1 ÿ1÷4 1 ÿ1 ÿ1 1
g1 g2 g3 g4 g5
÷5 2 ÿ2 0 0 0
gi g1 g2 g3 g4
|CG(gi)| 10 5 5 2
÷1 1 1 1 1÷2 2 (ÿ1�p5)=2 (ÿ1ÿp5)=2 0÷3 1 1 1 ÿ1÷4 2 (ÿ1ÿp5)=2 (ÿ1�p5)=2 0
Chapter 16 415
4. (a)P5
i�1÷i(g1)÷i(g2) � 0 gives 3� 3æ� 3æ � 0, andP5
i�1÷i(g2)÷i(g2) � 7gives 3� 2ææ � 7.Hence æ � (ÿ1� i
p7)=2.
(b) The column which corresponds to the conjugacy class containing gÿ12
has values which are the complex conjugates of those in the column ofg2 (see Proposition 13.9(3)); since æ is non-real, this is a differentcolumn of the character table of G.
5. Let g 2 G. By the column orthogonality relations applied to the columncorresponding to g, we have
Pki�1÷i(g)÷i(g) � jCG(g)j. This number is
equal to |G| if and only if CG(g) � G, which occurs if and only ifg 2 Z(G).
6. The matrix C is obtained from C by rearranging the columns (seeProposition 13.9(3)). Therefore det C � �det C; if det C � det C then det Cis real, and if det C � ÿdet C then det C is purely imaginary.
By the column orthogonality relations, CtC is the k 3 k diagonalmatrix whose diagonal entries are |CG(gi)| (1 < i < k). Hencejdet Cj2 � Q jCG(gi)j.
If G � C3 then det C � �i3p
3. (The sign depends upon the ordering ofrows and columns.)
Chapter 17
1. (a) The conjugacy classes of Q8 are f1g, fa2g, fa, a3g, fb, a2bg andfab, a3bg.
(b) G9 � f1, a2g and G=G9 � fG9, G9a, G9b, G9abg � C2 3 C2. Thecharacter table of C2 3 C2 is given in the solution to Exercise 16.1.Hence the linear characters of G are
(c) Using the column orthogonality relations, the last irreducible character ofG is
The character table of Q8 is the same as that of D8.
2. It is easy to see that a7 � b3 � 1. Use Proposition 12.13 to see quickly thatbÿ1ab � a2.(a) Using the relations, every element of G has the form ambn with
gi 1 a2 a b ab|CG(gi)| 8 8 4 4 4
÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1
1 a2 a b ab
÷5 2 ÿ2 0 0 0
416 Representations and characters of groups
0 < m < 6, 0 < n < 2; hence jGj < 21. But a has order 7 and b hasorder 3, so 21 divides jGj by Lagrange's Theorem. Therefore jGj � 21.
(b) The conjugacy classes of G are f1g, fa, a2, a4g, fa3, a5, a6g,famb: 0 < m < 6g and famb2: 0 < m < 6g.
(c) First, G9 � kal, so we get three linear characters of G:
where ù � e2ði=3. To ®nd the two remaining irreducible characters ÷4
and ÷5, we note that a is not conjugate to aÿ1; therefore for someirreducible character ÷, we have ÷(a) 6� ÷(a) (see Corollary 15.6). Hence÷4 and ÷5 must be complex conjugates of each other. Applying thecolumn orthogonality relations, we obtain
where á � (ÿ1� ip
7)=2.
3. The number of linear characters of G divides jGj by Theorem 17.11. Nowconsult the solution to Exercise 11.2 to see that there are 3, 4 or 12 linearcharacters. If there are 12, then G is abelian (see Proposition 9.18), so G iscertainly not simple. If G has 3 or 4 linear characters then jG=G9j � 3 or 4and again G is not simple as G9 v G.
4. In the character table below, we have ÷1 � 1G, ÷2 � ÷, ÷3 � ÷2, ÷4 � ÷2÷3,÷5 � ö, ÷6 � ö÷; all of these are irreducible characters by Proposition17.14. The centralizer orders are obtained by using the orthogonalityrelations, and the class sizes jgG
i j come from the equationsjGj � jCG(gi)jjgG
i j (Theorem 12.8).
gi 1 a a3 b b2
|CG(gi)| 21 7 7 3 3
÷1 1 1 1 1 1÷2 1 1 1 ù ù2
÷3 1 1 1 ù2 ù
1 a a3 b b2
÷4 3 á á 0 0÷5 3 á á 0 0
gi g1 g2 g3 g4 g5 g6
|CG(gi)| 12 4 4 6 6 12|gG
i | 1 3 3 2 2 1
÷1 1 1 1 1 1 1÷2 1 ÿi i 1 ÿ1 ÿ1÷3 1 ÿ1 ÿ1 1 1 1÷4 1 i ÿi 1 ÿ1 ÿ1÷5 2 0 0 ÿ1 ÿ1 2÷6 2 0 0 ÿ1 1 ÿ2
Chapter 17 417
5. The normal subgroups of D8 are
D8 � Ker ÷1, hai � Ker ÷2,
ha2, bi � Ker ÷3, ha2, abi � Ker ÷4,
ha2i � Ker ÷2 \ Ker ÷3, f1g � Ker ÷5:
6. (a) Check that the given matrices satisfy the relevant relations:
å 0
0 åÿ1
!2n
� 1 0
0 1
!,
å 0
0 åÿ1
!n
� 0 1
å n 0
!2
,
0 1
å n 0
!ÿ1å 0
0 åÿ1
!0 1
å n 0
!� å 0
0 åÿ1
!ÿ1
:
Hence we have representations of T4n (cf. Example 1.4).(b) The representations in part (a) are irreducible unless å � �1, by Exercise
8.4. For å � e2ðir=2n, with r � 1, 2, . . . , n ÿ 1, we get n ÿ 1 irreduciblerepresentations, no two of which are equivalent, since they have distinctcharacters. Moreover G9 � ka2l, so jG=G9j � 4 and there are fourrepresentations of degree 1 (see Theorem 17.11). Now the sum of thesquares of the degrees of the irreducible representations we have foundso far is
(nÿ 1) . 22 � 4 . 12 � 4n:
Hence we have found all the irreducible representations, by Theorem11.12. (For further details on the representations of degree 1, see thesolution to Exercise 18.3; note that the structure of G=G9 depends uponwhether n is even or odd.)
7. (a) Check that the given matrices satisfy the relevant relations.(b) The given representations, for å � e2ðik=2n with 0 < k < n ÿ 1, are
irreducible (by Exercise 8.4), and inequivalent (consider the charactervalues on a2). Also G9 � kbl, so jG=G9j � 2n and there are 2nrepresentations of degree 1. The sum of the squares of the degrees ofthe irreducible representations we have found is
n . 22 � 2n . 12 � 6n,
so we have obtained all the irreducible representations.
8. (a) Check that the given matrices satisfy the relevant relations.(b) The given representations, for å � e2ði j=n with 0 < j < n ÿ 1, are
irreducible (by Exercise 8.4) and inequivalent (their characters aredistinct). Note that b2 does not belong to the kernel of any of theserepresentations.We get further representations by
a! ç 0
0 çÿ1
� �, b! 0 1
1 0
� �,
where ç is any (2n)th root of unity in C. For ç � e2ði j=2n with1 < j < n ÿ 1, these representations are irreducible and inequivalent.Moreover, they are not equivalent to any representation found earlier,since b2 is in the kernel of each of these representations.
418 Representations and characters of groups
Finally, G9 � ka2, b2l and G=G9 � C2 3 C2, so we get fourrepresentations of degree 1.
We have now found all the irreducible representations, since the sumof the squares of the irreducible representations given above is
n . 22 � (nÿ 1) . 22 � 4 . 12 � 8n:
Chapter 18
1. The character table of D8 is as shown.
(See Example 16.3(3) or Section 18.3.) Regarding D8 as the symmetrygroup of a square, take b to be a re¯ection in a diagonal of the square.Then ð takes the following values:
Hence ð � ÷1 � ÷3 � ÷5. (Compare Example 14.28(2), where we took b tobe a different re¯ection.)
2. Let ù � e2ði=6. Then ù � ùÿ1 � 1, ù2 � ùÿ2 � ù4 � ùÿ4 � ÿ1. Hence,using Section 18.3, the character table of D12 is as shown.
Character table of D8
gi 1 a2 a b ab|CG(gi)| 8 8 4 4 4
÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1÷5 2 ÿ2 0 0 0
1 a2 a b ab
ð 4 0 0 2 0
Character table of D12
gi 1 a3 a a2 b ab|CG(gi)| 12 12 6 6 4 4
÷1 1 1 1 1 1 1÷2 1 1 1 1 ÿ1 ÿ1÷3 1 ÿ1 ÿ1 1 1 ÿ1÷4 1 ÿ1 ÿ1 1 ÿ1 1÷5 2 ÿ2 1 ÿ1 0 0÷6 2 2 ÿ1 ÿ1 0 0
Chapter 18 419
Seven normal subgroups of D12 are G � Ker ÷1, kal � Ker ÷2,ka2, bl � Ker ÷3, ka2, abl � Ker ÷4, ka2l � Ker ÷3 \ Ker ÷4, ka3l � Ker ÷6 and{1} � Ker ÷5.
3. The n � 3 conjugacy classes of G are
f1g, fang, far, aÿrg(1 < r < nÿ 1), fa2 jb: 0 < j < nÿ 1g,fa2 j�1b: 0 < j < nÿ 1g:
We get n ÿ 1 irreducible characters ø j (1 < j < n ÿ 1) of G fromExercise 17.6 as follows:
where ù � e2ði=2n.The remaining four irreducible characters of G are linear. If n is odd,
then G=G9 � hG9bi � C4 and the linear characters are
If n is even, then G=G9 � C2 3 C2 and the linear characters are
Note that T4 � C4, T8 � Q8 and T12 is the Example in Section 18.4.
4. The 3n conjugacy classes of G are, for 0 < r < n ÿ 1,
fa2rg, fa2r b, a2rb2g, fa2r�1, a2r�1b, a2r�1b2g:We have G9 � hbi and G=G9 � hG9ai � C2n. Hence we get 2n linearcharacters ÷ j (0 < j < 2n ÿ 1), as shown. Exercise 17.7 gives us nirreducible characters øk (0 < k < n ÿ 1).
gi 1 an ar (1 < r < n ÿ 1) b ab|CG(gi)| 4n 4n 2n 4 4
ø j 2 2(ÿ1) j ùrj � ùÿrj 0 0
gi 1 an ar (1 < r < n ÿ 1) b ab
÷1 1 1 1 1 1÷2 1 ÿ1 (ÿ1)r i ÿi÷3 1 1 1 ÿ1 ÿ1÷4 1 ÿ1 (ÿ1)r ÿi i
gi 1 an ar (1 < r < n ÿ 1) b ab
÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 (ÿ1)r 1 ÿ1÷4 1 1 (ÿ1)r ÿ1 1
420 Representations and characters of groups
Observe that U6 � D6, U12 � T12 and U18 � D6 3 C3.
5. The 2n � 3 conjugacy classes of G are
f1g, fb2g, fa2r�1, aÿ2rÿ1b2g(0 < r < nÿ 1),
fa2s, aÿ2sg, fa2sb2, aÿ2sb2g(1 < s < (nÿ 1)=2),
fajbk : j even, k � 1 or 3g, and
fajbk : j odd, k � 1 or 3g:
Using Exercise 17.8, we get four linear characters ÷1, . . . , ÷4, n charactersø j (0 < j < n ÿ 1) of degree 2, and a further n ÿ 1 characters ö j
(1 < j < n ÿ 1) of degree 2, as shown below. For example, the charactertable of V24 is given at the top of p. 422.
Character table of U6n
gi a2r a2rb a2r�1
|CG(gi)| 6n 3n 2n
÷ j ù2 jr ù2 jr ù j(2r�1)
(0 < j < 2n ÿ 1)øk 2ù2kr ÿù2kr 0(0 < k < n ÿ 1)
Note: ù � e2ði=2n.
Character table of V8n
gi 1 b2 a2r�1 a2s a2sb2 b ab(0 < r < n ÿ 1) (1 < s < (nÿ 1)=2)
|CG(gi)| 8n 8n 4n 4n 4n 4 4
÷1 1 1 1 1 1 1 1÷2 1 1 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 1 1 ÿ1÷4 1 1 ÿ1 1 1 ÿ1 1ø j 2 ÿ2 ù2 j(2r�1) ù4 js ÿù4 js 0 0(0 < j < n ÿ 1) ÿùÿ2 j(2r�1) �ùÿ4 js ÿùÿ4 js
ö j 2 2 ù j(2r�1) ù2 js ù2 js 0 0(1 < j < n ÿ 1) �ùÿ j(2r�1) �ùÿ2 js �ùÿ2 js
Note: ù � e2ði=2n
Chapter 18 421
Chapter 19
h÷ø, öi � 1
jGjXg2G
÷(g)ø(g)ö(g) � 1
jGjXg2G
÷(g)ø(g)ö(g) � h÷, øöi:1:
Similarly, k÷ø, öl � kø, ÷öl.
2. k÷ø, 1Gl � k÷, øl, by Exercise 1. The result now follows from Proposition13.15 and (14.13).
3. Let V be a CG-module with character ÷. Since ÷ is not faithful, there exists1 6� g 2 G with vg � v for all v 2 V. By Proposition 15.5 there is anirreducible character ø of G such that ø(g) 6� ø(1). Let n be an integerwith n > 0. Then wg � w for all w 2 V . . . V (n factors). Henceö(g) � ö(1) for all irreducible characters ö for which k÷ n, öl 6� 0.Therefore k÷ n, øl � 0.
4. Note that (1 2 3 4 5)2 is conjugate to (1 3 4 5 2) in A5. Using Proposition19.14 we obtain
Then
÷S � ø1 � ø2 � 2ø3,
÷A � ø2 � ø4 � ø5,
öS � ø1 � ø3,
öA � ø4:
Character table of V24
gi 1 b2 a a3 a5 a2 a2b2 b ab|CG(gi)| 24 24 12 12 12 12 12 4 4
÷1 1 1 1 1 1 1 1 1 1÷2 1 1 1 1 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 ÿ1 ÿ1 1 1 1 ÿ1÷4 1 1 ÿ1 ÿ1 ÿ1 1 1 ÿ1 1ø0 2 ÿ2 0 0 0 2 ÿ2 0 0ø1 2 ÿ2 i
p3 0 ÿi
p3 ÿ1 1 0 0
ø2 2 ÿ2 ÿip
3 0 ip
3 ÿ1 1 0 0ö1 2 2 1 ÿ2 1 ÿ1 ÿ1 0 0ö2 2 2 ÿ1 2 ÿ1 ÿ1 ÿ1 0 0
1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)
÷S 15 0 3 0 0÷A 10 1 ÿ2 0 0öS 6 0 2 1 1öA 3 0 ÿ1 (1�p5)=2 (1ÿp5)=2
422 Representations and characters of groups
5. We have recorded ÷ as ÷5, ÷S as ÷4 and ÷A as ÷2, below. Since k÷i, ÷il � 1for i � 2, 4, 5, these characters are irreducible. The table also records thetrivial character ÷1, ÷3 � ÷2, ÷6 � ÷5 and ÷7 � ÷2÷5; these are irreducible byPropositions 13.15 and 17.14. Since G has seven conjugacy classes, thecharacter table is complete.
6. Taking D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l, the character table ofD6 3 D6 is as shown.
Chapter 20
1. (a) Regard D8 as the subgroup of S4 which permutes the four corners of asquare, as in Example 1.1(3). Take b to be the re¯ection in the axisshown:
Character table of G (cf. Exercise 27.2)
gi g1 g2 g3 g4 g5 g6 g7
|CG(gi)| 24 24 4 6 6 6 6
÷1 1 1 1 1 1 1 1÷2 1 1 1 ù ù2 ù2 ù÷3 1 1 1 ù2 ù ù ù2
÷4 3 3 ÿ1 0 0 0 0÷5 2 ÿ2 0 ÿù2 ÿù ù ù2
÷6 2 ÿ2 0 ÿù ÿù2 ù2 ù÷7 2 ÿ2 0 ÿ1 ÿ1 1 1
Note: ù � e2ði=3
Character table of D6 3 D6
(gi, hj) (1, 1) (a, 1) (b, 1) (1, a) (a, a) (b, a) (1, b) (a, b) (b, b)|CG(gi, hj)| 36 18 12 18 9 6 12 6 4
÷1 3 ÷1 1 1 1 1 1 1 1 1 1÷2 3 ÷1 1 1 ÿ1 1 1 ÿ1 1 1 ÿ1÷3 3 ÷1 2 ÿ1 0 2 ÿ1 0 2 ÿ1 0÷1 3 ÷2 1 1 1 1 1 1 ÿ1 ÿ1 ÿ1÷2 3 ÷2 1 1 ÿ1 1 1 ÿ1 ÿ1 ÿ1 1÷3 3 ÷2 2 ÿ1 0 2 ÿ1 0 ÿ2 1 0÷1 3 ÷3 2 2 2 ÿ1 ÿ1 ÿ1 0 0 0÷2 3 ÷3 2 2 ÿ2 ÿ1 ÿ1 1 0 0 0÷3 3 ÷3 4 ÿ2 0 ÿ2 1 0 0 0 0
Chapter 20 423
Then a! (1 2 3 4), b! (1 3) gives the required isomorphism.(b) Take the irreducible characters ÷1, . . . , ÷5 of S4 as in Section 18.1, and
take the character table of H to be
(see Example 16.3(3) or Section 18.3). We obtain
÷1 # H � ø1, ÷2 # H � ø4, ÷3 # H � ø1 � ø4, ÷4 # H � ø3 � ø5,
÷5 # H � ø2 � ø5:
2. Let ÷1, . . . , ÷11 be the irreducible characters of S6, as in Example 19.17.Either by direct calculation, or using (20.13), we ®nd that the characters÷i # A6 (i � 1, 3, 5, 7, 9) are distinct irreducible characters of A6; thesegive the characters ø1, . . . , ø5 in our character table below. Also,k÷11 # A6, ÷11 # A6l � 2. Arguing as in Example 20.14, we obtain from÷11 # A6 the two irreducible characters which we have called ø6 and ø7.
Note: á � (1�p5)=2, â � (1ÿp5)=2
3. Let ø1, . . . , ør be the irreducible characters of H. Then ÷ # H � d1ø1 �. . . � drør for some non-negative integers di. Since each øi has degree 1,the inequality (20.6) gives
÷(1) � d1 � : : :� dr < d21 � : : :� d2
r < n:
4. The inequality k÷ # H, ÷ # Hl H < 3 follows at once from Proposition 20.5.Write d � k÷ # H, ÷ # Hl H . For examples with d � 1 or 2, take G � S3
and H a subgroup of order 2, and ÷ an irreducible character of G of degreed. For an example with d � 3, take G � A4, H � V4 and ÷ an irreduciblecharacter of G of degree 3 (see Section 18.2).
gi 1 (1 3)(2 4) (1 2 3 4) (1 3) (1 2)(3 4)|CH (gi)| 8 8 4 4 4
ø1 1 1 1 1 1ø2 1 1 1 ÿ1 ÿ1ø3 1 1 ÿ1 1 ÿ1ø4 1 1 ÿ1 ÿ1 1ø5 2 ÿ2 0 0 0
Character table of A6
gi 1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2) (1 2 3)(4 5 6) (1 2 3 4)(5 6)|CA6
(gi)| 360 9 8 5 5 9 4
ø1 1 1 1 1 1 1 1ø2 5 2 1 0 0 ÿ1 ÿ1ø3 10 1 ÿ2 0 0 1 0ø4 9 0 1 ÿ1 ÿ1 0 1ø5 5 ÿ1 1 0 0 2 ÿ1ø6 8 ÿ1 0 á â ÿ1 0ø7 8 ÿ1 0 â á ÿ1 0
424 Representations and characters of groups
5. See (20.13). Since 20 occurs only once in the list of degrees for S7, therestriction of the irreducible character of degree 20 to A7 must be the sumof two different irreducible characters of degree 10. From the remainingfourteen irreducible characters of S7, upon restriction to A7 we get at leastseven irreducible characters of A7; and we get precisely seven if and only ifthe restriction of each of the fourteen characters is irreducible. We are toldthat A7 has exactly nine conjugacy classes. Hence the irreducible charactersof A7 have degrees
1, 6, 14, 14, 15, 10, 10, 21, 35:
Chapter 21
1. (a) Let u � 1 ÿ a2 � b ÿ a2b. Then ua2 � ÿu and ub � u. Hence sp (u) is aCH-submodule of CH.
(b) Since every element of G belongs to H or to Ha, the elements u and uaform a basis of U " G.
(c) The character ø of U and the character ø " G of U " G are given by
Since hø " G, ø " Gi � 1, the induced module U " G is irreducible.
2. Label the characters of H as follows:
where ù � e2ði=3.(a) ÷1 # H � ÷2 # H � ø1, ÷3 # H � ø2 � ø3,
÷4 # H � ÷5 # H � ø1 � ø2 � ø3.(b) Using the Frobenius Reciprocity Theorem, we obtain
ø1 " G � ÷1 � ÷2 � ÷4 � ÷5, ø2 " G � ø3 " G � ÷3 � ÷4 � ÷5:
3. It is suf®cient to prove that if U is a CH-submodule of CH thendim (U " G) � jG : H jdim U . Let Hgj (1 < j < m) be the distinct right
1 a2 b a2b
ø 1 ÿ1 1 ÿ1
1 a2 a b ab
ø " G 2 ÿ2 0 0 0
1 (1 2 3) (1 3 2)
ø1 1 1 1ø2 1 ù ù2
ø3 1 ù2 ù
Chapter 21 425
cosets of H in G. Then U(CG) � Ug1 � . . . � Ugm, where Ugj � {ugj:u 2 U}. The sum Ug1 � . . . � Ugm is direct (since the elements in Ugj arelinear combinations of elements in the right coset Hgj). Also,dim (Ugj) � dim U (since u! ugj (u 2 U) is a vector space isomorphism).Hence dim(U " G) � dim(U (CG)) � m dim U .
4. Let ö be an irreducible character of G. By using the Frobenius ReciprocityTheorem twice, together with the result of Exercise 19.1 (also twice), weobtain
h(ø(÷ # H)) " G, öiG � hø(÷ # H), ö # HiH � hø, (÷ö) # HiH
� hø " G, ÷öiG � h(ø " G)÷, öiG:Since this holds for all irreducible characters ö of G, we deduce fromTheorem 14.17 that (ø(÷ # H)) " G � (ø " G)÷.
5. The values of ö " G and ø " G are given by Proposition 21.23. On elementsof cycle-shapes (1), (7) and (3, 3), the values are as follows, and on allother elements the values are zero.
6. We have |G: H|ø(1) � d1÷1(1) � . . . � dk÷k(1), where
di � hø " G, ÷iiG � hø, ÷i # HiH ,
by the Frobenius Reciprocity Theorem. Hence, since ø is irreducible,÷i # H � diø � â where either â is a character of H or â � 0. Thus÷i(1) > diø(1), and therefore
jG: H jø(1) > (d21 � : : :� d2
k)ø(1):
The required result follows.
7. By applying the result of Exercise 6, we deduce, as in the proof ofProposition 20.9, that either(1)ø " G is irreducible, or(2)ø " G is the sum of two different irreducible characters of the same
degree.Suppose ®rst that ø " G is irreducible, say ø " G � ÷. Then ÷(1) � 2ø(1)
and k÷ # H, øl H � 1, by the Frobenius Reciprocity Theorem; hence ÷ # H isreducible, say ÷ # H � ø � ö. Now suppose that ø9 is an irreduciblecharacter of H. We have
hø9 " G, ÷iG 6� 0, hø9, ÷ # HiH 6� 0, ø9 � ø or ö:
ThusIf ø " G is irreducible, then for precisely one other irreducible character
ö of H we have ø " G � ö " G. (Compare Proposition 20.11.)
Cycle-shape (1) (7) (3, 3)
ö " G 240 2 12ø " G 720 ÿ1 0
426 Representations and characters of groups
Suppose next that ø " G is reducible, say ø " G � ÷1 � ÷2. Then÷1(1) � ø(1) and k÷1 # H, øl H � 1; hence ÷1 # H � ø. Now suppose that ø9is an irreducible character of H. We have
hø9 " G, ÷1iG 6� 0, hø9, ÷1 # HiH 6� 0, ø9 � ø:
ThusIf ø " G � ÷1 � ÷2 where ÷1 and ÷2 are irreducible characters of G, and
ø9 is an irreducible character of H such that ø9 " G has ÷1 or ÷2 as aconstituent, then ø9 � ø. (Compare Proposition 20.12.)
Chapter 22
1. The number of linear characters divides 15 (Theorem 17.11), and the sumof the squares of the degrees of the irreducible characters is 15 (Theorem11.12); moreover, each degree divides 15 (Theorem 22.11). Hence everyirreducible character has degree 1, and so G is abelian by Proposition 9.18.
2. Use Theorems 11.12, 17.11 and 22.11 again. The degree of each irreduciblecharacter is 1 or 2, and if there are r characters of degree 1 and s ofdegree 2, then
r divides 16, and r . 12 � s . 22 � 16:
Hence r � 4 or 8 or 16, and r � s � 7 or 10 or 16.
3. (a) Since G is non-abelian, not every irreducible character has degree 1(Proposition 9.18). This time, Theorems 11.12, 17.11 and 22.11 showthat there are r irreducible characters of degree 1 and s irreduciblecharacters of degree q, where
r divides pq, 1 < s and r � sq2 � pq:
Hence r � q and s � ( p ÿ 1)/q.(b) jG9j � p by Theorem 17.11.(c) The number of conjugacy classes of G is r � s.(For more information on groups of order pq, see Chapter 25.)
4. (a) By hypothesis, there exist a, b 2 C such that ö(g) � a for all g 6� 1 andö(1) � a� bjGj. Then ö � a1G � b÷reg, since both sides of this equationtake the same values on all elements of G.
(b) We have
h1G, öi � 1
jGj (a� bjGj � (jGj ÿ 1)a) � a� b, and
h÷reg, öi � 1
jGj jGj(a� bjGj) � a� bjGj:
Since ö is a character, both k1G, öl and k÷reg, öl are integers.(c) If ÷ is a non-trivial irreducible character of G, then kö, ÷l 2 Z and
k1G, ÷l � 0. Hence kö ÿ a1G, ÷l 2 Z. But kö ÿ a1G, ÷l � kb÷reg, ÷l �bjGj÷(1)=jGj � b÷(1).
Chapter 22 427
(d) Since ÷(1) divides jGj, part (c) implies that b|G| is an integer. Therefore,by part (b), a, and hence also b, is an integer.
5. (a) If g 2 G then g has odd order, by Lagrange's Theorem. Therefore, ifg2 � 1 then g � 1.
(b) For all g 2 G, we have ÷(g)� ÷(gÿ1) � ÷(g)� ÷(g) � 2÷(g), and ÷(g)is an algebraic integer. Partition Gnf1g into subsets by putting eachelement with its inverse. Each such subset has size 2, by part (a). HenceX
g2G
÷(g) � ÷(1)� 2á
for some algebraic integer á. The stated result follows, since
h÷, 1Gi � 1
jGjXg2G
÷(g):
(c) If ÷ 6� 1G in part (b), then k÷, 1Gl � 0, and hence á � ÿ÷(1)=2. Butÿ÷(1)=2 is a rational number which is not an integer (since ÷(1) dividesjGj, hence is odd). This contradicts Proposition 22.5. Thus ÷ � 1G.
(Further information about the number of characters ÷ such that ÷ � ÷appears in Theorem 23.1 and Corollary 23.2.)
6. (a) By Theorem 22.16, ÷(g) is an integer for all characters ÷. Let ÷1, . . . ,÷7 be the irreducible characters of G, with ÷1 � 1G. By the columnorthogonality relations, we have
(I) 1�X7
i�2
(÷i(g))2 � 5, and (II) 1�X7
i�2
÷i(1)÷i(g) � 0:
From equation (I) we deduce that either ÷i(g) � 0, �1 for all i, or÷i(g) � �2 for exactly one i and ÷i(g) � 0 for all other i . 1; thesecond possibility is ruled out by equation (II).
(b) We deduce from part (a) that ÷i(g) � 0 for two values of i, say i � 2, 3,and ÷i(g) � �1 for 4 < i < 7. By Corollary 22.27,÷2(1) � ÷3(1) � 0 mod 5. Also
(III)X7
i�1
(÷i(1))2 � 120:
Since 52 � 102 . 120, we deduce that ÷2(1) � ÷3(1) � 5.(c) By Corollary 22.27, ÷i(1) � ÷i(g) mod 5 for all i, and from equation (III)
we have
(IV)X7
i�4
(÷i(1))2 � 69:
Hence the possibilities for the pairs of integers (÷i(1), ÷i(g)) with4 < i < 7 are (1, 1), (4, ÿ1), (6, 1). The only possibility which isconsistent with equation (IV) is that the values of ÷i(1) for 4 < i < 7 are1, 4, 4, 6 in some order.
(d) We now have the following part of the character table of G:
428 Representations and characters of groups
gi g1 g2 g3 g4 g5 g6 g7
Order of gi 1 2 2 3 4 6 5|CG(gi)| 120 12 8 6 4 6 5
÷1 1 1 1 1 1 1 1÷2 5 0÷3 5 0÷4 1 1÷5 4 ÿ1÷6 4 ÿ1÷7 6 1
We successively calculate the ®ve remaining columns of the charactertable. We are told that ÷i(gj) is an integer for all i, j.
(1) First, ÷i(g4) � ÷i(1) mod 3 andP
7i�1(÷i(g4))2 � 6. Hence the values of
÷i(g4) for 1 < i < 7 are 1, ÿ1, ÿ1, 1, 1, 1, 0, respectively.(2) Next, ÷i(g5) � ÷i(1) mod 2 and
P7i�1(÷i(g5))2 � 4. Hence ÷i(g5) � �1
for 1 < i < 4 and ÷i(g5) � 0 for 5 < i < 7. SinceP
7i�1÷i(1)÷i(g5) � 0,
we deduce that ÷4(g5) � ÿ1 and (without loss of generality)÷2(g5) � ÿ÷3(g5) � 1.
(3) Since ÷i(g3) � ÷i(1) mod 2 andP
7i�1(÷i(g3))2 � 8, we deduce that
÷i(g3) � �1 for 1 < i < 4 and that the values of ÷i(g3) for 5 < i < 7 are0, 0, �2 in some order. Also
P7i�1÷i(g3)÷i(gr) � 0 for r � 4, 7, from
which we see that ÷i(g3) � 1 for 1 < i < 4. From the relationP7i�1÷i(1)÷i(g3) � 0 we now deduce that the entries in column 3 are 1,
1, 1, 1, 0, 0, ÿ2 in order from the top.(4) We have ÷i(g6) � ÷i(g4) mod 2 and
P7i�1(÷i(g6))2 � 6. Therefore
÷i(g6) � �1 for 1 < i < 6 and ÷7(g6) � 0. By applying the columnorthogonality relations involving column 6 and columns 3, 4, 5 and 7 weobtain ÷2(g6) � ÿ÷3(g6) � ÷4(g6) � ÿ1 and (without loss of generality)÷5(g6) � ÿ÷6(g6) � 1.
(5) Only the entries in column 2 remain to be calculated. These can beobtained from the column orthogonality relations.
The character table of G is as shown.
gi g1 g2 g3 g4 g5 g6 g7
Order of gi 1 2 2 3 4 6 5|CG(gi)| 120 12 8 6 4 6 5
÷1 1 1 1 1 1 1 1÷2 5 ÿ1 1 ÿ1 1 ÿ1 0÷3 5 1 1 ÿ1 ÿ1 1 0÷4 1 ÿ1 1 1 ÿ1 ÿ1 1÷5 4 ÿ2 0 1 0 1 ÿ1÷6 4 2 0 1 0 ÿ1 ÿ1÷7 6 0 ÿ2 0 0 0 1
Chapter 22 429
7. Suppose ®rst that ë is an eigenvalue of an n 3 n matrix A, all of whoseentries are integers. Then det (A ÿ ëIn) � 0, and hence ë is a root of thepolynomial det (xIn ÿ A), which is of the form
xn � anÿ1xnÿ1 � : : :� a1x� a0 (ar 2 Z):
Conversely, assume that ë is a root of the polynomial p(x) � a0 � a1x �. . . � anÿ1x nÿ1 � x n (ar 2 Z). Let
A �
0 1 0 : : : 0
0 0 1 0
0 0 0 0
..
. ...
0 0 0 1
ÿa0 ÿa1 ÿa2 : : : ÿanÿ1
0BBBBBB@
1CCCCCCA:
Check that det (xIn ÿ A) � p(x). As p(ë) � 0, it follows that ë is aneigenvalue of A. Since A has integer entries, ë is therefore an algebraicinteger.
Chapter 23
1. Assume that x 2 G and x is real. Then gÿ1xg � xÿ1 for some g 2 G. Hencegÿ2xg2 � x, so g2 2 CG(x). Let m be the order of g. Since jGj is odd,m � 2n � 1 for some integer n, by Lagrange's Theorem. Then g �g2(n�1) 2 CG(x). Therefore xÿ1 � gÿ1xg � x. Since x2 � 1 and x has oddorder, it follows that x � 1.
2. Adopt the notation of Theorem 9.8. The character ÷ of G � Cn13 . . .
3 Cnrwhich is given by
÷(gi11 : : : gir
r ) � ëi11 : : : ëi r
r
is real if and only if ëi � �1 for all i with 1 < i < r. Now the nith root ofunity ëi can be ÿ1 if and only if ni is even. Hence the number of realirreducible characters is 2m, where m is the number of the integers n1, . . . ,nr which are even. However, the elements g of G which satisfy g2 � 1 areprecisely those elements gi1
1 . . . gi rr where for each j, either i j � 0 or n j is
even and i j � n j=2. The number of such elements is also 2m.
3. The elements g of D2n for which g2 � 1 are
1, aib (0 < i < nÿ 1) (and also an=2 if n is even):
This gives n � 1 elements if n is odd, and n � 2 elements if n is even.These numbers coincide with
P÷(1), summing over all the irreducible
characters ÷. Since é÷ < 1 for all ÷, it follows from the Frobenius±SchurCount of Involutions that we must have é÷ � 1 for all ÷.
4. Let ë1 and ë2 be the eigenvalues of gr. Then ÷A(g) �12((ë1 � ë2)2 ÿ (ë2
1 � ë22)) � ë1ë2 � det (gr) (see Proposition 19.14). Since
÷(1) � 2 we have ÷A(1) � 1. It now follows from the De®nition 23.13 of é÷that é÷ � ÿ1 if and only if ÷A � 1G. The result follows.
430 Representations and characters of groups
5. (a) First, it is easy to check that ÷(g) is real for all g 2 G, so é÷ � �1. Letr be the representation obtained by using the basis v1, v2 of V. Thendet (ar) � 1 and det (br) � ÿå n; hence det (gr) � 1 for all g 2 G if andonly if å n � ÿ1. The result now follows from Exercise 4.
(b) It is easy to check that if g � a or b and i, j 2 {1, 2} then
â(vig, v j) � â(vi, v jgÿ1):
For example, â(v1b, v1) � â(v2, v1) � å n � â(v1, å nv2) � â(v1, v1bÿ1).Hence â is G-invariant. The de®nition of â shows that â is symmetric ifå n � 1 and â is skew-symmetric if å n � ÿ1. The result now followsfrom Theorem 23.16.
(c) The elements of T4n are ai and aib (0 < i < 2n ÿ 1); a has order 2n andaib has order 4. Hence an is the only element of order 2.
(d) Refer the Exercise 18.3 for the characters ø j (1 < j < n ÿ 1) and ÷ j
(1 < j < 4) of T4n. Clearly é÷1 � é÷3 � 1, and é÷2 � é÷4 � 0 or 1,according to whether n is odd or even, respectively. By part (a) (or part(b)) and the construction of the characters ø j in Exercise 17.6, we getéø j � ÿ1 or 1, according to whether j is odd or even, respectively.Therefore
Pnÿ1j�1 (éø j)ø j(1) � 0 or ÿ2, according to whether n is odd or
even, respectively. HenceP
÷ (é÷)÷(1) � 2.
6. Let V be a CG-module with character ÷. Since é÷ � ÿ1, there exists a non-zero G-invariant skew-symmetric bilinear form â on V. As â is G-invariant,the subspace {u 2 V: â(u, v) � 0 for all v 2 V} is a CG-submodule of V; since V is irreducible it follows that
fu 2 V : â(u, v) � 0 for all v 2 Vg � f0g: (�)Pick a basis v1, . . . , vn of V and let A be the n 3 n matrix with ij-entryâ(vi, v j). Since â is skew-symmetric, we have At � ÿA. Thereforedet (At) � (ÿ1)ndet A, so det A � (ÿ1)n det A. Also A is invertible by (�), sodet A 6� 0. It follows that n is even; as n � ÷(1) the result is proved.
7. Choose a basis f1, . . . , fn of V and de®ne the symmetric n 3 n matricesA � (aij) and B � (bij) by
aij � â1( f i, f j), bij � â( f i, f j):
By applying the Gram±Schmidt orthogonalization process, we may constructa basis f 91, : : : , f 9n of V such that â1( f 9i, f 9j) � äij for all i, j. LetP � ( pij) be the n 3 n matrix which is given by
f 9i �X
j
pij f j:
Then PAPt � I n and PBPt is symmetric. By a well known property ofsymmetric matrices, there is an orthogonal matrix Q (i.e. QQt � I) suchthat Q(PBPt)Qÿ1 is diagonal. Write Q � (qij), and de®ne the basis e1, . . . ,en of V by
ei �X
j
qij f 9j:
Chapter 23 431
Then
â1(ei, ej) � äij, since QPAPtQt � In, and
â(ei, ej) � 0 if i 6� j, since QPBPtQt is diagonal:
8. (a) The proof is similar to that of part (1) of Schur's Lemma 9.1.(b) Let v1, . . . , vn be a basis of the RG-module V, and consider the vector
space V9 over C with basis v1, . . . , vn. Then V9 is a CG-module, whichwe are assuming to be an irreducible CG-module. By Schur's Lemmathere exists ë 2 C such that vW � ëv for all v 2 V9. But v1W � ëv1 2 V, soë 2 R.
(c) Let G � C3 � ka: a3 � 1l, and let V be the RG-submodule of the regularRG-module which is spanned by 1 ÿ a and 1 ÿ a2. Then V is anirreducible RG-module. De®ne W: V! V by vW � av (v 2 V).
9. r g is a permutation, as Hxg � Hyg) Hx � Hy, and r is a homomorphismas (Hx)(r gh) � Hxgh � (Hx)(r g)(rh). We have
g 2 ker r, Hxg � Hx, 8x 2 G,xgxÿ1 2 H , 8x 2 G, g 2 \x2G xÿ1Hx:
Finally, r is a homomorphism G! Sym(Ù) � Sn with kernelT
x2Gxÿ1Hx,which is contained in H.
10. Let c1, c2 be the columns of the character table of G corresponding to theclasses {1} and tG. By the orthogonality relation for c2 we haveP
÷i(t)2 � |CG(t)| � 2 (the sum over all irreducible characters ÷i, with÷1 � 1G), so we may take ÷1(t) � 1, ÷2(t) � �1 and ÷i(t) � 0 for i > 3.Now the orthogonality of c1 and c2 gives ÷1(1) � ÷2(1) � 1 and÷2(t) � ÿ1. Further, ÷1 and ÷2 are the only linear characters, since a linearcharacter must take the value �1 on t. Hence |G : G9| � 2 by Theorem17.11. For the last part, if G is simple then since G9 v G, we haveG9 � 1, i.e. G is abelian. Hence G � C2.
Chapter 25
1. It is clear that the given set of matrices has size p( p ÿ 1). Call it G. Forclosure, note that
1 y
0 x
� �1 y90 x9
� �� 1 y9� yx
0 xx9
� �;
associativity is a property of matrix multiplication, identity is
1 0
0 1
� �; inverse of
1 y
0 x
� �is
1 ÿyxÿ1
0 xÿ1
� �.
Therefore G is a group.
2. Let ç � e2ði=5 and å � e2ði=11, and write
á � å� å3 � å4 � å5 � å9, â � å2 � å6 � å7 � å8 � å10:
432 Representations and characters of groups
3. Recall that Z�p is cyclic, so by Exercise 1.6(c), there exists an integer msuch that um � v mod p. Also, m is coprime to q, since both u and v haveorder q modulo p. Hence bm has order q. Also, bÿmabm � aum � av. Letb9 � bm. Then
G1 � ha, b9: ap � b9q � 1, b9ÿ1ab9 � avi:Hence G1 � G2.
4. (a) Note that ÿ1 is the only element of order 2 in Z�p. Hence
um � ÿ1 mod p for some m
, the element u of Z�p has even order
, q is even
, p � 1 mod 4:
(b) By Proposition 25.9, aG � {aum
: m 2 Z}. Therefore
aÿ1 2 aG , um � ÿ1 mod p for some m, p � 1 mod 4:
(c) ÷i(a) � 1 for all the q linear characters ÷i of G. Hence
0 �X÷ irred
÷(1)÷(a) � q� qö1(a)� qö2(a):
Therefore ö1(a) � ö2(a) � ÿ1. Also, |CG(a)| � p, so
p �X÷
÷(a)÷(a) � q� ö1(a)ö1(a)� ö2(a)ö2(a):
Hence ö1(a)ö1(a)� ö2(a)ö2(a) � ( p� 1)=2.If p � 1 mod 4, then a is conjugate to aÿ1 and so ÷(a) is real for all
characters ÷, so (ö1(a))2 � (ö2(a))2 � ( p� 1)=2. Hence ö1(a) and ö2(a)are (ÿ1�p p)=2.
If p � ÿ1 mod 4, then a is not conjugate to aÿ1, and it follows fromCorollary 15.6 that ö1(a) and ö2(a) are not both real. Henceö2(a) � ö1(a). This time, 2ö1(a)ö1(a) � ( p� 1)=2, and we ®nd thatö1(a) and ö2(a) are (ÿ1� i
pp)=2.
Character table of F11, 5
gi 1 a a2 b b2 b3 b4
|CG(gi)| 55 11 11 5 5 5 5
÷1 1 1 1 1 1 1 1÷2 1 1 1 ç ç2 ç3 ç4
÷3 1 1 1 ç2 ç4 ç ç3
÷4 1 1 1 ç3 ç ç4 ç2
÷5 1 1 1 ç4 ç3 ç2 ç÷6 5 á â 0 0 0 0÷7 5 â á 0 0 0 0
Chapter 25 433
(d) By Theorem 25.10, ö1(a) �P( pÿ1)=2m�1 åu m
. Since Z�p is cyclic of orderp ÿ 1 and u has order ( pÿ 1)=2, it follows that {u, u2, . . . , u( pÿ1)=2} isprecisely the set of quadratic residues modulo p. The result now followsfrom part (c).
5. Let H � ka, bl. Then for all h 2 H, the conjugacy class hE consists of hand hÿ1. All the elements outside H form a single conjugacy class of E.Also, E9 � H, so E has exactly two linear characters, say ÷1 and ÷2.A typical non-trivial linear character of H is
where ù � e2ði=3. Then ÷ " E is the irreducible character ÷3 given in thetable which follows. The characters ÷4, ÷5 and ÷6 are obtained similarly.
6. Z(E) � {1}, and for all i with 1 < i < 6, there exist gi 2 E such that gi 6� 1but ÷i(gi) � ÷i(1) (so gi 2 Ker ÷i).
7. (a) F13,3 (see Theorem 25.10).(b) C2 3 F13,3 (see Theorem 19.18).(c) D6 3 F13,3 (see Theorem 19.18).
8. The conjugacy classes of G are
f1g, fa3, a6g, far: 3 B rg, farb2: 3 B rg, farb4: 3 B rg,farb2: r � 0, 3, 6g, farb4: r � 0, 3, 6g,farb: 0 < r < 8g, farb3: 0 < r < 8g, farb5: 0 < r < 8g:
Let H1 � kal. Then H1 v G and G=H1 � C6. Hence we get six linearcharacters ÷1, . . . , ÷6 of G, as shown.
Let H2 � ka3, b2l. Then H2 v G and G=H2 � hH2a, H2bi � D6. Lift theirreducible character of D6 of degree 2 to obtain ÷7 in the table below.Then ÷8 � ÷7÷2 and ÷9 � ÷7÷3 are also irreducible. The ®nal irreduciblecharacter ÷10 can be found by using the column orthogonality relations.
1 a a2 b b2 ab ab2 a2b a2b2
÷ 1 1 1 ù ù2 ù ù2 ù ù2
Character table of E
gi 1 a b ab a2b c|CG(gi)| 18 9 9 9 9 2
÷1 1 1 1 1 1 1÷2 1 1 1 1 1 ÿ1÷3 2 2 ÿ1 ÿ1 ÿ1 0÷4 2 ÿ1 2 ÿ1 ÿ1 0÷5 2 ÿ1 ÿ1 2 ÿ1 0÷6 2 ÿ1 ÿ1 ÿ1 2 0
434 Representations and characters of groups
Note: ù � e2ði=3
Chapter 26
1. Let ÷ be an irreducible character of G. Then k÷ # H, øl H 6� 0 for someirreducible character ø of H. Therefore h÷, ø " GiG 6� 0 by the FrobeniusReciprocity Theorem. But ø(1) � 1, since H is abelian; and (ø " G)(1) � p,by Corollary 21.20. Hence ÷(1) < p, and so ÷(1) � 1 or p by Theorem22.11. Assume that G has r linear characters and s irreducible characters ofdegree p. Then
r � pm for some m, by Theorem 17:11, and
r � sp2 � pn, by Theorem 11:12:
Since s � pnÿ2 ÿ pmÿ2 and s is an integer, m is at least 2.
2. {1}, {z} and {z2} are conjugacy classes of H. For all other elements h ofH, the conjugacy class hH � {h, hz, hz2}.
Character table of G � ka, b: a9 � b6 � 1, bÿ1ab � a2l
gi 1 a3 a ab2 ab4 b b2 b3 b4 b5
|CG(gi)| 54 27 9 9 9 6 18 6 18 6
÷1 1 1 1 1 1 1 1 1 1 1÷2 1 1 1 ù2 ù ÿù ù2 ÿ1 ù ÿù2
÷3 1 1 1 ù ù2 ù2 ù 1 ù2 ù÷4 1 1 1 1 1 ÿ1 1 ÿ1 1 ÿ1÷5 1 1 1 ù2 ù ù ù2 1 ù ù2
÷6 1 1 1 ù ù2 ÿù2 ù ÿ1 ù2 ÿù÷7 2 2 ÿ1 ÿ1 ÿ1 0 2 0 2 0÷8 2 2 ÿ1 ÿù2 ÿù 0 2ù2 0 2ù 0÷9 2 2 ÿ1 ÿù ÿù2 0 2ù 0 2ù2 0÷10 6 ÿ3 0 0 0 0 0 0 0 0
Character table of H (a non-abelian group of order 27)
gi 1 z z2 a a2 b ab a2b b2 ab2 a2b2
|CH (gi)| 27 27 27 9 9 9 9 9 9 9 9
÷00 1 1 1 1 1 1 1 1 1 1 1÷01 1 1 1 1 1 ù ù ù ù2 ù2 ù2
÷02 1 1 1 1 1 ù2 ù2 ù2 ù ù ù÷10 1 1 1 ù ù2 1 ù ù2 1 ù ù2
÷11 1 1 1 ù ù2 ù ù2 1 ù2 1 ù÷12 1 1 1 ù ù2 ù2 1 ù ù ù2 1÷20 1 1 1 ù2 ù 1 ù2 ù 1 ù2 ù÷21 1 1 1 ù2 ù ù 1 ù2 ù2 ù 1÷22 1 1 1 ù2 ù ù2 ù 1 ù 1 ù2
ö1 3 3ù 3ù2 0 0 0 0 0 0 0 0ö2 3 3ù2 3ù 0 0 0 0 0 0 0 0
Note: ù � e2ði=3
Chapter 26 435
3. G has 11 conjugacy classes: {1}, {a8}, {ar, aÿr} (1 < r < 7),{arb: r even}, {arb: r odd}. Here, the group K which appears in Theorem26.4 is {1, a8}, and G=K � D16. The character table of D16 is given inSection 26.8 (D16 � G1) and in Section 18.3. By lifting the irreduciblecharacters of D16, we obtain the characters ÷1, . . . , ÷7 of G as shown inthe table at the top of this page. Then the four characters ø j ( j � 1, 3, 5,7) come from inducing to G those linear characters ÷ of kal for which÷(a8) � ÿ1.
4. (a) Check that AB � ÿBA, AC � ÿCA, AD � DA, BC � CB, BD � ÿDB,CD � ÿDC. Hence Z 2 G, and G=hZi is abelian while G is non-abelian.Therefore G9 � hZi (see Proposition 17.10).
(b) A2 � ÿB2 � ÿC2 � D2 � I. Since G=hZi is abelian, it follows thatg2 2 hZi for all g 2 G. Hence every element of G has the formAi BjC k Dl Z m for some i, j, k, l, m 2 f0, 1g, so jGj < 32; also G is a2-group, since g4 � 1 for all g 2 G.
(c) A routine calculation shows that every matrix which commutes with eachof A, B, C and D has the form ëI for some ë 2 C. Hence by Corollary9.3, the given representation is irreducible.
(d) Since G has irreducible representations of degrees 1 and 4,jGj > 12 � 42 � 17. Combined with part (b), this shows that jGj � 32.Since G9 � hZi, G has precisely 16 representations of degree 1. Theseare as follows: for each (r, s, t, u) with r, s, t, u 2 {0, 1}, we get arepresentation
Ai BjC k Dl Z m ! (ÿ1)ir� js�kt� lu:
Together with the irreducible representation of degree 4, these are all theirreducible representations of G, by Theorem 11.12.
5. (a) Let å � e2ði=8. We obtain representations as follows:
Character table of G � ka, b: a16 � 1, b2 � a8, bÿ1ab � aÿ1l
gi 1 a8 a a2 a3 a4 a5 a6 a7 b ab|CG(gi)| 32 32 16 16 16 16 16 16 16 4 4
÷1 1 1 1 1 1 1 1 1 1 1 1÷2 1 1 1 1 1 1 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1 ÿ1 1÷5 2 2 0 ÿ2 0 2 0 ÿ2 0 0 0÷6 2 2
p2 0 ÿp2 ÿ2 ÿp2 0
p2 0 0
÷7 2 2 ÿp2 0p
2 ÿ2p
2 0 ÿp2 0 0ø j 2 ÿ2 cj c2 j c3 j c4 j c5 j c6 j c7 j 0 0( j � 1, 3, 5, 7)
Note: cm � e2ðim=16 � eÿ2ðim=16 � 2 cos (mð/8)
436 Representations and characters of groups
G1: a! å 0
0 åÿ1
!, b! 0 1
1 0
!;
G2: a! å 0
0 åÿ1
!, b! 0 1
ÿ1 0
!;
G3: a! å 0
0 å3
!, b! 0 1
1 0
!;
G4: a! å 0
0 å5
!, b! 0 1
1 0
!;
G5: a! 1 0
0 ÿ1
!, b! 0 1
1 0
!, z! i 0
0 i
!:
(b) If j � 5, 6, 7 or 8 then Z(Gj) � {1, a2, z, a2z} � C2 3 C2. Since Z(Gj)is not cyclic, Gj has no faithful irreducible representation, by Proposition9.16.
(c) Check that the matrices satisfy the required relations, so giverepresentations. It is easy to see that the matrices generate groups withmore than eight elements, so the representations are faithful.
(d) The following give faithful representations:
G7: a!i 0 0
0 ÿi 0
0 0 1
0B@1CA, b!
0 1 0
1 0 0
0 0 1
0B@1CA, z!
1 0 0
0 1 0
0 0 ÿ1
0B@1CA;
G8: a!i 0 0
0 ÿi 0
0 0 1
0B@1CA, b!
0 1 0
ÿ1 0 0
0 0 1
0B@1CA, z!
1 0 0
0 1 0
0 0 ÿ1
0B@1CA:
6. The following table records the numbers of elements of orders 1, 2, 4 and 8in G1, . . . , G9:
Therefore no two of G1, . . . , G9 are isomorphic, except possibly G5 andG8. But G5=G59 � C2 3 C4, while G8=G89 � C2 3 C2 3 C2, so G5 6� G8.
7. (a) By Lemma 26.1(1) we have {1} 6� Z(G) 6� G. Also jG=Z(G)j 6� p byLemma 26.1(2). Therefore jZ(G)j � p or p2. Assume that jZ(G)j � p2.If g =2 Z(G) then Z(G) < CG(g) 6� G, and g 2 CG(g). Hence
G1 G2 G3 G4 G5 G6 G7 G8 G9
Order 1 1 1 1 1 1 1 1 1 1Order 2 9 1 5 3 3 7 11 3 5Order 4 2 10 6 4 12 8 4 12 10Order 8 4 4 4 8 0 0 0 0 0
Chapter 26 437
jCG(g)j � p3 and jgGj � p. Therefore, G has p2 � ( p4 ÿ p2)=pconjugacy classes.
(b) Assume that G has r irreducible characters of degree 1 and s irreduciblecharacters of degree p. Since
P÷(1)2 � p4 (Theorem 11.12), there are
no irreducible characters of degree greater than p, so r � sp2 � p4.Therefore jG=G9j � r � p2 or p3, and if r � p2 then r � s � 2 p2 ÿ 1.Part (b) follows, as r � s is equal to the number of conjugacy classes ofG.
(c) Note that G9 \ Z(G) 6� {1} by Lemma 26.1(1). By parts (a) and (b), if|Z(G)| � p2 then |G9| � p; and if |G9| � p2 then |Z(G)| � p. Hence|G9 \ Z(G)| � p.
8. (a) Let Z � Z(G), and assume that
G=Z � haZ, bZi, with a4 2 Z, a2 Z � b2 Z, bÿ1abZ � aÿ1 Z:
Then a2 � b2z for some z 2 Z, and hence ba2 � b3z � b2zb � a2b. Sincea2 commutes with a, b and all elements in Z, we have a2 2 Z.Therefore G=Z 6� Q8.
(b) If G is a non-abelian group of order 16, then by Exercise 7, eitherG9 \ Z(G) � G9, in which case G=(G9 \ Z(G)) is abelian, orG9 \ Z(G) � Z(G), in which case G=(G9 \ Z(G)) 6� Q8 by part (a).
Chapter 27
1. Assume that
z � a b
c d
� �2 Z(SL (2, p)):
Then
z1 1
0 1
� �� 1 1
0 1
� �z) c � 0, and
z0 1
ÿ1 0
� �� 0 1
ÿ1 0
� �z) c � ÿb, a � d:
Therefore z � aI; and since z 2 SL (2, p), we have a2 � 1, so a � �1.
2. Check that
1 1
0 1
� �and
1 ÿ1
0 1
� �are elements of G � SL (2, 3) of order 3 which are not conjugate to eachother. The element
ÿ1 0
0 ÿ1
� �lies in Z(G), and
438 Representations and characters of groups
0 1
ÿ1 0
� �
has order 4. Hence the following are conjugacy class representatives:
g1 g2 g3 g4
1 0
0 1
� � ÿ1 0
0 ÿ1
� �0 1
ÿ1 0
� �1 1
0 1
� �
Order of gi 1 2 4 3|CG(gi)| 24 24 4 6
g5 g6 g7
1 ÿ1
0 1
� � ÿ1 1
0 ÿ1
� � ÿ1 ÿ1
0 ÿ1
� �
Order of gi 3 6 6|CG(gi)| 6 6 6
We now describe how to construct the character table of G, which is givenbelow.
First observe that the vector space (Z3)2 has exactly four 1-dimensionalsubspaces, namely the spans of the vectors (0, 1), (1, 1), (2, 1) and (1, 0).The group G permutes these subspaces among themselves, so we obtain ahomomorphism ö: G! S4. Check that Kerö � {�I}. HenceG=f�Ig � Imö, a subgroup of S4 of order 12; therefore G=f�Ig � A4.The characters ÷1, ÷2, ÷3, ÷4 of G are obtained by lifting to G theirreducible characters of A4 (which are given in Section 18.2).
The values of ÷5, ÷6, ÷7 on the elements g1, g2, g3 can be deduced fromthe column orthogonality relations. Note that G has three real conjugacyclasses, so by Theorem 23.1, one of ÷5, ÷6 and ÷7 must be real. Assume,without loss of generality, that ÷5 is real. Then ÷5(g4) � á, where á is real.Also á 6� 0, by Corollary 22.27. Since ÷5÷2 and ÷5÷3 are irreduciblecharacters of G of degree 2, whose values on g4 are áù and áù2, theymust be ÷6 and ÷7 in some order, say ÷5÷2 � ÷6 and ÷5÷3 � ÷7. Theequation
Pj÷ j(g4)÷ j(g4) � 6 gives áá � 1. Since á is real, á � �1. Then
á � ÿ1 since ÷5(g4) � ÷5(1) mod 3. Now note that for j � 5, 6, 7, Exercise13.5 implies that ÷ j(g7) � ÿ÷ j(g4). Finally, ÷(g5) � ÷(g4) and ÷(g6) � ÷(g7)for all ÷.
Chapter 27 439
3. Apply Proposition 17.6.
4. (a) For the character table of T, notice that T is isomorphic to the group oforder 21 whose character table is found in Exercise 17.2 and Example21.25. Representatives of the conjugacy classes of T are h1, . . . , h5,where
h1 �1 0
0 1
!Z, h2 �
2 0
0 4
!Z, h3 �
4 0
0 2
!Z,
h4 �1 1
0 1
!Z, h5 �
1 ÿ1
0 1
!Z:
Two of the linear characters of T are
where ù � e2ði=3. The values of 1T : G and ë : G are as follows (seeProposition 21.23):
We ®nd that k1T : G, 1T : Gl � 2 and k1T : G, 1Gl � 1. Hence1T : G � 1G � ÷, where ÷ is an irreducible character of G. Also, kë : G,ë : Gl � 1, so ë : G is irreducible; write ö � ë : G.
Character table of SL (2, 3)
gi g1 g2 g3 g4 g5 g6 g7
|CG(gi)| 24 24 4 6 6 6 6
÷1 1 1 1 1 1 1 1÷2 1 1 1 ù ù2 ù2 ù÷3 1 1 1 ù2 ù ù ù2
÷4 3 3 ÿ1 0 0 0 0÷5 2 ÿ2 0 ÿ1 ÿ1 1 1÷6 2 ÿ2 0 ÿù ÿù2 ù2 ù÷7 2 ÿ2 0 ÿù2 ÿù ù ù2
Note: ù � e2ði=3
hi h1 h2 h3 h4 h5
|CT (hi)| 21 3 3 7 7
1T 1 1 1 1 1ë 1 ù ù2 1 1
gi g1 g2 g3 g4 g5 g6
|CG(gi)| 168 8 4 3 7 7
1T " G 8 0 0 2 1 1ë " G 8 0 0 ÿ1 1 1
440 Representations and characters of groups
(c) The values of ÷ and ÷S are as shown below (see Proposition 19.14):
We ®nd that k÷S , 1Gl � k÷S , öl � k÷S , ÷l � 1. Hence there is a characteræ of G such that
÷S � 1G � ö� ÷� æ:
The values of æ are as shown above. We calculate that kæ, æl � 4, soeither æ � 2ø for some irreducible character ø, or æ is the sum of fourdistinct irreducible characters (cf. Exercise 14.7).Now 1G, ö and ÷ are three of the six irreducible characters of G, andnone is a constituent of æ. Since there are only six irreducible charactersin all, æ cannot therefore be the sum of four irreducible characters, andso æ � 2ø with ø irreducible. The values of ø are as shown above.
(d) The characters 1G, ö, ÷ and ø are the characters ÷1, ÷3, ÷2 and ÷6,respectively, in the character table of G given at the end of Chapter 27.The remaining irreducible characters ÷4, ÷5 can readily be calculatedusing the column orthogonality relations (noting that gi is real if andonly if 1 < i < 4).
5. (a) Compare the proof of Lemma 27.1. Note that because g2 lies in Z(G),gi and gig2 have the same centralizer for all i.
(b) By lifting, we obtain the characters ÷1, . . . , ÷6 in the character tableshown below.
(c) Use Exercise 13.5 (noting that ÷ j(ÿI) 6� ÷ j(I) for 7 < j < 11, since ÿI isnot in the kernel of these characters).
(d) SinceP
11j�1 ÷ j(g3)÷ j(g3) � 8, we deduce that ÷ j(g3) � 0 for 7 < j < 11.
(Alternatively, apply part (c).) Also, by Corollary 22.27, ÷ j(1) is even.(e) By Theorem 22.16, ÷(g6) 2 Z for all characters ÷. SinceP
11j�1 (÷ j(g6))2 � 6, the values of ÷ j(g6) for 7 < j < 11 must be �1,
�1, �1, 0, 0, in some order. By Corollary 22.27 again, two of ÷7, . . . ,÷11, say ÷9 and ÷10, have degrees divisible by 6. Further,P
11j�1 (÷ j(1))2 � 168, and 122 � 62 . 168, so ÷9(1) � ÷10(1) � 6.Next, ÷7(1)2 � ÷8(1)2 � ÷11(1)2 � 96. The only possibility is that two
of ÷7(1), ÷8(1), ÷11(1), say ÷7(1) and ÷8(1), are equal to 4, and÷11(1) � 8. The congruences ÷(1) � ÷(g6) mod 3 now give the remainingvalues on g6. Use part (c) to ®ll in the values on g2 and g7.
(f ) By Proposition 19.14, øA has the following values on g1, g2, g3 and g6:
gi g1 g2 g3 g4 g5 g6
|CG(gi)| 168 8 4 3 7 7
÷ 7 ÿ1 ÿ1 1 0 0÷S 28 4 0 1 0 0æ 12 4 0 0 ÿ2 ÿ2ø 6 2 0 0 ÿ1 ÿ1
g1 g2 g3 g6
øA 6 6 2 0
Chapter 27 441
The values of øA on g1 and g2 show that øA is a linear combination of÷1, ÷4, ÷5 and ÷6. Then the value of øA on g6 shows that ÷1 is not aconstituent of øA; ®nally, the value on g3 forces øA � ÷6.Now g2
4 is conjugate to g3. Hence(ø(g4)2 ÿ ø(g3))=2 � øA(g4) � ÷6(g4) � 0, and therefore, ø(g4) � 0.Similarly, ø(g5) � 0.Let x � ø(g8). Since g2
8 is conjugate to g8, we get(x2 ÿ x)=2 � øA(g8) � ÿ1. Therefore x � (1� i
p7)=2. Say
÷7(g8) � (1ÿ ip
7)=2. Then ÷8 � ÷7. For all ÷, ÷(g10) � ÷(g8); using thisfact and part (c), we ®ll in the values of ÷7 and ÷8.
(g) For i 6� 6, we haveP11
j�1 ÷ j(gi)÷ j(g6) � 0. This allows us to ®ll in thevalues of ÷11. Since g4 is conjugate to gÿ1
4 , ÷(g4) is real for all ÷. ThenP11j�1÷ j(g1)÷ j(g4) � 0 and
P11j�1÷ j(g4)2 � 8 imply that
÷9(g4) � ÿ÷10(g4) � �p2. Say ÷9(g4) � p2. The column orthogonalityrelations now let us ®nd the remaining values of ÷9 and ÷10, therebycompleting the character table of G.
6. Let Z � {�I} and de®ne the subgroup T of G, of order 55, by
T � a b
0 aÿ1
� �Z: a 2 Z�11, b 2 Z11
� �:
Then T is generated by
x � 1 1
0 1
� �Z and y � 2 0
0 6
� �Z,
and T has ®ve linear characters æ j (0 < j < 4), where
æ j: xuyv ! e2ði jv=5:
Character table of SL (2, 7)
gi g1 g2 g3 g4 g5 g6 g7 g8 g9 g10 g11
Order of gi 1 2 4 8 8 3 6 7 14 7 14|CG(gi)| 336 336 8 8 8 6 6 14 14 14 14
÷1 1 1 1 1 1 1 1 1 1 1 1÷2 7 7 ÿ1 ÿ1 ÿ1 1 1 0 0 0 0÷3 8 8 0 0 0 ÿ1 ÿ1 1 1 1 1÷4 3 3 ÿ1 1 1 0 0 á á á á÷5 3 3 ÿ1 1 1 0 0 á á á á÷6 6 6 2 0 0 0 0 ÿ1 ÿ1 ÿ1 ÿ1÷7 4 ÿ4 0 0 0 1 ÿ1 ÿá á ÿá á÷8 4 ÿ4 0 0 0 1 ÿ1 ÿá á ÿá á÷9 6 ÿ6 0
p2 ÿp2 0 0 ÿ1 1 ÿ1 1
÷10 6 ÿ6 0 ÿp2p
2 0 0 ÿ1 1 ÿ1 1÷11 8 ÿ8 0 0 0 ÿ1 1 1 ÿ1 1 ÿ1
Note: á � (ÿ1� ip
7)=2
442 Representations and characters of groups
The characters æ1 " G and æ2 " G are irreducible; they are ÷3 and ÷4 in thetable. (In calculating ÷3(g5), note that e2ði=5 � eÿ2ði=5 � (ÿ1�p5)=2:) Let÷1 � 1G. We have hæ0 " G, ÷1i � 1 and hæ0 " G, æ0 " Gi � 2. Henceæ0 " G � ÷1 � ÷2 for an irreducible character ÷2 of G.
We have now found four of the eight irreducible characters of G, namely÷1, ÷2, ÷3, ÷4.
SinceP
8j�1 ÷ j(g5)÷ j(g5) � 5, we deduce that the remaining irreducible
characters ÷5, ÷6, ÷7, ÷8 take the value 0 on g5. By Corollary 22.27,÷ j(1) � 0 mod 5 for 5 < j < 8. But
P8j�5 (÷ j(1))2 � 250; hence, without loss
of generality, ÷5(1), ÷6(1), ÷7(1), ÷8(1) are 10, 10, 5, 5, respectively.Note that ÷(gj) is an integer for 1 < j < 4 and all characters ÷, by
Theorem 22.16.Since ÷(1) � ÷(g3) mod 3 for all characters ÷, and
P8j�1 ÷ j(g3)2 � 6, the
values of ÷ j(g3) (5 < j < 8) are as shown.Now ÷(g4) � ÷(g3) mod 2 for all ÷, and
P8j�1 ÷ j(g4)2 � 6, so ÷ j(g4) � �1
for 5 < j < 8. Next, ÷(g2) � ÷(g4) mod 3 for all ÷, andP
8j�1 ÷ j(g2)2 � 12;
hence |÷(g2)| , 3 for all irreducible ÷. We may now conclude from the factsthat ÷(g2) � ÷(g1) mod 2 and
P8j�1 ÷ j(g2)2 � 12, that ÷ j(g2) � �2 for j � 5,
6 and ÷ j(g2) � �1 for j � 7, 8. By consideringP
8j�1 ÷ j(1)÷ j(g2) � 0, we
see that ÷7(g2) � ÷8(g2) � 1, and ÷5(g2), ÷6(g2) have opposite signs;without loss of generality, ÷5(g2) � 2 � ÿ÷6(g2).
We have now completed columns 1, 2, 3 and 5 of the character table.Since ÷(g4) � ÷(g2) mod 3 for all characters ÷, and
P8j�1 ÷ j(g4)2 � 6, we
can complete column 4.The column orthogonality relations now enable us to ®nish the character
table.
Character table of PSL (2, 11)
gi g1 g2 g3 g4 g5 g6 g7 g8
Order of gi 1 2 3 6 5 5 11 11|CG(gi)| 660 12 6 6 5 5 11 11
÷1 1 1 1 1 1 1 1 1÷2 11 ÿ1 ÿ1 ÿ1 1 1 0 0÷3 12 0 0 0 á â 1 1÷4 12 0 0 0 â á 1 1÷5 10 2 1 ÿ1 0 0 ÿ1 ÿ1÷6 10 ÿ2 1 1 0 0 ÿ1 ÿ1÷7 5 1 ÿ1 1 0 0 ã ã÷8 5 1 ÿ1 1 0 0 ã ã
Note: á � (ÿ1�p5)=2, â � (ÿ1ÿp5)=2 and ã � (ÿ1� ip
11)=2
Chapter 27 443
Chapter 28
1. We take g1, : : : , g8 as representatives of the conjugacy classes, where
g1 �1 0
0 1
!g2 �
2 0
0 2
!g3 �
1 1
0 1
!g4 �
2 1
0 2
!
g5 �1 0
0 2
!g6 �
0 1
2 0
!g7 �
0 1
1 2
!g8 �
0 1
1 1
!:
The character table of GL(2, 3) is then as follows.
gi g1 g2 g3 g4 g5 g6 g7 g8
|CG(gi)| 48 48 6 6 4 8 8 8
ë0 1 1 1 1 1 1 1 1ë1 1 1 1 1 ÿ1 1 ÿ1 ÿ1ø0 3 3 0 0 1 ÿ1 ÿ1 ÿ1ø1 3 3 0 0 ÿ1 ÿ1 1 1ø0,1 4 ÿ4 1 ÿ1 0 0 0 0÷1 2 ÿ2 ÿ1 1 0 0 i
���2p ÿi
���2p
÷2 2 2 ÿ1 ÿ1 0 2 0 0÷4 2 ÿ2 ÿ1 1 0 0 ÿi
���2p
i���2p
2. Every element r of Fq can be expressed as a square, since r � rq and q iseven.
Suppose thata b
c d
� �2 GL(2, q). We may write ad ÿ bc as s2 for some
s in F�q . Then
a b
c d
� �� s 0
0 s
� �a=s b=s
c=s d=s
� �:
The ®rst matrix in the product is sI and the second belongs to SL(2, q). Itnow follows easily that GL(2, q) � Z 3 SL(2, q) where Z � fsI : s 2 F�q g.
You should have no dif®culty in proving that the conjugacy classes ofSL(2, q) have representatives as follows.(a) The identity I has centralizer of order q3 ÿ q.
(b) The matrix u1 � 1 1
0 1
� �has centralizer of order q.
(c) There are (qÿ 2)=2 conjugacy classes with representatives
ds,sÿ1 � s 0
0 sÿ1
� �, indexed by unordered pairs fs, sÿ1g of elements
from FqnF2. Each such element has centralizer of order qÿ 1.
444 Representations and characters of groups
(d) There are q=2 conjugacy classes with representatives
vr � 0 1
1 r � rÿ1
� �, indexed by unordered pairs fr, rÿ1g of elements
from Fq2nFq such that r1�q � 1. Each such element has centralizer oforder q� 1.
By restricting characters from GL(2, q) to SL(2, q) you will quicklybe able to prove that the character table of SL(2, q) is as follows.
Here, we have used the function r! r de®ned in (28.3). The subscriptsfor ø0,i satisfy 1 < i < (qÿ 2)=2, and the subscripts for ÷i satisfy1 < i < q=2.
If q 6� 2 then the kernel of every non-trivial character is the identitysubgroup, and therefore SL(2, q) is simple.
3. Note ®rst that PSL(2, 8) � SL(2, 8).The polynomial x3 � x� 1 is irreducible over F2. Hence we may write
F8 � fa� bç� cç2 : a, b, c 2 F2 and ç3 � 1� çg:The pairs fs, sÿ1g of elements from F8nF2 are
fç, 1� ç2g, fç2, 1� ç� ç2g, f1� ç, ç� ç2g:These give us the conjugacy class representatives g3, g4, g5 below.
The irreducible monic quadratics over F8 with constant term 1 are
x2 � x� 1, x2 � çx� 1, x2 � ç2x� 1, x2 � (ç� ç2)x� 1:
The companion matrices for these quadratics give use the conjugacy classrepresentatives g6, g7, g8, g9 below
We can now list representatives g1, : : : , g9 of the conjugacy classes ofSL(2, 8), as follows.
g1 �1 0
0 1
!g2 �
1 1
0 1
!
g3 �ç 0
0 1� ç2
!g4 �
ç2 0
0 1� ç� ç2
!g5 �
1� ç 0
0 ç� ç2
!
g6 �0 1
1 1
!g7 �
0 1
1 ç
!g8 �
0 1
1 ç2
!g9 �
0 1
1 ç� ç2
!:
We may choose a generator å of F�64 so that å7 � åÿ7 � ç. Then
I u1 ds,sÿ1 vr
ë0 1 1 1 1ø0 q 0 1 ÿ1ø0,i q� 1 1 si � sÿi 0÷i qÿ 1 ÿ1 0 ÿ(r i � rÿi)
Chapter 28 445
å14 � åÿ14 � ç2, å21 � åÿ21 � 1 and å28 � åÿ28 � ç4 � ç� ç2. Thecharacter table of SL(2, 8) is then as follows.
Here, the 3 3 3 submatrices A and B are given by
A �2 cos(2ð=7) 2 cos(4ð=7) 2 cos(6ð=7)
2 cos(4ð=7) 2 cos(6ð=7) 2 cos(2ð=7)
2 cos(6ð=7) 2 cos(2ð=7) 2 cos(4ð=7)
0B@1CA
B �ÿ2 cos(2ð=9) ÿ2 cos(4ð=9) ÿ2 cos(8ð=9)
ÿ2 cos(4ð=9) ÿ2 cos(8ð=9) ÿ2 cos(2ð=9)
ÿ2 cos(8ð=9) ÿ2 cos(2ð=9) ÿ2 cos(4ð=9)
0B@1CA:
Chapter 29
1. (a) It is straightforward to check that ö is a homomorphism. For x, y 2 G,the element (x, y) 2 G 3 G sends x to y, so that action is transitive.
(b) (G 3 G)1 � f(g, g) : g 2 Gg, and kerö � f(z, z) : z 2 Z(G)g.(c) Every orbit of G 3 G on G 3 G contains an ordered pair of the form
(1, x), and if (g, h) sends (1, x) to (1, y) then g � h and y � gÿ1xg.Hence if C1, : : : , Ck are the conjugacy classes of G, and xi 2 Ci, then(1, xi) (1 < i < k) are orbit representatives for the action of G 3 G onG 3 G, and so the rank is equal to k.
Since x((g, h)ö) � x if and only if xhxÿ1 � g, we see that ð(g, h) �jfixG(g, h)j is equal to 0 if g is not conjugate to h, and is equal tojCG(x)j if g is conjugate to h (since in the latter case, if xhxÿ1 � g thenan arbitrary element y 2 G such that yhyÿ1 � g is of the form y � xcwith c 2 CG(x)). Hence using the column orthogonality relations we seethat ð �P÷ 3 ÷.
2. There are q2 ÿ 1 non-zero vectors in V, and each 1-dimensional subspace
gi g1 g2 g3 g4 g5 g6 g7 g8 g9
|CG(gi)| 504 8 7 7 7 9 9 9 9
ë0 1 1 1 1 1 1 1 1 1ø0 8 0 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1ø0,1 9 1 0 0 0 0ø0,2 9 1 A 0 0 0 0ø0,3 9 1 0 0 0 0÷3 7 ÿ1 0 0 0 ÿ2 1 1 1÷1 7 ÿ1 0 0 0 1÷2 7 ÿ1 0 0 0 1 B÷4 7 ÿ1 0 0 0 1
446 Representations and characters of groups
contains qÿ 1 of them; also two 1-dimensional subspaces have no non-zerovectors in common. Hence jÙj � (q2 ÿ 1)=(qÿ 1) � q� 1.
3. Use the notation for the conjugacy classes and irreducible characters ofGL(2, q) given in Proposition 28.4 and Theorem 28.5. It is easy to checkthat ð takes the values q2 ÿ 1, qÿ 1, qÿ 1 on the classes withrepresentatives I, u1, d1, t respectively, and takes the value 0 on all otherclasses. Taking inner products we ®nd
hð, 1Gi � hð, ø0i � hð, ø0, ji � 1 (1 < j < qÿ 2):
As 1G � ø0 �Pqÿ2
1 ø0, j has degree q2 ÿ 1 � ð(1), we conclude that
ð � 1G � ø0 �Pqÿ2
1 ø0, j.
4. Observe that the coset H1x is ®xed by g if and only if xgxÿ1 2 H1. If G isabelian this amounts to g 2 H1, and hence we see that ð1(g) � 0 if g =2 H1
and ð1(g) � jG : H1j if g 2 H1. Thus H1 � fg 2 G : ð1(g) 6� 0g. Likewisefor H2; since ð1 � ð2 we deduce that H1 � H2.
As a counterexample for G non-abelian, takeG � D8 � ha, b : a4 � b2 � 1, bÿ1ab � aÿ1i with H1 � hbi, H2 � ha2bi.Then ð1 � ð2 but H1 6� H2.
5. By Proposition 29.4 we have 1 � 1jGjP
g2GjfixÙ(g)j, hencePjfixÙ(g)j �
jGj. Since jfixÙ(g)j is a non-negative integer for each g, andjfixÙ(1)j � jÙj. 1, we must have jfixÙ(g)j � 0 for some g.
6. Write ð � ð(nÿ2,1,1). Calculating inner products using Proposition 29.6, as inthe proof of Theorem 29.13, we ®nd
hð, ði � 7, hð, 1i � 1, hð, ð(nÿ1,1)i � 3, hð, ð(nÿ2,2)i � 4:
Using Theorem 29.13 it follows that ð(nÿ2,1,1) � 1� 2÷(nÿ1,1) � ÷(nÿ2,2) � ÷with ÷ irreducible. Hence ÷ � 1� ðÿ ð(nÿ1,1) ÿ ð(nÿ2,2), from which it iseasy to calculate that ÷(1) � 1
2(nÿ 1)(nÿ 2),
÷(12) � 12(nÿ 2)(nÿ 5), ÷(123) � 1
2(nÿ 4)(nÿ 5). For n � 6, ÷(4,1,1) is the
character ÷5 in Example 19.17.
Chapter 30
1. By Theorem 30.4, a245 � 168/(8´3) � 7. Hence, by (30.3), PSL (2, 7)contains elements a and b such that a has order 2, b has order 3 and abhas order 7.
2. No: a225 � (1� (ÿ1� ip
7)=6� (ÿ1ÿ ip
7)=6ÿ 4=6)168=(8:8) � 0, andsimilarly a226 � 0. Hence PSL (2,7) does not contain two involutions whoseproduct has order 7.
3. Yes. Number the conjugacy classes of PSL (2, 11) as in the solution toExercise 27.6. Then
a235 � 660
12:61� 1
11
� �� 10:
Therefore PSL (2, 11) contains elements x and y such that x, y and xy haveorders 2, 3 and 5, respectively. Let H be the subgroup kx, yl of PSL (2, 11).
Chapter 29 447
There is a homomorphism W from A5 onto H (W sends a! x, b! y).Since Ker W 3 A5 and A5 is simple, we deduce that H � A5.
4. Suppose that G is a group whose character table iswhere á � (1�p5)=2, â � (1ÿp5)=2.
For 1 < i < 5 we have jCG(gi)j �P5
j�1÷ j(gi)÷ j(gi). Therefore thecentralizers of g1, g2, g3, g4, g5 have orders 60, 3, 4, 5, 5, respectively.Hence the orders of g2, g4 and g5 are 3, 5 and 5; also the order of g3
must be 2, since for no other i (except i � 1) is |CG(gi)| even.Now a324 � 60=(4 . 3). Therefore G contains elements x and y such that x
has order 2, y has order 3 and xy has order 5. As in the solution toExercise 3, G has a subgroup H with H � A5. Since jGj � 60, we haveG � A5.
5. (a) Using the fact thatP
7j�1÷ j(gi)÷ j(gi) � |CG(gi)|, we ®nd that the
centralizer orders and class sizes are as follows:
Hence jGj � 360. Also G is simple, by Proposition 17.6.(b) By the Frobenius±Schur Count of Involutions (Corollary 23.17), the
number of involutions t in G is bounded by
1� t <X7
j�1
÷ j(1) � 46:
By considering jCG(gi)j, we see that gi has even order only for i � 2and 3. Since t < 45, only g2 can be an involution. Hence g3 has order4. (This information about the orders of g2 and g3 can also be deducedusing Sylow's Theorem.)
(c) Clearly g6 and g7 have order 5, and at least one of g4 and g5 has order3. If j � 4 or 5 and k � 6 or 7 then
g1 g2 g3 g4 g5
÷1 1 1 1 1 1÷2 4 1 0 ÿ1 ÿ1÷3 5 ÿ1 1 0 0÷4 3 0 ÿ1 á â÷5 3 0 ÿ1 â á
g1 g2 g3 g4 g5 g6 g7
|CG(gi)| 360 8 4 9 9 5 5|gG
i | 1 45 90 40 40 72 72
448 Representations and characters of groups
a2 jk � jGjjCG(g2)j jCG(gj)j
X÷
÷(g2)÷(gj)÷ (gk)
÷(1)
� 360
8 . 9� 5:
Therefore G contains elements x and y such that x has order 2, y hasorder 3 and xy has order 5. As in the solution to Exercise 3, thesubgroup H � kx, yl of G is isomorphic to A5.
(d) If g, h 2 G then
(gh)r: Hx! Hxgh, and
(gr)(hr): Hx! Hxg ! Hxgh:
Hence r is a homomorphism.(e) Since G is simple, Ker r � {1}. Hence G is isomorphic to a subgroup K
of S6. Since jS6: Kj � 6!=360 � 2, K must be A6.
6. Consider the ®gure in Example 30.6(3). We shall explain how to label thevertices by elements of G.Choose a vertex and label it 1. Label the vertices according to thefollowing inductive rule. Assume that v is a vertex and an adjacent vertex uis labelled by g. Then label v by
ga if the edge uv has no arrow,
gb if the edge uv has an arrow from u to v,
gbÿ1 if the edge uv has an arrow from v to u:
For example, if you decided to label the bottom left-hand vertex by 1, thenpart of the labelling would be
The relation a2 � 1 ensures that the labelling is consistent along unmarkededges; since b3 � 1, the labelling is consistent around triangles; and therelation abababab � 1 deals with the octagons.
Every element in G has the form given by the label of one of the 24vertices, so jGj < 24.
7. The conjugacy classes of PSL(2, 7) are given in Lemma 27.1 The elementg2 is an involution with centralizer of order 8 given in the proof of
Lemma 27.1. Letting a � 2 2
ÿ2 2
� �, b � 2 4
4 ÿ2
� �, we see that the
centralizer is generated by a and b, and a4 � b2 � 1, bÿ1ab � aÿ1ÿ, hencethe centralizer is isomorphic to D8.
As in Exercises 12.3 and 12.4, we see that CA6((12)(34)) has order 8 and
is generated by (1324) and (13)(24), hence as above is isomorphic to D8.
Chapter 30 449
8. Let G be the simple group PSL(2, 17). In the ®eld Z17 the element 4 is a
fourth root of unity, so t � 4 0
0 ÿ4
� �Z is an involution. Calculate that
CG(t) is generated by the group of diagonal matrices together with
b � 0 1
ÿ1 0
� �Z, hence is generated by b and a � 3 0
0 6
� �Z. As
a8 � b2 � 1 and bÿ1ab � aÿ1, we have CG(t) � D16.
Chapter 31
1. Assume that G has an abelian subgroup H of index pr ( p prime), and that|G| . p. If H � {1} then |G| � pr and so G is not simple (see Lemma26.1(1)). So assume that H 6� {1}; pick 1 6� h 2 H. Then H < CG(h) as His abelian, so |G:CG(h)| is a power of p. If |G:CG(h)| � 1 then khl v G andG is not simple. And if |G:CG(h)| . 1, then G is not simple by Theorem31.3.
2. By Burnside's Theorem, jGj is divisible by at least three distinct primes.Since 3 . 5 . 7 . 80, jGj is even. Then by Exercise 13.8, |G| is divisible by4. Since 4 . 3 . 7 . 80, the only possibility is that jGj � 4 . 3 . 5 � 60.
Chapter 32
1. (a) The fact that BBt � I follows from the observation that for all i, j,
d(eib, ejb) � d(ei, ej) � äij:
Since 1 � det I � (det B)(det Bt) � (det B)2, we have det B � �1.(b) (i) The eigenvalues of C are the roots of det (C ÿ xI), which is a cubic
polynomial over R. Therefore, C has one or three real eigenvalues. (ii)Moreover, the product of the eigenvalues of C is det C � 1. If C hasthree real eigenvalues, then they cannot all be negative; and if C has onereal eigenvalue ë and a pair of conjugate non-real eigenvalues ì, ì, thenëìì � 1, and hence ë . 0. Therefore C has a real positive eigenvalue,say ë. (iii) Let v be an eigenvector for ë. Then
d(v, v) � d(vC, vC) � d(ëv, ëv) � ë2d(v, v),
and so ë � 1.(c) Let c be the isometry v! vC. By (b), c ®xes a vector v; it is now easy
to convince yourself that c must be a rotation about the axis through v.The required result for b follows from the de®nition of c.
(d) Take three orthogonal axes, one of which is the axis of the rotation b.With respect to these axes, the matrix of b is
1 0 0
0 cosö sinö0 ÿsinö cosö
0@ 1A:Hence tr B � 1 � 2 cosö.
450 Representations and characters of groups
2. We regard G as a subgroup of O(R3). It is easy to see that the translationsubmodule T (which consists of all the translation modes) is isomorphic tothe RG-module given by the natural action of G on R3. Hence by part (d)of Exercise 1,
÷T (g) �
1� 2 cosö, if g is a rotation through ö, about some
axis,
ÿ(1� 2 cosö), if the element ÿ g of O(R3) is a rotation
through ö:
8>>>>>><>>>>>>:Now consider the rotation submodule R, which consists of all the rotationmodes. A rotation mode is speci®ed by a 3-dimensional vector öv, where vis a unit vector along the axis of the rotation and ö denotes the angle ofrotation, taken positive in the right-hand screw sense. Let g 2 G, andconsider g acting on öv. It sends v to vg, and if g is a rotation, itpreserves the sense of the rotation; however, if g is a re¯ection then ittransforms a right-hand screw to a left-hand screw, and hence sends öv to(ÿö)(vg). Therefore
÷R(g) �÷T (g), if g is a rotation,
ÿ÷T (g), if g is not a rotation,
(and so ÷T � ÷R has the required form.
3. The matrix A is
k
m
ÿ3=2 0 3=4 ÿp3=4 3=4p
3=4
0 ÿ1=2 ÿp3=4 1=4p
3=4 1=4
3=4 ÿp3=4 ÿ3=4p
3=4 0 0
ÿp3=4 1=4p
3=4 ÿ5=4 0 1
3=4p
3=4 0 0 ÿ3=4 ÿp3=4p3=4 1=4 0 1 ÿp3=4 ÿ5=4
0BBBBBB@
1CCCCCCA:4. A simpler basis is given by
12(r1 � r2) � (v12 � v21)ÿ (v34 � v43),
12(r1 � r3) � (v13 � v31)ÿ (v24 � v42),
12(r1 � r4) � (v14 � v41)ÿ (v23 � v32):
We chose r1, r2, r3, r4 to be the images of w1, w2, w3, w4 under an RG-isomorphism. (Compare the construction of the matrix B, towards the endof Example 32.20.)
5. (a) This is a routine geometrical exercise.(b) Let the displaced positions of the atoms be 09, 19, 29, 39, 49. The
distance of 19 from the plane through 1 perpendicular to 12 isx12 � 1
2(x13 � x14). Similarly, the distance of 29 from the plane through 2
perpendicular to 12 is x21 � 12(x23 � x24). Therefore 12 has decreased by
x12 � x21� 12(x13 � x14 � x23 � x24). The other calculations can be done in
the same way.(c) We express the force at each atom as a vector, and then write this vector
Chapter 32 451
as a linear combination of our three chosen unit vectors at the initialposition of the atom. Let dij denote the decrease in the length ij, ascalculated in part (b). Then, for example, at vertex 1, the contributionsto the component of the force vector in the direction 12 are as follows:ÿk1d12 from the force between atoms 1 and 2; zero from the forcebetween atoms 1 and 3 and from the force between atoms 1 and 4; andÿk2d10 sec (/012) from the force between atoms 1 and 0. Hence
m1�x12 � ÿk1d12 ÿ 13
p(3=2)k2d10:
Upon substituting for d12 and d10 from part (b), we obtain the givenexpression for �x12.
The other accelerations are calculated in the same way.(d) The entries in the 15 3 15 matrix A are the coef®cients which appear in
the equations of motion �x � xA. If you write down the matrix A, thenyou will easily verify that the given vectors are eigenvectors of A (witheigenvalues
ÿ(4k1 � k2)=m1, 0, 0, 0, ÿk1=m1, ÿk1=m1,
respectively).(e) The matrix B is
ÿ(6k1 � k2)=3m1 ÿ2k2=3m1 ÿ4k2
p2=(m2
p3)
ÿ(3k1 � k2)=3m1 ÿ2k2=3m1 ÿ4k2
p2=(m2
p3)
ÿk2
p3=(9m1
p2) ÿ2k2
p3=(9m1
p2) ÿ4k2=3m2
0@ 1A:(f ) You will ®nd that (1, ÿ2,
p6) is an eigenvector of B, with eigenvalue 0.
This agrees with the statement in Example 32.20 that the translationvector
r1 ÿ 2s1 � 3 cos Ww1
is an eigenvector of A.
6. (a) Assign coordinate axes along the edges of the square, as shown below.
The symmetry group is G � D8. Let a denote the rotation sendingP! Q! R! S! P, and let b denote the re¯ection in the axis PR.The character ÷ of the RG-module R8 is
1 a2 a b ab
÷ 8 0 0 0 0
452 Representations and characters of groups
We refer to the character table of D8 which is given in Example 16.3(3),and see that
÷ � ÷1 � ÷2 � ÷3 � ÷4 � 2÷5:
The rotation mode is (t � â)v, where
v � (1, ÿ1, 1, ÿ1, 1, ÿ1, 1, ÿ1) 2 V÷2:
The translation modes are (t � â)v, where v is in the span of v1, v2
and
v1 � (1, 0, 0, 1, ÿ1, 0, 0, ÿ1), v2 � (0, 1, ÿ1, 0, 0, ÿ1, 1, 0):
The homogeneous components V÷1, V÷3
and V÷4are spanned by the
vectors (1, 1, 1, 1, 1, 1, 1, 1), (1, 1, ÿ1, ÿ1, 1, 1, ÿ1, ÿ1) and (1, ÿ1,ÿ1, 1, 1, ÿ1, ÿ1, 1), respectively. The ®nal set of eigen-vectors is givenby V÷5
\ R8vib which is spanned by
(1, 0, 0, ÿ1, ÿ1, 0, 0, 1) and (0, 1, 1, 0, 0, ÿ1, ÿ1, 0):
(b) The matrix A is
ÿ k
m
1 0 0 0 0 0 0 1
0 1 1 0 0 0 0 0
0 1 1 0 0 0 0 0
0 0 0 1 1 0 0 0
0 0 0 1 1 0 0 0
0 0 0 0 0 1 1 0
0 0 0 0 0 1 1 0
1 0 0 0 0 0 0 1
0BBBBBBBBBB@
1CCCCCCCCCCA:
7. (a, b) Let åi (1 < i < m) be the projection which is given by
åi: u1 � : : :� um ! ui
(where uk 2 Uk for all k). Then
w! wAå jWÿ1j Wi (w 2 Ui)
gives an RG-homomorphism from Ui to Ui. By Exercise 23.8, there existëij 2 R such that for all w 2 Ui,
wAå j � ëijwWÿ1i W j:
SinceP
mj�1å j is the identity endomorphism of U1 � . . . � Um, we have
wA �Xm
j�1
ëijwWÿ1i W j for all w 2 Ui:
Now take in turn w � uWi and w � vWi to obtain the results of parts (a)and (b) of the question.
(c) Take a basis u1, . . . , un of U1. Assume that the eigenvectors of Au areknown. For all k with 1 < k < n, the eigenvectors of A insp (ukW1, . . . , ukWm) are given by the eigenvectors of Au (see part (b)).Hence we know all the eigenvectors of A in U1 � . . . � Um.
Chapter 32 453
454
Bibliography
Books mentioned in the text
H. S. M. Coxeter and W. J. O. Moser, Generators and Relations for DiscreteGroups (Fourth Edition), Springer-Verlag, 1980.
J. B. Fraleigh, A First Course in Abstract Algebra (Third Edition), Addison-Wesley, 1982.
D. Gorenstein, Finite Simple Groups: An Introduction to their Classi®cation,Plenum Press, New York, 1982.
G. D. James, The Representation Theory of the Symmetric Groups, Lecture Notesin Mathematics No. 682, Spring-Verlag, 1978.
D. S. Passman, Permutation Groups, Benjamin, 1968.H. Pollard and H. G. Diamond, The Theory of Algebraic Numbers (Second
Edition), Carus Mathematical Monographs No. 9, Mathematical Associationof America, 1975.
J. J. Rotman, An Introduction to the Theory of Groups (Third Edition), Allyn andBacon, 1984.
D. S. Schonland, Molecular Symmetry ± an Introduction to Group Theory and itsuses in Chemistry, Van Nostrand, 1965.
Suggestions for further reading
M. J. Collins, Representations and Characters of Finite Groups, CambridgeUniversity Press, 1990.
C. W. Curtis and I. Reiner, Methods of Representation Theory with Applicationsto Finite Groups and Orders, Volume I, Wiley-Interscience, 1981.
W. Feit, Characters of Finite Groups, Benjamin, 1967.I. M. Isaacs, Character Theory of Finite Groups, Academic Press, 1976.W. Ledermann, Introduction to Group Characters (Second Edition), Cambridge
University Press, 1987.J. P. Serre, Linear Representations of Finite Groups, Springer-Verlag, 1977.
455
Index
A4, 112, 130, 136, 181, 308A5, 10, 112, 116, 220, 312, 359A6, 116, 222, 354, 360A7, 223An, 5, 9, 11, 111, 343abelian group, 3, 11, 81, 82action, 337algebra, 55, 56algebraic integer, 244, 362algebraic number, 361alternating group, 5, 9, 11, 111antisymmetric part, 196, 273associative, 2
basis, 15natural, 45, 54
bijection, 6bilinear form, 269
symmetric, 269skew-symmetric, 269
Brauer±Fowler Theroem, 278Burnside's Lemma, 340Burnside's Theorem, 363, 364
C, 2Cn, 2, 82, 88centralizer, 106centre
of group, 85, 107, 116, 298of group algebra, 83, 114, 153
change of basis, 24character, 118
degree, 122, 247faithful, 125, 195generalized, 355induced, 230, 234, 236integer-valued, 253irreducible, 119kernel of, 125linear, 122, 172, 174
permutation, 129product, 176, 192real, 263realized over R, 265reducible, 119regular, 127, 128, 150trivial, 122
character table, 159A4, 181A5, 221, 359A6, 359, 424C2, 160C3, 160C4, 412C2 3 C2, 415Cn, 82D6, 160D8, 161D10, 415D12 � S3 3 C2, 207, 419D2n (n odd), 182D2n (n even), 183D6 3 D6, 423E, of order 18, 434F7,3, 240, 417F11,5, 433Fp,q, 291GL(2,q), 327PSL(2,7), 318PSL(2,8), 445PSL(2,11), 443Q8, 416S4, 180S5, 201, 262S6, 205SL(2,3), 440SL(2,7), 442SL(2,q), 445T12, 186T4n, 420
U6n, 421V24, 422V8n, 421direct product, 206order 16, 305, 306, 307order 27, 435order , 32, 308order p3, 301order pq, 291p-group, 300
class algebra constants, 349class equation, 107class function, 152class sum, 114Clifford's Theorem, 216complete set, 101completely reducible, 74composition, 2composition factor, 90
common, 91, 96congruences, 259conjugacy class, 104conjugate, 104, 361constituent, 143, 213coset, 8cycle notation, 8cycle-shape, 109cyclic group, 2, 4, 12, 82, 88
D2n, 2, 12, 107, 181degree, 30, 122, 249derived subgroup, 173diagonalization, 83dicyclic group T4n, 178, 187, 281, 420dihedral group, 2, 12, 107, 181dimension, 15direct product, 5, 206direct sum, 17, 66
external, 18
eigenvalue, 24eigenvector, 24endomorphism, 20equivalent, 32, 46even permutation, 5expansion±contraction mode, 381external direct sum, 18
F � R or C, 3Fn, 15Fp,q, 290FG, 53factor group, 9faithful character, 125, 195faithful module, 44, 56, 85faithful representation, 34FG-module, see module
FG-submodule, 49FG-homomorphism, 61FG-isomorphism, 63Frobenius group, 290Frobenius Reciprocity Theorem, 232Frobenius±Schur Count of Involutions, 277function, 6
bijective, 6injective, 6invertible, 6surjective, 6
GL(n,F), 3GL(2,q), 324, 343general linear group, 3group, 1
abelian, 3, 11, 81, 82alternating, 5, 9, 11, 111cyclic, 2, 4, 12, 82, 88dicyclic, 178dihedral, 2, 12, 107, 181factor, 9®nite, 2general linear, 3order, 2order p3, 302, 304orthogonal, 367projective special linear, 312quaternion, 5rotation, 368simple, 10, 250, 278, 311, 318, 353,
363, 364soluble, 365special linear, 311symmetric, 3, 109, 116, 175, 254symmetry, 368
group algebra, 55
H < G, 3H v G, 9HomCG(V, W), 95, 96homogeneous component, 376homomorphism, 6, 10, 61
ideal, 256maximal, 257proper, 257
index of subgroup, 9indicator function, 273induced character, 230, 234, 236induced module, 226, 228inner product, 134involution, 277, 353irreducible character, 119irreducible module, 50, 74, 79, 91irreducible representation, 50, 79isomorphism, 7, 20, 63
456 Representations and characters of groups
kernel, 10, 19, 34, 124, 125
Lagrange's Theorem, 9lift, 169linear character, 122, 173, 174linear transformation, 18linearly dependent, 15linearly independent, 15
Maschke's Theorem, 70, 76matrix, 21
change of basis, 24diagonal, 26identity, 3, 21invertible, 23permutation, 45
methane, 384minimal polynomial, 361module, 39
completely reducible, 74faithful, 44, 56, 85irreducible, 50, 79, 85permutation, 45, 62reducible, 50regular, 56trivial, 43
natural basis, 45, 54normal modes of vibration, 372, 373normal p-complement, 251normal subgroup, 9, 113, 171, 215, 216,
217
odd permutation, 5orbit, 338order of G, 2order of g, 4orthogonal group, 367orthogonality relations, 161
PSL(2,7), 312, 318, 319, 354, 359, 360PSL(2,11), 321, 359PSL(2, p), 312p-group, 298p9-part, 256, 258permutation, 3, 5
even, 5odd, 5
permutation module, 45, 62, 340permutation character, 129, 340permutation matrix, 45powers of characters, 193presentation, 3primitive root, 284product of characters, 176, 192
projection, 27, 67projective special linear group, 312
Q8, 5, 116, 177, 278, 416quaternion group, 5, 116, 177, 278, 416
R, 3rank, 342Rank±Nullity Theorem, 19real character, 263real conjugacy class, 263real element, 263reducible character, 119reducible module, 50reducible representation, 50regular character, 127, 128, 150regular module, 56regular representation, 56representation, 30
degree, 30, 249equivalent, 32, 46faithful, 34irreducible, 50, 79kernel of, 34, 124reducible, 50regular, 56trivial, 34
representatives, 105restriction, 210rotation group, 368rotation mode, 379rotation submodule, 380, 394
S4, 44, 110, 113, 180, 275S5, 111, 201, 262S6, 116, 205S7, 223Sn, 3, 109, 116, 175, 254, 343, 344SL(2,3), 319, 440SL(2,7), 320, 442SL(2, p), 311SL(2,q), 336, 445Schur's Lemma, 78simple group, 10, 250, 278, 311, 318,
363, 364skew-symmetric bilinear form, 269special linear group, 311stabilizer, 339subgroup, 3, 4
cyclic, 4derived, 173generated, 4normal, 9, 113, 171, 215, 216, 217
submodule, 49irreducible, 74
Sylow's Theorem, 354, 365symmetric bilinear form, 269
Index 457
symmetric group, 3, 109, 116, 175, 254symmetric part, 196, 273symmetry group, 368
T4n, 178, 187, 281, 420tensor product module, 190tensor product space, 188trace, 117transitive, 338, 341transitivity of induction, 229translation mode, 379translation submodule, 380, 394transposition, 5
trivial character, 122module, 43representation, 34
U6n, 178, 187, 421
V8n, 178, 187, 421Vandermonde matrix, 194vibratory modes, 381
water, 369, 374
Z, 2
458 Representations and characters of groups