GROUPS

SIMON WADSLEY

Contents

1. Examples of groups 21.1. A motivating example 21.2. Some initial definitions 41.3. Further geometric examples 71.4. Subgroups and homomorphisms 91.5. The Mobius Group 132. Lagrange’s Theorem 172.1. Cosets 172.2. Lagrange’s Theorem 182.3. Groups of order at most 8 192.4. The Quaternions 222.5. Fermat–Euler theorem 223. Group Actions 233.1. Definitions and examples 233.2. Orbits and Stabilisers 243.3. Conjugacy classes 273.4. Cayley’s Theorem 283.5. Cauchy’s Theorem 294. Quotient Groups 294.1. Normal subgroups 294.2. The isomorphism theorem 325. Matrix groups 335.1. The general and special linear groups 345.2. Mobius maps as projective linear transformations 355.3. Change of basis 375.4. The orthogonal and special orthogonal groups 395.5. Reflections 416. Permutations 446.1. Permutations as products of disjoint cycles 446.2. Permuations as products of transpositions 476.3. Conjugacy in Sn and in An 486.4. Simple groups 50

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2 SIMON WADSLEY

Purpose of notes. Please note that these are not notes of the lectures but notesmade by the lecturer in preparation for the lectures. This means they may notexactly correspond to what was said and/or written during the lectures.

Lecture 1

1. Examples of groups

Groups are fundamentally about symmetry. More precisely they are an algebraictool designed to abstract the notion of symmetry. Symmetry arises all over mathe-matics; which is to say that groups arise all over mathematics. Roughly speaking asymmetry is a transformation of an object that preserves certain properties of theobject.

As I understand it, the purpose of this course is two-fold. First to introducegroups so that those who follow the course will be familiar with them and be betterequipped to study symmetry in any mathematical context that they encounter.Second as an introduction to abstraction in mathematics, and to proving thingsabout abstract mathematical objects.

It is perfectly possible to study groups in a purely abstract manner withoutgeometric motivation. But this seems to both miss the point of why groups areinteresting and make getting used to reasoning about abstract objects more difficult.So we will try to keep remembering that groups are about symmetry. So what dowe mean by that?

1.1. A motivating example.

Question. What are the distance preserving functions from the integers to theintegers? That is what are the members of the set

Isom(Z) := {f : Z→ Z such that |f(n)− f(m)| = |n−m| for all n,m ∈ Z}?

These functions might reasonably be called the symmetries of the integers; theydescribe all the ways of ‘rearranging’ the integers that preserve the distance betweenany pair.

Let’s begin to answer our question by giving some examples of such functions.Suppose that a ∈ Z is an integer. We can define the function ‘translation by a’ by

ta : n 7→ n+ a for n ∈ Z.

For any choice of m,n ∈ Z

|ta(n)− ta(m)| = |(n+ a)− (m+ a)| = |n−m|.

Thus ta is an element of Isom(Z). We might observe that if a and b are both integersthen

(ta ◦ tb)(n) = ta(b+ n) = a+ b+ n = ta+b(n)

for every integer n, that is that ta+b = ta ◦ tb1. Moreover t0 is the identity or ‘donothing’ function id: Z → Z that maps every integer n to itself. Thus for everya ∈ Z, t−a is the inverse of ta, that is ta ◦ t−a = id = t−a ◦ ta.

1This is because two functions f, g : X → Y are the same function, i.e. f = g, precisely ifthey have the same effect on every element of the set they are defined on, i.e. f(x) = g(x) for all

x ∈ X.

GROUPS 3

Suppose now that f ∈ Isom(Z) is a symmetry of the integers. Consider thefunction g := t−f(0) ◦ f . Then for n,m ∈ Z,

|g(n)− g(m)| = |(f(n)− f(0))− (f(m)− f(0))| = |f(n)− f(m)| = |n−m|,so g ∈ Isom(Z) is also a symmetry of the integers.

Moreover g(0) = t−f(0)(f(0)) = f(0)− f(0) = 0, i.e. g fixes the integer 0. Whatdoes this tell us about g? For example, what does it tell us about g(1)? Since g isa symmetry and g(0) = 0 it must be the case that

|g(1)| = |g(1)− 0| = |g(1)− g(0)| = |1− 0| = 1.

That is g(1) = ±1.If g(1) = 1, what else can we say? For any n ∈ Z,

|g(n)| = |g(n)− g(0)| = |n− 0| = |n|i.e. g(n) = ±n. But also,

|g(n)− 1| = |g(n)− g(1)| = |n− 1|i.e. g(n) = 1± (n−1). These two conditions together force g(n) = n and so g = id.Now in this case

tf(0) = tf(0) ◦ id = tf(0) ◦ (t−f(0) ◦ f) = (tf(0) ◦ t−f(0)) ◦ f = id ◦f = f.

Thus f is translation by f(0) in this case.What about the case when g(1) = −1? In this case we still must have g(n) = ±n

for every integer n but now also

|g(n) + 1| = |g(n)− g(1)| = |n− 1|i.e. g(n) = −1± (n− 1). These two conditions together force g(n) = −n and so gis the ‘reflection about 0’-function

s : n 7→ −n for all n ∈ Z.Now we’ve seen that s = g = t−f(0) ◦ f in this case. It follows that f = tf(0) ◦ s.

We’ve now proven that every element of Isom(Z) is either a translation ta or ofthe form ta ◦ s (with a ∈ Z in either case). That is all symmtries of Z are of theform n 7→ n+ a or of the form n 7→ a− n.

It is worth reflecting at this point on some key facts we’ve used in the argumentabove which is sometimes known as a ‘nailing to the wall argument’.

(1) We’ve used that the composition of two symmetries of the integers is itselfa symmetry of the integers. In fact, we’ve only used this for some specialcases but it is true in general since if f, g ∈ Isom(Z) and n,m ∈ Z then

|f(g(n))− f(g(m))| = |g(n)− g(m)| = |n−m|.We might note that for a, n ∈ Z,

s(ta(n)) = s(n+ a) = −a− n = t−a(s(n))

and so s ◦ ta = t−a ◦ s. Thus order of composition matters.(2) We’ve used that there is a ‘do nothing’ symmetry of the integers id and

that for any other symmetry f , f ◦ id = f = id ◦f .(3) We’ve used that symmetries are ‘undo-able’, that is that given any sym-

metry f there is a symmetry g such that g ◦ f = id = f ◦ g (in fact we’veonly used this for f = ta and f = s and only that there is a g such thatg ◦ f = id but again it is true as stated. (Why?).

4 SIMON WADSLEY

(4) We’ve used that composition of symmetries is associative, that is that forsymmetries f, g and h, (f ◦ g) ◦ h = f ◦ (g ◦ h).

We’ll see that these properties say precisely that Isom(Z) is a group.

1.2. Some initial definitions. First we need to make some definitions.

Definition. Suppose that S is a set. A binary operation on S is a function

◦ : S × S → S; (x, y) 7→ x ◦ y.

This defintion means that a binary operation is something that takes an orderedpair of elements of S and uses them to produce an element of S. If x ◦ y = y ◦ xthen we say that x and y commute (with respect to ◦). We say ◦ is commutativeif every pair of elements of S commute.

Examples.

(1) Composition of functions is a non-commutative binary operation on Isom(Z).(2) Addition, multiplication, and subtraction are all binary operations on Z.

Note that addition and multiplication are both commutative operations onZ but distinct integers never commute with respect to subtraction.

(3) Addition and multiplication are also binary operations on N := {1, 2, 3, . . .}.Subtraction is not a binary operation on N since 2− 3 6∈ N.

(4) Exponentiation: (a, b) 7→ ba is a binary operation on N.(5) If X is any set and S = {f : X → X} is the set of all functions from X to

itself then composition of functions is a binary operation on S.

Definition. A binary operation ◦ on a set S is associative if (x ◦ y) ◦ z = x ◦ (y ◦ z)for all x, y, z ∈ S.

This means that when ◦ is associative there is a well-defined element x◦y◦z ∈ Si.e. it doesn’t matter which of the two ◦ we use first. It will be instructive toconvince yourself that if ◦ is an associative binary operation on S and w, x, y, z ∈ Sthen

w ◦ (x ◦ y ◦ z) = (w ◦ x) ◦ (y ◦ z) = (w ◦ x ◦ y) ◦ z.Having done this you should also convince yourself that there is nothing specialabout four and the obvious generalisation holds for any (finite) number of elementsof S whenever ◦ is associative. This means that whenever ◦ is an associative binaryoperation we may (and will!) omit brackets, writing for example w◦x◦y◦z withoutambiguity. If it is clear what operation we have in mind we will often omit it too,writing wxyz, for example.

Examples.

(1) Addition and multiplication are associative when viewed as binary opera-tions on Z or N. Subtraction is not associative on Z since ((0−1)−2) = −3but 0− (1− 2) = 1 6= −3.

(2) Exponentiation (a, b) 7→ ba is not associative on N since 232

= 29 but(23)2 = 26 6= 29.

(3) Composition is always an associative operation on the set of functions fromX to X since if f, g and h are three such functions and x ∈ X then

((f ◦ g) ◦ h)(x) = f(g(h(x))) = (f ◦ (g ◦ h))(x).

GROUPS 5

Definition. A binary operation ◦ on a set S has an identity if there is some elemente ∈ S such that for all x ∈ S, e ◦ x = x = x ◦ e.

Examples.

(1) 0 is an identity for addition on Z but addition has no identity on N. 1 isan identity for multiplication on both these sets. Subtraction on Z doesnot have an identity since if e − x = x for all x ∈ Z then e = 2x for allx ∈ Z and this is absurd. Note however that x − 0 = x for all x ∈ Z. Wesometimes say that 0 is a right identity for subtraction to d9escribe this.

(2) (a, b) 7→ ba does not have an identity but 1 is a left identity in the obvioussense.

(3) If X is any set then the identity function id: X → X; s 7→ s is an identityfor composition of functions from X to X.

Lemma. If a binary operation ◦ on a set S has an identity then it is unique.

Proof. Suppose that e and e′ are both identities on S. Then e ◦ e′ = e′ = e′ ◦ esince e is an identity. But also e′ ◦ e = e = e ◦ e′ since e′ is an identity. Thus e = e′

as required. �

Lecture 2

Definition. If a binary operation ◦ on a set S has an identity e then we say thatit has inverses if for every x ∈ S there is some y ∈ S such that x ◦ y = e = y ◦ x.

Examples.

(1) + on Z has inverses since for every n ∈ Z, n + (−n) = 0 = (−n) + n.Multiplication does not have inverses on N or Z since there is no integer(and therefore no natural number) n such that 2n = 1.

(2) Multiplication defines an associative binary operation on the rationals Qwith an identity (1) but it still does not have inverses. Although for everynon-zero rational q, 1/q is also rational and q · 1/q = 1 = 1/q · q, 0 is alsorational and there is no rational r such that r·0 = 1. However multiplicationdoes have inverses on the set Q\{0}.

(3) In general composition on the set of functions X → X does not have in-verses. For example the function f : Z → Z;n 7→ 0 has no inverse since ifg : Z→ Z were an inverse then we’d have f(g(n)) = n for all n ∈ Z but infact however g is defined f(g(1)) = 0. This idea can be adapted to showthat whenever |X| > 1 there is a function f : X → X that has no inverse.

Definition. A set G equipped with a binary operation ◦ is a group if

(i) the operation ◦ is associative;(ii) the operation ◦ has an identity;

(iii) the operation ◦ has inverses.

Examples.

(1) Isom(Z) is a group (under composition).(2) (Z,+) is a group since + is associative and has an identity and inverses.(3) (N,+) is not a group since it does not have an identity.

6 SIMON WADSLEY

(4) (Z,−) is not a group since − is not associative.2

(5) (Z, ·) is not a group since it does not have inverses but (Q\{0}, ·) is a group.(6) If X is a set with more than one element then the set of functions X → X

is not a group under composition of functions since not all such functionshave inverses.

We will sometimes say that G is a group without specifying the operation ◦.This is laziness and the operation will always be implicit and either clear what it is(in concrete settings) or unimportant what it is (in abstract settings). We’ll nearlyalways call the identity of a group e (or eG if we want to be clear which group it isthe identity for) if we don’t know it by some other name.

Definition. We say that a group G is abelian if any pair of elements of G commute.

Definition. We say that a group G is finite if it has finitely many elements as aset. We call the number of elements of a finite group G the order of G written |G|.

Example. For every integer n > 1 we can define a group that is the set Zn :={0, 1, . . . , n − 1} equipped with the operation +n where x +n y is the remainderafter dividing x+ y by n3. It is straightforward to see that Zn is an abelian groupof order n.

Lemma. Suppose that G is a group.

(i) inverses are unique i.e. if g ∈ G there is precisely one element g−1 in G suchthat g−1g = e = gg−1;

(ii) for all g ∈ G, (g−1)−1 = g;(iii) for all g, h ∈ G, (gh)−1 = h−1g−1 (the shoes and socks lemma).

Proof. (i) By assumption every element g ∈ G has at least one inverse. We mustshow that there is at most one. Suppose that h, k ∈ G such that gh = e = hg andgk = e = kg. Then

h = eh = kgh = ke = k

i.e. h and k are the same element of G.(ii) Given g ∈ G, g satisfies the equations for the inverse of g−1 i.e.

gg−1 = e = gg−1

so by (i), (g−1)−1 = g.(iii) Given g, h ∈ G,

(gh)(h−1g−1) = ghh−1g−1 = geg−1 = gg−1 = e

and similarly (h−1g−1)(gh) = e. �

Notation. For each element g in a group G and natural number n we define gn

recursively by g1 := g and gn := ggn−1 for n > 1. We’ll also write g0 := e andgn := (g−1)−n for integers n < 0. It follows that gagb = ga+b for all a, b ∈ Z.

Definition. If G is a group then we say that g ∈ G has finite order if there isa natural number n such that gn = e. If g has finite order, we call the smallestnatural number n such that gn = e the order of g and write o(g) = n.

2recall that it also does not have an identity but to see that it is not a group it suffices to see

that any one of the three properties fails.3Z12 is familiar from everyday life. When is it used?

GROUPS 7

1.3. Further geometric examples.

1.3.1. Symmetry groups of regular polygons. Suppose we want to consider the setD2n of all symmetries of a regular polygon P with n vertices (for n > 3) livingin the complex plane C. By symmetry of P we will mean a distance preservingtransformation of the plane that maps P to itself. We might as well assume thatthe centre of P is at the origin 0 and that one of the vertices is the point 1 = 1+0i4.

Proposition. D2n is a group of order 2n under composition.

Lecture 3

Proof. First we observe that if g, h ∈ D2n then for z, w ∈ C,

|g(h(z)− g(h(w))| = |h(z)− h(w)| = |z − w|,

and if p ∈ P then g(h(p)) ∈ P since h(p) ∈ P . Thus composition of functions definesa binary operation on D2n. Since composition of functions is always associative thisoperation on D2n is associative. Moreover the identity function id is obviously asymmetry of P . Thus to see that D2n is a group it remains to show that everyelement of D2n is invertible. We could do this directly but instead we’ll use a nailingto the wall argument as we did for Isom(Z).

Let r : C→ C; z 7→ e2πi/nz denote rotation anticlockwise about 0 by 2π/n. Thenr does preserve distances in C as

|r(z)− r(w)| = |e2πi/n||z − w| = |z − w|

and r(P ) = P i.e. r ∈ D2n. Moreover rm ∈ D2n for all m ∈ N and rn = id; i.e.r−1 = rn−1.

Similarly let s : C→ C; z 7→ z denote complex conjugation; i.e. reflection in thereal axis. Once again s preserves distances in C and s(P ) = P . Moreover s2 = id;i.e. s−1 = s.

We will try to write down all the elements of D2n using r and s.Consider the set V := {e2πik/n | k = 0, 1, . . . n− 1} ⊂ C of nth roots of unity in

the complex plane, the set of vertices of our regular n-gon. Any element of D2n willpreserve V ; that is to say if g ∈ D2n then g(V ) = V 5. So if we pick any g ∈ D2n

then g(1) = e2πik/n for some k = 0, 1, . . . , n− 1. Thus

rn−kg(1) = e2πi(n−k)e2πik/n = 1.

Now e2πi/n and e−2πi/n are the only two elements of V of distance |1 − e2πi/n|from 1. Thus rn−kg(e2πi/n) = e±2πi/n.

If rn−kg(e2πi/n) = e2πi/n then rn−kg is a distance preserving transformation ofC fixing 0, 1 and e2πi/n. Thus rn−kg = id and g = rk.

If rn−kg(e2πi/n) = e−2πi/n then srn−kg is a distance preserving transformationof C fixing 0, 1 and e2πi/n. Thus srn−kg = id and g = rks.

So we have computed that

D2n = {rk, rks|k = 0, . . . , n− 1}.

4we’ll be able to make precise why this assumption is reasonable later but it should at least

seem reasonable already.5Recall that g(V ) := {g(v) | v ∈ V }.

8 SIMON WADSLEY

Thus it has 2n elements as required. We can compute for z ∈ C,

srk(z) = s(e2πik/nz) = e−2πik/nz = r−ks(z).

Thus rksrks = rkr−kss = e and rks is self-inverse. Geometrically rks denotereflection in the line through 0 and the point eπik/2n. To show this we can compute

rks(0) = 0,

rks(1) = e2πik/n and

rks(eπik/n) = eπik/n.

More generally if we wish to multiply elements in this standard form,

rk · rl = rk+nl,

rk · rls = rk+nls,

rks · rl = rk+n(−l)s and

rks · rls = rk+n(−l).

In particular sr = rn−1s = r−1s. It would be instructive to reflect on the geometricmeaning of these equations. �

1.3.2. Symmetry groups of regular solids. Suppose that X is a regular solid in R3.We can consider Sym(X), the group of distance preserving transformations ρ ofR3 such that ρ(X) = X. These form a group. We will consider the cases X atetrahedron and X a cube later in the course.

1.3.3. The Symmetric group. We might hope that given any set X the set of in-vertible functions from X to X forms a group under composition; that is the set offunctions f : X → X such that there is some g : X → X such that f ◦g = id = g◦f .This is true but not immediate: we need to check that composition of functions isa binary operation on this set; that is that the composition of two invertible func-tions is invertible. Some people would say that we need to check that the binaryoperation is closed but ‘closure’ is built into our definition of binary operation.

Lemma. Suppose that f1, f2 : X → X are invertible. Then f1 ◦ f2 : X → X isinvertible.

Proof. There are functions g1, g2 : X → X such that f1 ◦ g1 = id = g1 ◦ f1 andf2 ◦ g2 = id = g2 ◦ f2. Then since ◦ is associative

(f1 ◦ f2) ◦ (g2 ◦ g1) = f1 ◦ id ◦g1 = id

and

(g2 ◦ g1) ◦ (f1 ◦ f2) = g2 ◦ id ◦f2 = id .

�

It follows that for every set X, the set S(X) = {f : X → X | f is invertible} is agroup under the composition of functions. It is called the symmetric group on X6.We call elements of the symmetric group permutations of X. If X = {1, . . . , n} wewrite Sn instead on S(X). We will return to the groups Sn later in the course.

6The name comes from the fact that it can be viewed as the set of symmetries of the set X.This is quite a subtle idea but you might like to think further about it when you come to revise

the course

GROUPS 9

1.4. Subgroups and homomorphisms. Sometimes when considering the sym-metries of an object we want to restrict ourselves to considering symmetries thatpreserve certain additional properties of the object. In fact we’ve already seen this,the sets of distance preserving transformations of C and of R3 are both groups ofsymmetries under composition. The groups D2n and Sym(X) for X a regular solidare defined to consist of those symmetries that preserve a certain subset of thewhole space. Similary, instead of considering D2n, the group of all symmetries ofa regular n-gon we might want to restrict only to those symmetries that preserveorientation, that is the rotations. This idea leads us to the notion of subgroup.

Definition. If (G, ◦) is a group then a subset H ⊂ G is a subgroup if ◦ restricts toa binary operation on H7 and (H, ◦) is a group. We write H 6 G to denote thatH is a subgroup of G.

Examples.

(1) Isom(Z) 6 S(Z).(2) D2n 6 Isom(C) 6 S(C).(3) Isom+(Z) := {f : Z→ Z | f(n)− f(m) = n−m for all n,m ∈ Z} 6 Isom(Z).(4) Z is a subgroup of (Q,+).(5) If H ⊂ D2n consists of all rotations of the n-gon then H is a subgroup.(6) For any n ∈ Z, nZ := {an ∈ Z | a ∈ Z} is a subgroup of (Z,+).(7) For every group G, {e} 6 G (the trivial subgroup) and G 6 G (we call a

subgroup H of G with H 6= G a proper subgroup).

Lecture 4

Lemma (Subgroup criteria). A subset H of a group G is a subgroup if and only ifthe following conditions hold

(i) for every pair of elements h1, h2 ∈ H, h1h2 ∈ H;(ii) the identity e ∈ H;

(iii) for every h ∈ H, h−1 ∈ H.

Proof. We must show that if conditions (i)-(iii) hold then H is a group. Condition(i) tells us that the binary operation on G restricts to one on H. That this oper-ation is associative follows immediately from the associativity of its extension toG. Condition (ii) tells us that it has an identity8 and condition (iii) tells us H hasinverses9.

We must also show that if H is a subgroup of G then conditions (i)-(iii) hold.That condition (i) holds is immediate from the definition of subgroup. If eH isthe identity in H then eHe = eH = eeH since e is the identity in G. But alsoeHeH = eH since eH is the identity in H. Thus eHe = eHeH . But eH has aninverse in G and we see that e−1H eHe = e−1H eHeH ie e = eH ∈ H i.e. condition (ii)holds. That condition (iii) holds follows from uniqueness of inverses in G since forh ∈ H the inverse of h in H will also be an inverse of h in G. �

Remark. Our subgroup criteria contain no mention of associativity since as notedin the proof it is immediate from the associativity of the operation on G.

7precisely h1 ◦ h2 ∈ H for all h1, h2 ∈ H8the same identity as G since if h ∈ H then h ∈ G so eh = h = he9the inverse h−1 of h ∈ H ⊂ G in G is also an inverse in H since hh−1 = e = h−1h.

10 SIMON WADSLEY

There is an even shorter set of criteria for a subset to be a subgroup.

Corollary. A subset H of G is a subgroup precisely if it is non-empty and h−11 h2 ∈H for all h1, h2 ∈ H.

Proof. Suppose that conditions (i)-(iii) of the last lemma hold. Condition (ii) tellsus that H is non-empty and conditions (iii) and (i) combined give that h−11 h2 ∈ Hfor all h1, h2 ∈ H.

For the converse, since H is non-empty, there is some h ∈ H. By assumption e =h−1h ∈ H so condition (ii) holds. Now for h ∈ H, h−1e = h−1 ∈ H by assumptioni.e. condition (iii) holds. Finally, for h1, h2 ∈ H, h1h2 = (h−11 )−1h2 ∈ H. �

Example. The set H = {f ∈ Isom(Z)|f(0) = 0} is a subgroup of Isom(Z). We cansee this using the corollary. Certainly id(0) = 0 so H 6= ∅. Moreover if h1, h2 ∈ Hthen

h−11 h2(0) = h−11 (0) = h−11 h1(0) = id(0) = 0.

Note that this argument isn’t much simpler than verifying conditions (i)-(iii) of thelemma in practice.

We will also be interested in maps between groups. However we won’t typicallybe interested an arbitary functions between two groups but only those that respectthe structure of the two groups. More precisely we make the following definition.

Definition. If (G, ◦) and (H, ∗) are two groups then θ : H → G is a group homo-morphism (or just homomorphism) precisely if θ(h1 ∗ h2) = θ(h1) ◦ θ(h2) for allh1, h2 ∈ H.

Definition. A group homomorphism θ : H → G is an isomorphism if θ is invertibleas a function; ie if there is a function θ−1 : G → H such that θ ◦ θ−1 = idG andθ−1 ◦ θ = idH .

Examples.(1) If H 6 G then the inclusion map ι : H → G;h 7→ h is a group homomorphism.

It is not an isomorphism unless H = G.(2) The function θ : Z→ Zn such that θ(a) is the remainder after dividing a by n

is always a homomorphism from (Z,+) to (Zn,+n) but never an isomorphism.(3) If G is any group and g ∈ G is any element then θ : Z → G; n 7→ gn is a

homomorphism from (Z,+) to G. Indeed every homomorphism from (Z,+) toG arises in this way.

(4) θ : Z→ Isom+(Z); n 7→ tn10 is an isomorphism.

(5) The exponential function defines an isomorphism

exp: (R,+)→ ({r ∈ R | r > 0}, ·); a 7→ ea.

The inverse map is given by log = loge.

If you are alert you will be asking why we don’t require homomorphisms θ : H →G to satisfy θ(eH) = eG and θ(h−1) = θ(h)−1 for all h ∈ H. The following lemmashows that this is because these properties follow from our definition.

Lemma. Suppose that θ : H → G is a group homomorphism.

(i) θ(eH) = eG.(ii) For all h ∈ H, θ(h−1) = θ(h)−1.

10recall tn denotes translation by n

GROUPS 11

Proof. (i) Since θ is a homomorphism θ(eH) = θ(eHeH) = θ(eH)θ(eH). ThuseG = θ(eH)−1θ(eH) = θ(eH)−1θ(eH)θ(eH) = θ(eH) as required.

(ii) Pick h ∈ H. Then θ(h)θ(h−1) = θ(hh−1) = θ(eH) = eG (by (i)). Similarlyθ(h−1)θ(h) = θ(h−1h) = θ(eH) = eG. Thus θ(h−1) = θ(h)−1 as required. �

Definition. If θ : H → G is a group homomorphism then the kernel of θ is definedby

ker θ := {h ∈ H | θ(h) = eG}and the image of θ is defined by

Im θ := θ(H).

Proposition. If θ : H → G is a homomorphism then ker θ is a subgroup of H andIm θ is a subgroup of G.

Proof. We’ve seen θ(eH) = eG so eH ∈ ker θ and eG ∈ Im θ.Suppose that k1, k2 ∈ ker θ. Then θ(k1k2) = θ(k1)θ(k2) = eGeG = eG, i.e.

k1k2 ∈ ker θ. Similarly θ(k−11 ) = θ(k1)−1 = e−1G = eG, i.e. k−11 ∈ ker θ . Soker θ 6 H.

Suppose now that θ(h1), θ(h2) ∈ Im θ. Then θ(h1)θ(h2) = θ(h1h2) so θ(h1)θ(h2) ∈Im θ. Similarly θ(h1)−1 = θ(h−11 ), so θ(h1)−1 ∈ Im θ. So Im θ 6 G. �

Lecture 5

Theorem (Special case of the isomorphism theorem). A group homomorphismθ : H → G is an isomorphism if and only if ker θ = {eH} and Im θ = G. In thiscase, θ−1 : G→ H is a group homomorphism (and so also an isomorphism).

Proof. Suppose that θ has an inverse. Certainly eH ∈ ker θ and so θ−1(eG) =θ−1θ(eH) = eH . Thus if k ∈ ker θ then k = θ−1θ(k) = θ−1(eG) = eH and soker θ = {eH}. Moreover for each g ∈ G, θ−1(g) ∈ H and g = θ(θ−1(g)). ThusG = Im θ.

Conversely, suppose that ker θ = eH and Im θ = G. By the latter propertyfor each g ∈ G we may choose an element hg ∈ H such that θ(hg) = g. Now ifs : G → H is the function g 7→ hg then θs(g) = θ(hg) = g for all g ∈ G. It followsthat for all h ∈ H,

θ(s(θ(h))h−1) = θ(s(θ(h)))θ(h−1) = θ(h)θ(h−1) = θ(hh−1) = θ(eH) = eG,

i.e. s(θ(h))h−1 ∈ ker θ = {eH}. It follows that sθ(h) = h for all h ∈ H and so s isan inverse of θ.

Finally, suppose that θ is an isomorphism and g1, g2 ∈ G. Then

θ(θ−1(g1g2)) = g1g2

andθ(θ−1(g1)θ−1(g2)) = θ(θ−1(g1))θ(θ−1(g2)) = g1g2

since θ is a homomorphism. Thus

θ−1(g1g2) = θ−1(θ(θ−1(g1g2))) = θ−1(θ(θ−1(g1)θ(θ−1(g2)))) = θ−1(g1)θ−1(g2)

as required. �

Lemma. The composite of two group homomorphisms is a group homomorphism.In particular the composite of two isomorphisms is an isomorphism.

12 SIMON WADSLEY

Proof. Suppose θ1 : H → G and θ2 : K → H are homomorpisms and k1, k2 ∈ Kthen θ1(θ2(k1k2)) = θ1(θ2(k1)θ2(k2)) = θ1θ2(k1)θ1θ2(k2) ie θ1θ2 is a homomor-phism. Finally, if θ1 and θ2 are invertible then θ−12 θ−11 is an inverse of θ1θ2. �

Definition. We say that a group G is cyclic if there is a homomorphism f : Z→ Gsuch that Im f = G. Given such a homomorphism f we call f(1) a generator of G.

Note that G is cyclic with generator g if and only if every element of G is of theform gi with i ∈ Z. More generally we say that a subset S of G generates G ifevery element of G is a product of elements of S and their inverses — that is if Gthe unique smallest subgroup of G containing S11.

Examples.

(1) The identity map id: Z→ Z; n 7→ n and the ‘reflection about 0’ map s : Z→ Z;n 7→ −n are both homomorphisms with image Z. Thus Z is cyclic and both 1and −1 are generators. No other element generates Z.

(2) Zn is cyclic. In Numbers and Sets it is proven that an element of {0, 1, . . . , n−1}generates Zn if and only if it is coprime to n12. The ‘if’ part is a conseqeuncefrom Euclid’s algorithm; the only if part is elementary.

Lemma. Suppose that G is a group containing an element g with gn = e. There isa unique group homomorphism f : Zn → G such that f(1) = g. In particular everygroup of order n with an element of order n is isomorphic to Zn.

Proof. Suppose that f : Zn → G is a homomorphism such that f(1) = g. Then fora = 0, 1, . . . n− 1, f(a+n 1) = f(a)f(1) = f(a)g. Thus we can see inductively thatf(a) = ga for all a ∈ Zn. Thus if f exists then it is unique.

We now see how to construct f . We define f(a) = ga for all a ∈ Zn and wemust prove that this defines a homomorphism. That is we must show that ga+nb =f(a+n b) and ga+b = f(a)f(b) are equal for all a, b ∈ Zn. Since a+b−(a+n b) = knfor some integer k and gn = e, we see that

ga+bg−(a+nb) = (gn)k = ek = e.

Thus ga+b = ga+nb as claimed.Suppose now that G has order n and g ∈ G has order n. By the previous part

there is a homomorphism f : Zn → G such that f(a) = ga for each a ∈ Zn. Supposethat f(a) = f(b) for a, b ∈ Zn. Then f(b−n a) = ga−nb = e thus a = b else g wouldhave order strictly smaller than n. It follows that ker f = {e} and | Im f | has nelements and so must be the whole of G. Thus f is an isomorphism. �

Notation. We’ll write Cn for any group that is cyclic of order n. We’ve verifiedthat any two such groups are isomorphic.

11The curious will be reflecting on why G should have a unique smallest subgroup containing

S. Their reflections will do them good12Recall that non-negative integers a, b are coprime if and only if their only common factor is

1.

GROUPS 13

Recall that we showed that D2n = {ri, ris | i = 0, 1, . . . , n− 1} where r denotesa rotation by 2π/n and s denotes a reflection. And that

rk · rl = rk+nl,

rk · rls = rk+nls,

rks · rl = rk+n(−l)s and

rks · rls = rk+n(−l).

Lemma. Let n > 2 and suppose that G is a group containing elements g, h suchthat gn = e, h2 = e and hgh−1 = g−1. There is a unique group homomorphismf : D2n → G such that f(r) = g and f(s) = h. Moreover if o(g) = n and |G| = 2nthen f is an isomorphism.

Proof. This is similar to the last proof. Any homomorphsim f : D2n → G suchthat f(r) = g and f(s) = h must satisfy f(ri) = gi and f(ris) = gih. So we mustshow that this does define a homomorphism i.e. show that f(xy) = f(x)f(y) whenx, y ∈ {ri, ris} = D2n. We’ll do the most difficult case and leave the rest as anexercise. We can compute

f(rks)f(rls) = gkhglh = gk(hgh−1)l = gkg−l = gk+n(−l) = f(rksrls)

as required.For the last part suppose that o(g) = n and |G| = 2n. If ri ∈ ker f with

i ∈ {0, 1, . . . n − 1} then gi = e so as o(g) = n, i = 0. If ris ∈ ker f thengih = e so h = gi. Then g−1 = hgh−1 = gigg−i = g and so g2 = e contradictingo(g) = n. Thus ker f = {e}. This means that f can’t take the same value twice:if f(x) = f(y) then f(xy−1) = e so x = y. That Im f = G now follows from thepigeonhole principle. �

Lecture 6

1.5. The Mobius Group. Informally, a Mobius transformation is a functionf : C→ C of the form

f : z 7→ az + b

cz + dwith a, b, c, d ∈ C and ad− bc 6= 0. The reason for the condition ad− bc 6= 0 is thatfor such a function if z, w ∈ C then

f(z)− f(w) =az + b

cz + d− aw + b

cw + d= (ad− bc) (z − w)

(cz + d)(cw + d)

so f would be constant if ad− bc were 0.13

Unfortunately the function is not well defined if z = −d/c since we may notdivide by zero in the complex numbers. This makes composition of Mobius trans-formations problematic since the image of one Mobius transformation may notcoincide with the domain of defintion of another. We will fix this by adjoining anadditional point to C called ∞.14

Notation. Let C∞ := C ∪ {∞} which we call the extended complex plane.

13This is bad because we’re really interested in invertible functions14And pronounced ‘infinity’.

14 SIMON WADSLEY

Definition. Given (a, b, c, d) ∈ C4 such that ad− bc 6= 0 we can define a functionf : C∞ → C∞ as follows:

if c 6= 0 then

f(z) :=

az+bcz+d if z ∈ C\{−d/c};∞ if z = −d/c;a/c if z =∞;

if c = 0 then

f(z) :=

{az+bcz+d if z ∈ C∞ if z =∞.

We call all functions from C∞ to C∞ that arise in this way Mobius transformationsand let

M := {f : C∞ → C∞ | f is a Mobius transformation}.

We’ll see another way to interpret this definition later in the course involvingprojective geometry. But for now we’ll work with it as it stands and also take forgranted a result that will will prove later.15

Theorem. The set M defines a subgroup of S(C∞).

Lemma. Suppose that f ∈M such that f(0) = 0, f(1) = 1 and f(∞) =∞. Thenf = id.

Proof. If f = az+bcz+d then f(∞) =∞ forces c = 0 and f(0) = 0 forces b = 0 and then

f(1) = 1 forces a/d = 1 and so f : z 7→ az+00z+a = z for all z ∈ C as required. �

Theorem (Strict triple transitivity of Mobius transformations). If (z1, z2, z3) and(w1, w2, w3) are two sets of three distinct points then there is a unique f ∈M suchthat f(zi) = wi for i = 1, 2 and 3.

Proof. First we consider the case w1 = 0, w2 = 1 and w3 =∞.

If none of z1, z2, z3 are ∞ then f(z) = (z2−z3)(z−z1)(z2−z1)(z−z3) does the job.

If some zi = ∞ then choose z4 ∈ C∞\{z1, z2, z3} and let s(z) = 1z−z4 then

(s(z1), s(z2), s(z3)) does not contain ∞ so by the above we may find g ∈ M suchthat g(s(z1)) = 0, g(s(z2)) = 1 and g(s(z3)) =∞. Then f = gs does the job.

In the general case we can find g, h ∈ M such that g(z1) = 0, g(z2) = 1,g(z3) =∞ and h(w1) = 0, h(w2) = 1 and h(w3) =∞. Then f = h−1g(zi) = wi fori = 1, 2, 3 and we’ve shown existence in general. Moreover if k is another such mapthen hkg−1 fixes 0, 1 and ∞ so by the lemma is the identity. Thus k = h−1g = fand we’ve shown uniqueness. �

Definition. Given distinct points z1, z2, z3, z4 ∈ C∞ the cross-ratio of z1, z2, z3, z4written [z1, z2, z3, z4] := f(z4) where f is the unique Mobius transformation suchthat f(z1) = 0, f(z2) = 1 and f(z3) =∞.16

Lemma. If z1, z2, z3, z4 ∈ C then [z1, z2, z3, z4] = (z4−z1)(z2−z3)(z2−z1)(z4−z3) .17

15There is a straight-forward if slightly fiddly way to prove it directly that demands care with

the point ∞. We will give a slightly more sophisticated but less fiddly proof.16There are 6 essentially different definitions of cross-ratio depending on how we order 0, 1

and ∞ in this definition. It doesn’t really matter which we choose as long as we are consistent.17We could’ve defined the cross-ratio by this formula but we’d need to be more careful when

some zi =∞.

GROUPS 15

Proof. We’ve already computed that f(z) = (z2−z3)(z−z1)(z2−z1)(z−z3) is the unique Mobius

transformation such that f(z1) = 0, f(z2) = 1 and f(z3) =∞ and so

[z1, z2, z3, z4] = f(z4)

is given by the required formula. �

Theorem (Invariance of Cross-Ratio). For all z1, z2, z3, z4 ∈ C∞ and g ∈ M,[g(z1), g(z2), g(z3), g(z4)] = [z1, z2, z3, z4].

Proof. Suppose that f(z1) = 0, f(z2) = 1 and f(z3) =∞. Then (fg−1)(g(z1)) = 0,(fg−1)(g(z2)) = 1 and (fg−1)(g(z3)) =∞. Thus

[g(z1), g(z2), g(z3), g(z4)] = fg−1(g(z4)) = f(z4) = [z1, z2, z3, z4]

as required. �

Proposition. Every element of M is a composite of Mobius transformations ofthe following forms.

(a) Da : z 7→ az = az+00z+1 with a ∈ C\{0} (rotation/dilations);

(b) Tb : z 7→ z + b = 1z+b0z+1 with b ∈ C (translations);

(c) S : z 7→ 1/z = 0z+11z+0 (inversion).

Lecture 7

Proof. We’ll give two proofs. First a ‘nailing to the wall’ type argument.Suppose f ∈M. If f(∞) ∈ C then ST−f(∞)f(∞) = S(0) =∞. So, by replacing

f by ST−f(∞)f if necessary, without loss of generality f(∞) =∞18.Now since f(∞) = ∞, f(0) 6= ∞. Thus T−f(0) exists, T−f(0)f(0) = 0 and

T−f(0)(∞) =∞ so we may assume f fixes both 0 and ∞.Now since f(∞) = ∞ and f(0) = 0, f(1) ∈ C\{0}. Thus D1/f(1) exists and

D1/f(1)(1) = 1, D1/f(1)(0) = 0 and D1/f(1)(∞) = ∞. Thus f = Df(1) and we’redone.

Alternatively if c 6= 0 then az+bcz+d = a

c + (bc−ad)c(cz+d) so f = Ta/cD(bc−ad)/cSTdDc. If

c = 0 then f = Tb/dDa/d. �

Definition. A circle in C∞ is a subset that is either of the form {z ∈ C | |z−a| = r}for some a ∈ C and r > 019 or of the form {z ∈ C | aRe(z) + bIm(z) = c} ∪ {∞}for some a, b, c ∈ R with (a, b) 6= (0, 0)20

It follows that any three distinct points in C∞ determine a unique circle in C∞.

Lemma. The general equation of a circle in C∞ is

Azz +Bz + Bz + C = 0

with A,C ∈ R, B ∈ C and AC < |B|2.21

18Since if we can prove that ST−f(∞)f is a product of the special maps then f can be obtained

by postcomposing this product with Tf(∞)S19i.e. a usual circle in C20i.e. a line in C together with ∞21where ∞ is understood to be a solution of this equation precisely if A = 0

16 SIMON WADSLEY

Proof. First consider the case A 6= 0. Then

Azz +Bz + Bz + C = 0 ⇐⇒ |z +B/A|2 = |B|2/A2 − C/A.This latter defines a usual circle if and only if |B|2/A2 > C/A i.e. |B|2 > AC.

Next consider the case A = 0. Then

Bz + Bz + C = 0 ⇐⇒ 2Re(B)Re(z) + 2Im(B)Im(z) = −C.The latter defines a line in C if and only if B 6= 0 ie |B|2 > AC = 0. �

Theorem (Preservation of circles). If f ∈ M and C is a circle in C∞ then f(C)is a circle in C∞.

Proof. If f is a rotation/dilation or translation the result is easy. Thus in light ofthe last proposition we may assume that f : z 7→ 1/z.

But z is a solution to Azz+Bz+ Bz+C if and only if w = 1/z is a solution toCww +Bw + Bw +A. �

Corollary. Four distinct points z1, z2, z3 and z4 in C∞ lie on a circle if and onlyif [z1, z2, z3, z4] ∈ R.

Proof. Using triple transitivity we can find f ∈ M such that f(z1) = 0, f(z2) = 1and f(z3) =∞. By preservation of circles z1, z2, z3 and z4 lie on a circle if and onlyif f(z1), f(z2), f(z3) and f(z4) lie on a circle i.e. if and only if f(z4) is real. But[z1, z2, z3, z4] = f(z4). �

Remark. It is possible to prove the corollary directly and then use a similar argu-ment to deduce that Mobius transformations preserve of circles from it.

Definition. Given two elements x, y of a group G we say y is conjugate to x ifthere is some g ∈ G such that y = gxg−1.

Note that if y = gxg−1 then x = g−1y(g−1)−1 so the notion of being conjugateis symmetric in x and y. Morever if also z = hyh−1 then z = (gh)x(gh)−1 so if zis conjugate to y and y is conjugate to x then z is conjugate to x.

Proposition. Every Mobius transformation f except the identity has precisely oneor two fixed points. If f has precisely one fixed point it is conjugate to the translationz 7→ z + 1. If f has precisely two fixed points it is conjugate to a map of the formz 7→ az with a ∈ C\{0}.

Proof. Let f be the map extending z 7→ az+bcz+d and assume that f 6= id.

Suppose first that c = 0. In this case ∞ is a fixed point. Moreover for z ∈ C, ffixes z if and only if dz = az+ b. The possible sets of solutions to this equation are{b/(d− a)} (when d 6= a), C (when d = a and b = 0), or ∅ (when d = a and b 6= 0).Thus when c = 0 and f has either one or two fixed points.

Next suppose that c 6= 0 then f does not fix ∞. For z ∈ C, z = f(z) if and onlyif cz2 + dz − az − b = 0. This is a quadratic equation and the coefficient of z2 isnon-zero so it has one or two solutions (two unless there is a repeated root). Thusagain f has 1 or 2 fixed points in this case.

Now suppose that f has precisely one fixed point w and pick x ∈ C that is notfixed by f . Then f(x) 6∈ {x,w}.22 Let g be the unique Mobius transformation suchthat g(w) = ∞, g(x) = 0 and g(f(x)) = 1. Now for z ∈ C∞, gfg−1(z) = z if and

22If f(x) = w then x = f−1(w) = w

GROUPS 17

only if fg−1(z) = g−1(z), i.e. if and only if g−1(z) is fixed by f . Since w is the onlyfixed point of f , ∞ = g(w) is the only fixed point of gfg−1. Thus by the discussionabove gfg−1 extends z 7→ az+b

0z+a = z + b/a for some a, b ∈ C\{0}. Since

gfg−1(0) = gf(x) = 1

we see that b/a = 1 as required.Suppose instead that f has two distinct fixed points z1 and z2. Let g be any

Mobius transformation such that g(z1) = 0 and g(z2) =∞. Then

gfg−1(0) = gf(z1) = g(z1) = 0

and

gfg−1(∞) = gf(z2) = g(z2) =∞.Thus gfg−1(z) = az for some a ∈ C∞. �

Remark. Suppose that f ∈M. If gfg−1 : z 7→ z + 1 then for each n > 0,

gfng−1 = (gfg−1)n : z 7→ z + n

so fn(z) = g−1(g(z)+n) for z ∈ C∞ not fixed by f . Similarly if gfg−1 : z 7→ az thenfor n > 0, fn(z) = g−1(ang(z)) for z not fixed by f . Thus we can use conjugationto compute iterates of f ∈M in a simple manner.

Lecture 8

2. Lagrange’s Theorem

2.1. Cosets.

Definition. Suppose that (G, ◦) is a group and H is a subgroup. A left coset ofH in G is a set of the form g ◦H := {g ◦ h | h ∈ H} for some g ∈ G. Similarly aright coset of H in G is a set of the form H ◦ g := {hg | h ∈ H} for some g ∈ G.We write G/H to denote the set of left cosets of H in G and H\G to denote theset of right cosets of H in G23.

As usual we will often suppress the ◦ and write gH or Hg.

Examples.

(1) Suppose n ∈ Z, so that nZ := {an | a ∈ Z} is a subgroup of (Z,+). Then0 + nZ = nZ = n + nZ. 1 + nZ = {1 + an | a ∈ Z} = (1 − n) + nZ. Moregenerally, b+ nZ is the set of integers x such that x− b is a multiple of n.24

(2) Suppose that G = D6 = {e, r, r2, s, rs, r2s} and H = {e, s} then

eH = {e, s} = sH

rH = {r, rs} = rsH

r2H = {r2, r2s} = r2sH.

23This latter is a little unfortunate in the \ is normally used to denote set-theoretic difference

but this should not cause confusion. Why?24Note that because the operation on Z is addition we don’t suppress it when we name cosets,

i.e. we write a+ nZ rather than anZ because the latter would create confusion.

18 SIMON WADSLEY

However,

He = {e, s} = Hs

Hr = {r, r2s} = Hr2s

Hr2 = {r2, rs} = Hrs.

Thus left cosets and right cosets need not agree when the group is not abelian.However if K = {e, r, r2} 6 D6 then K = eK = rK = r2K = Ke = Kr = Kr2

and {s, rs, r2s} = sK = rsK = r2sK = Ks = Krs = Kr2s. So in this casethe left and right cosets are the same.

(3) Suppose that M is the Mobius group and H = {f ∈M | f(0) = 0}. Then, forg ∈M,

gH = {f ∈M | f(0) = g(0)} whereas

Hg = {f ∈M | f−1(0) = g−1(0)}.We’ll return to this idea later in the course.

2.2. Lagrange’s Theorem.

Theorem (Lagrange’s Theorem). Suppose that G is a group and H is a subgroupof G then the left cosets of H in G partition G. In particular if G is finite then |H|divides |G|.Proof. To show that the left cosets of H in G partition G we must show (i) thatevery element of G lives inside some left coset; and (ii) if two left cosets have acommon element then they are equal.

For (i) notice that for all g ∈ G, g ∈ gH since e ∈ H and ge = g.For (ii) suppose that g ∈ g1H ∩ g2H. Then we may find h1, h2 ∈ H such that

g = g1h1 = g2h2. Then for all h ∈ H, g1h = gh−11 h = g2h2h−11 h. Since h2h

−11 h ∈

H, we see that g1H ⊂ g2H. By symmetry we may conclude that g2H ⊂ g1H andso g1H = g2H as required.

Now for the last part we observe that when G is finite, H must also be finitesince it is a subset of G. Moreover, |G| =

∑gH∈G/H |gH|. Now for each left

coset gH there is an invertible function lg : H → gH; h 7→ gh — the functionl−1g is given by gh 7→ g−1gh = h. Thus |gH| = |H| for every left coset gH and|G| = |G/H| · |H|. �

Remark. By a very similar argument the right cosets of H in G also partition G.

Corollary. If G is a finite group, then every element of G has order dividing |G|.Proof. Suppose g ∈ G and let f : Z→ G be the homomorphism defined by f(n) =gn. We claim that the order of g is precisely | Im f | which divides |G| by Lagrange.To prove the claim we first observe that Im f = {e, g, g2, . . . , go(g)−1} since if n ∈ Zthere are integers q, r with 0 6 r < o(g) and n = qo(g) + r and then gn =(go(g))qgr = gr. Moreover the elements e, g, . . . , go(g)−1 are distinct since if gi = gj

with 0 6 i < j < o(g) then gj−i = e contradicting the definition of o(g). �

Proposition. Suppose that p is prime. Then every group of order p is isomorphicto Cp.

Proof. Let G be a group of order p and g ∈ G\{e}. Since |G| = p and o(g) divides|G| and is not 1 we must have o(g) = p. Thus g generates G by counting and anearlier lemma tells us that G is isomorphic to Cp. �

GROUPS 19

2.3. Groups of order at most 8. In this section we will classify all groups oforder at most 8 under the perspective that two groups are the same precisely ifthey are isomorphic. We’ve already seen that every group of order 2, 3, 5 or 7 isisomorphic to the cyclic group of the same order. It is evident that the trivial groupis the only group of order 1 up to isomorphism.

Before we begin this we’ll need the following construction that enables us tobuild new groups from old ones.

Example. Suppose that G and H are groups. We can define a binary operation onG ×H via (g1, h1)(g2, h2) = (g1g2, h1h2) for g1, g2 ∈ G and h1, h2 ∈ H. We claimthat this makes G×H into a group.

Proof of claim. Since

((g1, h1)(g2, h2))(g3, h3) = (g1g2g3, h1h2h3) = (g1, h1)((g2, h2)(g3, h3))

for all g1, g2, g3 ∈ G and h1, h2, h3 ∈ H, the operation on G×H is associative.Since (eG, eH)(g, h) = (g, h) = (g, h)(eG, eH), (eG, eH) is an identity for the

operation on G×H.Finally since (g−1, h−1)(g, h) = (eG, eH) = (g, h)(g−1, h−1) the operation on

G×H has inverses. �

Exercise. Show that if G1, G2 and G3 are groups then G1 × G2 is isomorphic toG2 ×G1 and (G1 ×G2)×G3 is isomorphic to G1 × (G2 ×G3).

Lecture 9

Theorem (Direct Product Theorem). Suppose that H1, H2 6 G such that

(i) H1 ∩H2 = {e};(ii) if h1 ∈ H1 and h2 ∈ H2 then h1 and h2 commute;

(iii) for all g ∈ G there are h1 ∈ H1 and h2 ∈ H2 such that g = h1h2.

Then there is an isomorphism H1 ×H2 → G.

Proof. Let f : H1 ×H2 → G be given by f(h1, h2) = h1h2. Now if h1, k1 ∈ H1 andh2, k2 ∈ H2 then

f((h1, k1)(h2, k2)) = f((h1h2, k1k2)) = h1h2k1k2

and

f((h1, k1))f((h2, k2)) = h1k1h2k2.

Since k1 and h2 commute (by (ii)), we see that f is a homomorphism. Now if(h1, h2) ∈ ker f then h1h2 = e so h1 = h−12 ∈ H1 ∩ H2 = {e} (by (i)). Thusker f = {e}. Finally Im f = {h1h2 | h1 ∈ H1, h2 ∈ H2} = G (by (iii)) and so f isthe required isomorphism. �

We also need the following result that also appeared on the first example sheet.

Lemma. If G is a group such that every non-identity element has order two25 thenG is abelian.

25Of course in any group the identity has order 1

20 SIMON WADSLEY

Proof. Suppose that x, y ∈ G. We must show that xy = yx.Note that for all g ∈ G, g2 = e. So by uniqueness of inverses every element of

G is self-inverse, i.e. g = g−1. In particular (xy)−1 = xy. By the shoes and sockslemma it follows that y−1x−1 = xy. Since x−1 = x and y−1 = y, yx = xy asrequired. �

We also recall that every a group of order n with an element of order n isisomorphism to Cn and that every group of order 2n that has an element g of ordern and an element h of order 2 such that hg = g−1h is isomorphic to D2n.

Proposition. Every group of order 4 is isomorphic to precisely one of C4 andC2 × C2.

Proof. First we’ll show that C4 and C2 × C2 are not isomorphic. Suppose (g, h) ∈C2×C2. Then (g, h)2 = (e, e). So every element of C2×C2 has order 2 but C4 hasan element of order 4 thus they cannot be isomorphic.26

Now suppose that G is any group of order 4. If G contains an element of order 4then it is isomorphic to C4 so suppose not. By Lagrange every non-identity elementof G has order 2. Let g, h be two distinct such elements. Let H1 = {e, g} ' C2 andH2 = {e, h} ' C2. It suffices to show that the three conditions of the direct producttheorem hold. Condition (i) is immediate since g 6= h. Condition (ii) follows fromthe last lemma since it tells us that G must be abelian. Finally, since g 6∈ H2,gH2 6= H2 and G is partitioned by H2 and gH2 by Lagrange. Thus condition (iii)holds. �

Proposition. Every group of order 6 is isomorphic to precisely one of C6 or D6.

Proof. We can easily see that C6 and D6 are not isomorphic since C6 is abelianand D6 is not27.

Now suppose that G is any group of order 6. If G contains an element of order 6then it is isomorphic to C6 so suppose not. By Lagrange every non-identity elementof G has order 2 or 3. If there were no element of order 3 then any two non-identityelements would generate a subgroup of order 4 contradicting Lagrange. Thus Gcontains an element g of order 3. Let K = {e, g, g2} be the subgroup of G generatedby g which is isomorphic to C3. By Lagrange for any h 6∈ K, G is partitioned byK and hK i.e. G = {e, g, g2, h, hg, hg2}. Now consider h2: h2 6∈ hK else h ∈ K28.Thus h2 ∈ K. If h2 is g or g2 then h has order 6 contradicting our assumption thatG has no elements of order 6. Thus h2 = e.

By the right coset version of Lagrange, G is also partitioned by K and Kh thusKh = hK and {h, gh, g2h} = {h, hg, hg2}. So gh ∈ {hg, hg2}. If gh = hg then(gh)2 = g2 and (gh)3 = h. Thus gh does not have order 1, 2 or 3, a contradiction29.Thus gh = hg−1 And we can deduce that G ' D6. �

Example. The following set of matrices form an non-abelian group Q8 of order 8{±(

1 00 1

),±(i 00 −i

),±(

0 1−1 0

),±(

0 ii 0

)}.

26If f : Z4 → C2×C2 were an isomorphism then f(2) = f(1)2 = e giving f a non-trivial kernel.27If f : D6 → C6 were an isomorphism then for all g1, g2 ∈ D6, f(g1g2) = f(g1)f(g2) =

f(g2)f(g1) = f(g2g1) so g2g1 = g1g2.28if h2 = hk then h = k29had we not made the assumption about no elements of order 6 then in this case we’d get

that gk has order 6 so G ' C6

GROUPS 21

It is common to write

1 =

(1 00 1

), i =

(i 00 −i

), j =

(0 1−1 0

)and k =

(0 ii 0

).

Then Q8 = {±1,±i,±j,±k}. Then we can compute that 1 is an identity, −1 hasorder 2 commutes with everything and multiplies as you’d expect given the notation.Morever i, j and k all have order 4 (all of them square to −1), ij = k = −ji,ki = j = ik and jk = i = −kj.

Exercise. Verify that Q8 is a group.

Lecture 10

Proposition. Every group of order 8 is isomorphic to precisely one of C8, C4×C2,C2 × C2 × C2, D8 or Q8.

Proof. Let G be a group of order 8.If G has an element of order 8 then G ' C8.If every non-identity element of G has order 2 then G is abelian. Moreover any

two non-identity elements g1, g2 generate a subgroup H1 of G of order 4 necessarilyisomorphic to C2 ×C2.30 If g3 ∈ G\H1 then H2 = {e, g3} is a subgroup of order 2.One can easily verify that G ' H1×H2 ' C2×C2×C2 in a similar fashion to theargument above for C2 × C2.31

So we’re reduced to classifying groups G of order 8 with no element of order 8and at least one element g of order 4. In this setting the set K = {e, g, g2, g3} is asubgroup of G isomorphic to C4.

Let h ∈ G\K. ThenG = K∪hK by Lagrange soG = {e, g, g2, g3, h, hg, hg2, hg3}.Moreover h2 ∈ K else h2 ∈ hK whence h ∈ K.

If h2 = g or h2 = g−1 then h has order 8 contradicting our assumption that Ghas no such elements. So there are two cases remaining: case A where h2 = e andcase B where h2 = g2.

Suppose that we’re in case A and h2 = e. Then consider the product gh. ByLagrange again, gh ∈ hK = {h, hg, hg2, hg3}. If gh = hgi then h−1gh = gi and

g = h−2gh2 = h−1(h−1gh)h = h−1gih = (h−1gh)i = (gi)i = gi2

.32

Thus i = 1 or 3. If i = 1 then G is abelian and we can use the direct producttheorem to show that G ' K × {e, h} ' C4 × C2. If instead i = 3 then G has anelement g of order 4 and an element h of order 2 such that gh = hg−1 so as G hasorder 8 we know that G ' D8.

Finally suppose we’re in case B and h2 = g2. Again consider the product gh ∈{h, gh, g2h, g3h} = Kh = hK = {h, hg, hg2, hg3}. Since g2h = h3 = hg2, gh = hgor gh = hg3. If gh = hg then

(gh)2 = g2h2 = g4 = e

so {e, gh} 6 G and G ' K × {e, gh} ' C4 × C2 by the direct product theorem.

30the elements are e, g1, g2 and g1g2 it is easy to check that these are distinct and form a

subgroup.31See also Example Sheet 1 Q1432since h2 = e

22 SIMON WADSLEY

If gh = hg3 then do some renaming. Write 1 := e, −1 := h2 = g2, i := g, j := h,k := gh. Then

G = {±1,±i,±j,±k}and we can verify that multiplication is as for Q8.

Exercise. Complete the proof by showing that no two of the listed groups areisomorphic. �

2.4. The Quaternions.

Definition. The quaternions are the set of matrices

H := {a1 + bi + cj + dk | a, b, c, d ∈ R} ⊂ Mat2(C)

where as before

1 =

(1 00 1

), i =

(i 00 −i

), j =

(0 1−1 0

)and k =

(0 ii 0

).

The sum or product of two elements of H lives in H and + and · obey thesame associativity and distributivity laws as Q,R and C with identities 0 and 1respectively. Although the multiplication in H is not commutative (since ij = −ji),(H,+) is an abelian group and (H\{0}, ·) is a (non-abelian) group.33 To see thelatter we can define ‘quaternionic conjugation’ by

(a1 + bi + cj + dk)∗ = a1− bi− cj− dkand then verifying that if x = a1 + bi + cj + dk then

xx∗ = x∗x = (a2 + b2 + c2 + d2)1

so x−1 = 1a2+b2+c2+d2x

∗ for x 6= 0.

2.5. Fermat–Euler theorem. We can define a multiplication operation on theset Zn = {0, 1, . . . , n− 1} by setting a ·n b to be the remainder after dividing ab byn.

Definition. Let Un := {a ∈ Zn | ∃b ∈ Zn s.t. a ·n b = 1} be the set of invertibleelements of Zn with respect to ·n.

It is a result from Numbers and Sets that follows from Euclid’s algorithm that|Un| = ϕ(n) where ϕ(n) denotes the number of elements of Zn coprime to n. IndeedUn = {a ∈ Zn | (a, n) = 1}.

Lemma. (Un, ·n) is an abelian group.

Proof. ·n defines an associative and commutative operation on Zn with an identity1.34 If a1, a2 ∈ Un then there are b1, b2 ∈ Un such that ai ·n bi = 1. Then

(a1 ·n a2) ·n (b1 ·n b2) = (a1 ·n b1) ·n (a2 ·n b2) = 1 ·n 1 = 1

so ·n restricts to an associative binary operation on Un. 1 ∈ Un is an identity andif a ∈ Un then there is b ∈ Zn with a ·n b = b ·n a = 1. Then b ∈ Un and b = a−1.So Un has inverses. �

Theorem (Fermat–Euler Theorem). If (a, n) = 1 then aϕ(n) ≡ 1 mod n.

33We say that H is a division algebra. R,C and H are the only division algebras that are finite

dimensional as vector spaces over R.34for a, b, c ∈ Zn, a ·n (b ·n c) ≡ abc ≡ (a ·n b) ·n c mod n

GROUPS 23

Proof. Since (Un, ·n) is a group of order ϕ(n) it is consequence of Lagrange’s the-orem that every element of (Un, ·n) has order dividing ϕ(n). The result followsimmediately. �

Lecture 11

3. Group Actions

We started the course by saying that groups are fundamentally about symmetrybut the connection has been opaque for the last three lectures. In this chapter wewill discuss how to recover the notion of symmetry from the group axioms.

3.1. Definitions and examples.

Definition. An action of a group G on a set X is a function

· : G×X → X; (g, x) 7→ g · xsuch that for all x ∈ X

(i) e · x = x;(ii) g · (h · x) = (gh) · x for all g, h ∈ G.

Examples.

(1) Isom(Z) acts on Z via f · n = f(n).(2) The Mobius groupM acts on the extended complex plane C∞ via f · z = f(z).(3) Generalising both the examples above, if H 6 S(X) then H acts on X via

h · x = h(x). We call this the natural action of H on X.(4) M also acts on the set of circles in C∞. We proved in §1.5 that if f ∈ M

and C ⊂ C∞ is a circle then f(C) ⊂ C∞ is also a circle so (f, C) 7→ f(C) is afunction. Moreover for all circles C the conditions id(C) = C and f(g(C)) =(fg)(C) for f, g ∈M are both clear.

(5) D2n acts on the set of points a regular n-gon. D2n also acts on the set ofvertices of a regular n-gon and on the set of edges of a regular n-gon.

(6) If X is a regular solid then Sym(X) acts on the set of points (and on the setsof vertices/edges/faces) of X.

(7) If H 6 G then G acts on G/H, the set of left cosets of G in H via g ·kH = gkHfor g, k ∈ G. To see this we need to check that if kH = k′H then gkH = gk′H.But if kH = k′H then k′ = kh for some h ∈ H so gk′ = gkh ∈ gk′H ∩gkH andgkH = gk′H by Lagrange. Given this we see that for all k ∈ G, ekH = kHand g1(g2kH) = (g1g2)kH for all g1, g2 ∈ G.

(8) For any group G and set X we can define the trivial action via g · x = x for allg ∈ G and x ∈ X.

Theorem. For every group G and set X there is a 1− 1 correspondence

{actions of G on X} ←→ {θ : G→ S(X) | θ is a homomorphism}such that an action · : G×X → X corresponds to the homomorphism θ : G→ S(X)given by θ(g)(x) = g · x.

Proof. First we must show that if · : G × X → X is a action then the formulaθ(g)(x) = g · x does define a homomorphism G→ S(X).

For each g ∈ G, define θ(g) : X → X via θ(g)(x) = g · x. So that

θ : G→ {f : X → X}.

24 SIMON WADSLEY

Then for g, h ∈ G we can compute

θ(gh)(x) = (gh) · x = g · (h · x) = θ(g) ◦ θ(h)(x)

ie θ(gh) = θ(g)θ(h). So to show that θ defines a homomorphism G → S(X) itsuffices to show that θ(g) ∈ S(X) for each g ∈ G; i.e. that each θ(g) is invertible.But

θ(g)θ(g−1) = θ(e) = θ(g−1)θ(g)

so to show this it suffices to show that θ(e) is idX . But θ(e)(x) = e·x = x = idX(x).So we’re done.

To finish we must show that every homomorphism θ : G→ S(X) arises like thisin precisely one way. So suppose that θ is any such homomorphism, θ correspondsto an action · : G × X → X precisely if g · x = θ(g)(x) defines an action. So itremains to check that e · x = x and g · (h · x) = (gh) · x for all x ∈ X and g, h ∈ G.But

e · x = θ(e)(x) = idX(x) = x35

andg · (h · x) = θ(g)(θ(h)(x)) = θ(gh)(x) = (gh) · x

as required. �

Definition. We say that an action of G on X is faithful if the kernel of the corre-sponding homomorphism G→ S(X) is the trivial group.

3.2. Orbits and Stabilisers.

Definition. Suppose a group G acts on a set X and that x ∈ X. The orbit of xunder the action is given by

OrbG(x) := {g · x | g ∈ G} ⊂ X.The stabiliser of x under the action is given by

StabG(x) := {g ∈ G | g · x = x} ⊂ G.Thus an action is faithful precisely if

⋂x∈X StabG(x) = {e}.

Examples.(1) Under the natural action of Isom(Z) on Z, for all n ∈ Z

OrbIsom(Z)(n) = Zand

StabIsom(Z) = {id,m 7→ 2n−m}(2) Under the natural action of M on C∞, for all z ∈ C∞

OrbM(z) = C∞and

StabM(∞) =

{z 7→ az + b

0c+ d| ad 6= 0

}.

(3) Under the action of D2n on the set of points of a regular n-gon the orbit of avertex of the n-gon is the set of all vertices of the n-gon and the stabliser of avertex consists of the identity and reflection in the line through the centre ofthe n-gon and the vertex.36

35Since θ is a homomorphism it must send the identity of G to the identity of S(X).36It would be instructive to think about what the orbits and stabilisers of other points of n-gon

are under the action of D2n.

GROUPS 25

(4) For the left coset action of G on G/H defined earlier

OrbG(eH) = G/H

andStabG(eH) = {g ∈ G | gH = eH} = H.

More generally

StabG(kH) = {g ∈ G | gkH = kH} = {g ∈ G | k−1gkH = H} = kHk−1.

(5) For the trivial action of G on X and any x ∈ X,

OrbG(x) = {x} and StabG(x) = G.

Lemma. Suppose that G is a group acting on a set X.

(i) Each stabiliser StabG(x) is a subgroup of G.(ii) The orbits OrbG(x) partition X. In particular if X is finite and the distinct

orbits are O1, . . . ,Om then

|X| =m∑i=1

|Oi|

Lecture 12

Proof. (i) Let x ∈ X. Since e · x = x, e ∈ StabG(x). Suppose that g, h ∈ StabG(x).Then

(gh) · x = g · (h · x) = g · x = x

so gh ∈ StabG(x) and

x = e · x = (g−1g) · x = g−1 · (g · x) = g−1 · xso g−1 ∈ StabG(x). Thus StabG(x) 6 G as required.37

(ii) Since for any x ∈ X, x = e · x so x ∈ OrbG(x), it suffices to prove thatfor any x, y ∈ X either OrbG(x) = OrbG(y) or OrbG(x) ∩ OrbG(y) = ∅. Supposethat z ∈ OrbG(x) ∩ OrbG(y). Then there are some elements g, h ∈ G such thatg · x = z = h · y. Thus x = (g−1h) · y. Now if w ∈ OrbG(x) then there is somef ∈ G such that w = f · x. Whence we can compute w = (fg−1h) · y and soOrbG(x) ⊂ OrbG(y). By symmetry OrbG(y) ⊂ OrbG(x). Now we can see thatOrbG(x) = OrbG(y) as required.38 �

Definition. We say that an action of G on X is transitive if there is only one orbiti.e. if X = OrbG(x) for any x ∈ X.

Theorem (Orbit-Stabiliser Theorem). Suppose a group G acts on a set X andx ∈ X. There is a (natural) invertible function

G/StabG(x)→ OrbG(x).

In particular if G is finite

|G| = |OrbG(x)| · | StabG(x)|.37We used the same argument for a special case of this in Lecture 4 when we showed that

StabIsom(Z)(0) 6 Isom(Z).38You should spend some time thinking about the relationship between this proof and the

proof of Lagrange’s Theorem. Can you find an action of H on G making Lagrange a special case

of this result?

26 SIMON WADSLEY

Proof. For y ∈ OrbG(x), {g ∈ G | g · x = y} is a left coset of StabG(x) in G sinceit is non-empty and if g · x = y then for h ∈ G

h · x = y ⇐⇒ g−1 · (h · x) = x

⇐⇒ g−1h ∈ StabG(x)

⇐⇒ h ∈ g StabG(x).

Thus there is a function G/StabG(x) → OrbG(x) given by hStabG(x) 7→ h · xwith inverse OrbG(x)→ G/StabG(x) given by y 7→ {g ∈ G | g · x = y}.

Now if G is finite then |OrbG(x)| = |G/StabG(x)| = |G|/|StabG(x)| by La-grange.39 �

Examples.(1) For the natural action of Isom(Z) on Z the set of left cosets of StabIsom(Z)(0) ={e, n 7→ −n} in Isom(Z) is in bijection with Z. We secretly used this fact whenwe computed Isom(Z) in the first lecture.

(2) For the usual action of D2n on the vertices of the n-gon and v such a vertex wesee that |D2n| = |StabD2n

(v)||OrbD2n(v)| = 2n. Again we secretly used this

when we computed |D2n| = 2n.(3) The symmetric group Sn acts on X = {1, 2, . . . , n} via the natural action

f ·x = f(x). Then OrbSn(n) = X since for each i ∈ X the function fi : X → X;fi(i) = n, fi(n) = i, fi(x) = x for x 6∈ {i, n} is an element of Sn. Thus|Sn| = nStabSn(n). But StabSn(n) is isomorphic to Sn−1 by restricting f ∈ Snthat fixes n to a permutation of {1, . . . , n − 1}. Thus |Sn| = n|Sn−1|. Since|S1| = 140 we deduce that |Sn| = n!.

Lecture 13

Fact. If f : R3 → R3 is an isometry that fixes 4 non-coplanar points then f isthe identity.

(4) Let X be a regular tetrahedron. Then Sym(X) acts transitively on the set of4 vertices of X and the stabiliser of a vertex v ∈ X consists of three rotationsand three reflections. Thus |Sym(X)| = 6 · 4 = 24.

This calculation enables us to prove that Sym(X) ' S4: if we label thevertices by the numbers 1, 2, 3, 4 then the action of Sym(X) on the verticesdefines a homomorphism θ : Sym(X)→ S4. Since any isometry of R3 fixing allfour vertices is the identity we can conclude that ker θ = {id}. By counting wecan deduce Im θ = S4.

(5) Let X be a cube. Then Sym(X) acts transitively on the set of 6 faces of X andthe stabiliser H := StabSym(X)(F ) of a face F acts transitively on the set of 4

vertices contained in it41. If v is one of these vertices and w is the diagonallyopposite vertex in F then

StabH(v) = {e, reflection in plane containing v, w and the centre of X}42.

39You might like to think about whether you can do this last part without recourse to Lagrange.40Or if you prefer |S0| = 141This can be seen by considering rotations about an axis through the centre of F and the

centre of its opposite face.42Since if an isometry of R3 fixes all vertices of F and the centre of X then it is the identity.

GROUPS 27

Thus

Sym(X) = 6|H| = 6 · |OrbH(v)||StabH(v)| = 6 · 4 · 2 = 48.

3.3. Conjugacy classes.

Definition. If G is a group then the conjugation action of G on itself is given by· : G×G→ G; g · x = gxg−1.

Note that the conjugation action is indeed an action since for g, h, x ∈ G,

e · x = exe−1 = x

and

g · (h · x) = g(hxh−1)g−1 = (gh)x(gh)−1 = (gh) · x.

Definition. The orbits of G on itself under the conjugation action are called theconjugacy classes of G: the orbit of x ∈ G will be denoted ccl(x); i.e.

ccl(x) = {gxg−1 | g ∈ G}.43

The stabliser of x ∈ G under this action is called the centraliser of x and will bedenoted CG(x).

Examples.(1) Suppose G = Isom(Z) = {ta : n 7→ a + n, sa : n 7→ a − n | a ∈ Z}. Let

H := {ta | a ∈ Z} denote the subgroup of translations We know that for anya, b ∈ Z,

tbtat−1b = ta

and for n ∈ Z,

sbtas−1b (n) = sbta(b− n) = sb(a+ b− n) = n− a

ie

sbtas−1b = t−a

so for a 6= 0

CG(ta) = H and ccl(ta) = {ta, t−a}44.Similarly for n ∈ Z

tbsat−1b (n) = tbsa(n− b) = tb(a− (n− b)) = (2b+ a− n) = s2b+a(n)

and

sbsas−1b (n) = sbsa(b−n) = sb(a− (b−n)) = b− (a+n− b) = 2b−a−n = s2b−a(n)

so as a ≡ −a mod 2

CG(sa) = {t0, sa} and ccl(sa) = {sa+2b | b ∈ Z}.

That is there are two conjugacy classes of reflections ccl(s0) and ccl(s1).45

43Since conjugacy classes are orbits of an action they partition G; that is every element of G

lies in precisely one conjugacy class.44Of course CG(t0) = G and ccl(t0) = {t0}.45What is the geometric meaning of this?

28 SIMON WADSLEY

(2) We saw in section 1.5 that in the Mobius group M the conjugacy class ofz 7→ z+ 1 consists of all Mobius transformations with precisely one fixed pointi.e.

ccl(z 7→ z + 1) = {f ∈M | f has precisely one fixed point}and that every Mobius transformation with precisely two fixed points is in thesame conjugacy class as a Mobius transformation of the form z 7→ az. We willreturn later to the question of when ccl(z 7→ az) = ccl(z 7→ bz) and what thecentralisers of these elements are.46 Of course ccl(id) = {id} and CM(id) =M.

Definition. The kernel of the homomorphism G→ S(G) given by the conjugationaction of G on itself is called the centre of G and written Z(G).

Lemma. Suppose that G is a group.

(a) For x ∈ G, CG(x) = {g ∈ G | xg = gx}.(b) Z(G) = {g ∈ G | gx = xg for all x ∈ G} =

⋂x∈G CG(x).

(c) Z(G) = {g ∈ G | |ccl(g)| = 1}.

Proof. (a) For g, x ∈ G, g ∈ CG(x) if and only if gxg−1 = x that is if and only ifgx = xg.

(b) For g ∈ G, g ∈ Z(G) if and only if gxg−1 = x for all x ∈ G i.e. if and only ifgx = xg for all x ∈ G. The other equality follows immediately from (a).

(c) This follows easily from (b): g ∈ Z(G) if and only if gx = xg for all x ∈ Gi.e. if and only if xgx−1 = g for all x ∈ G. �

3.4. Cayley’s Theorem. Cayley’s Theorem will tell us that every group is iso-morphic to a subgroup of a symmetric group.

Definition. If G is a group then the left regular action of G on itself is given bythe function · : G×G→ G; g · x = gx.

Example. The left regular action of Z on itself is by translations. i.e. the corre-sponding homomorphism Z→ S(Z) is given by n 7→ tn.47

Lemma. The left regular action of G on G is an action that is both transitive andfaithful.

Proof. First we should check that the left regular action is indeed an action: for allg, h, x ∈ G, e · x = ex = x and g · (h · x) = ghx = (gh) · x.

Next we observe that OrbG(e) = G and StabG(e) = {e} since for all g ∈ G,g · e = g. Thus the left regular action is transitive and faithful. �

Lecture 14

Theorem (Cayley’s Theorem). If G is a group then G is isomorphic to a subgroupof S(G).

46Spoiler: ccl(z 7→ az) = ccl(z 7→ bz) if and only if b ∈ {a, 1/a}, CG(z 7→ z + 1) ={translations in M} and, for a 6= 1, CG(z 7→ az) = {dilations/rotations in M}. Can you provethese facts now? Hint: if g(z 7→ az)g−1 = z 7→ bz for g ∈ M what can you say about g(0) and

g(∞)?47recall tn denotes translation by n

GROUPS 29

Proof. The left-regular action defines a homomorphism θ : G → S(G). Since Im θis a subgroup of G we may view this as a homomorphism G→ Im θ whose image isstill Im θ. Since the action is faithful, ker θ = {e} and we see that G is isomorphicto Im θ. �

It perhaps should be said that this theorem is simultaneously deep and almostuseless. Deep because it tells us that anything satisfying our abstract definition of agroup can be viewed as symmetries of something. Almost useless because knowingthis doesn’t really help prove things about groups.

3.5. Cauchy’s Theorem.

Theorem (Cauchy’s Theorem). Supppose that p is a prime and G is a finite groupwhose order is a multiple of p. Then G contains an element of order p.

Proof. Consider the action of Zp on Gp via

i · (g0, g1, . . . , gp−1) = (gi, g1+pi, . . . , g(p−1)+pi)

i.e. by cyclic permutation. This is an action since 0 · x = x for all x ∈ Gp and

(i+p j) · (g0, g1, . . . , gp−1) = (gi+pj , g1+pi+pj , . . . , g(p−1)+pi+pj)

= i · (j · (g0, g1, . . . , gp−1))

for all i, j ∈ Zp and g0, . . . , gp−1 ∈ G.Let X = {(g0, . . . , gp−1) ∈ Gp | g0g1 · · · gp−1 = e} ⊂ Gp. Then |X| = |G|p−1

since however we choose g0, . . . , gp−2 ∈ G there is precisely one choice of gp−1 ∈ Gsuch g0g1 · · · gp−1 = e48. Moreover the ‘constant tuple’ (g, g, . . . , g) is in X if andonly gp = e — and if g 6= e this is equivalent to o(g) = p.

Now the cyclic permutation action of Zp on Gp restricts to an action on X sinceif g0g1 · · · gp−1 = e and i ∈ Zp then

gig1+pi · · · g(p−1)+pi = (g0 · · · gi−1)−1(g0 · · · gp−1)(g0 · · · gi−1) = e.

Since Zp has prime order the orbit-stabiliser theorem tells us that every orbitOrbZp(x) has order 1 or p for x ∈ X. Since the orbits partition X and p is a factorof |X| it follows that the number of orbits of size 1 in X is a multiple of p.

Since the constant tuple (e, e, . . . , e) is an orbit in X of size 1 there must beat least p − 1 other such orbits. If (g0, . . . , gp−1) is one such orbit then, sincei · (g0, . . . , gp−1) = (g0, . . . , gp−1) for all i ∈ Zp, we can conclude that gi = g0 for alli ∈ Zp and so gp0 = e. �

4. Quotient Groups

4.1. Normal subgroups. Suppose that G is a group. Let P(G) denote the set ofsubsets of G, i.e. the power set of G. There is a natural binary operation on P(G)given by

AB := {ab | a ∈ A, b ∈ B}.

Examples.(1) If A ∈P(G) then A∅ = ∅ = ∅A. If A is non-empty then AG = G = GA.(2) If H 6 G then the binary operation on P(G) restricts to a binary operation

on P(H).(3) If H 6 G then the sets {g}H are precisely the left cosets gH of H in G.

48That is gp−1 = (g0g1 · · · gp−2)−1.

30 SIMON WADSLEY

Lemma. This operation on P(G) is associative and has an identity but does nothave inverses.

Proof. Suppose that A,B,C ∈P(G) then

(AB)C = {(ab)c | a ∈ A, b ∈ B, c ∈ C}= {a(bc) | a ∈ A, b ∈ B, c ∈ C}= A(BC)

If A is any subset of G then {e}A = A so {e} is an identity. Also always A∅ = ∅,so ∅ has no inverse. 49 �

We’ll be particularly interested in the product of two cosets under this operation— in particular if H 6 G we’d like to use it to put a group structure on the set ofleft cosets G/H of H in G. If G is abelian then this is straightforward:

g1Hg2H = {g1h1g2h2 | h1, h2 ∈ H} = {g1g2h1h2 | h1, h2 ∈ H} = g1g2H

and one can easily50 show that this does define a group structure on G/H. Howeverin general things are not so straightforward.

Example. Consider G = D6 = {e, r, r2, s, rs, r2s} where r denotes a non-trivialrotation in the group and s a reflection.

If H is the subgroup of rotations {e, r, r2} then the cosets of H in G are H andsH. We can compute

HH = H

HsH = sH

sHH = sH and

sHsH = H.

So G/H with this operation is isomorphic to C2.However if K is the subgroup {e, s} of G then

rKr2K = {r, rs}{r2, r2s} = {e, r2s, s, r2}which is not a left coset of K in G.

Proposition. Suppose H 6 G. The product of two left cosets of H in G is alwaysa left coset of H in G if and only if gHg−1 = H51 for all g ∈ G. In this caseg1Hg2H = g1g2H for all g1, g2 ∈ G.

Lecture 15

Proof. For g1, g2 ∈ G we compute

g1Hg2H = {g1h1g2h2 | h1, h2 ∈ H} = {g1g2g−12 h1g2h2 | h1, h2 ∈ H}.

Thus if g−12 Hg2 = H then g1Hg2H = g1g2H as claimed.Moreover in general g1g2H ⊂ g1Hg2H since we may take h1 = e above. Thus

for g1Hg2H to be a left coset we must have g1Hg2H = g1g2H as claimed. For this

49We note in passing that we didn’t use that G has inverses so a version of this result is still

true for G any set with an associative binary operation with an identity such as (Z, ·) or (N0,+).50and we will later51Here gHg−1 means {g}H{g−1}

GROUPS 31

we need g−12 h1g2 ∈ H for all h1 ∈ H since (g1g2)g−12 h1g2e ∈ g1Hg2H. So if theproduct of any two left cosets is a left coset then g−1Hg ⊂ H for all g ∈ G. Then

H = g(g−1Hg)g−1 ⊂ gHg−1 ⊂ H

so we have equalities throughout. �

Remark. Notice that along the way we proved that whenever gHg−1 ⊂ H for allg ∈ G, in fact gHg−1 = H for all g ∈ G.

Definition. We say that a subgroup H of a group G is normal if gHg−1 = H forall g ∈ G.

Warning. To show that a subset of G is a normal subgroup we must showthat it is a subgroup as well as that it satisfies the above conditon.

Examples.(1) If G is abelian then every subgroup is normal.(2) The group Isom+(Z) is normal in Isom(Z) but the subgroup {idZ, s : n 7→ −n}

is not normal is Isom(Z).(3) The subgroup of rotations in D2n is normal in D2n but no subgroup generated

by a reflection is normal in D2n.(4) StabM(∞) is not a normal subgroup of M.

Lemma. A subgroup H of a group G is normal if and only if every left coset is aright coset.52

Proof. Suppose that gHg−1 = H. Then gH = gHg−1g = Hg so the left coset gHis a right coset.

Conversely suppose that every left coset is a right coset and g ∈ G. ThengH = Hk for some k ∈ G. So g = hk for some h ∈ H. Thus

gH = Hk = Hkg−1g = Hh−1g = Hg

and gHg−1 = H as required. �

Proposition. If H is a normal subgroup of G then the restriction of the binaryoperation on P(G) makes G/H into a group such that g1Hg2H = g1g2H.

Definition. We call G/H the quotient group of G by H.

Proof of Proposition. We’ve already seen that the binary operation on P(G) re-stricts to an associative binary operation on G/H when H is normal and moreoverthat g1Hg2H = g1g2H.

Suppose that gH ∈ G/H. Then eHgH = gH = gHeH so eH = H is an identityin G/H. Finally gHg−1H = eH = g−1HgH so G/H has inverses. �

52We’ll often just say coset in this case.

32 SIMON WADSLEY

4.2. The isomorphism theorem.

Theorem (The (first) isomorphism theorem). Suppose that f : G→ H is a grouphomomorphism. Then ker f is a normal subgroup of G, Im f is a subgroup of Hand f induces an isomorphism

f : G/ ker f'−→ Im f

given by f(g ker f) = f(g).

Proof. We’ve already seen that ker f 6 G and Im f 6 H. So first we must showthat g(ker f)g−1 = ker f for all g ∈ G.

Suppose that k ∈ ker f and g ∈ G. Then

f(gkg−1) = f(g)f(k)f(g−1) = f(g)ef(g)−1 = e

since f(k) = e. Thus g(ker f)g−1 ⊂ ker f . As remarked above this suffices to seethat ker f is normal.

Now if g ker f = g′ ker f then g−1g′ ∈ ker f and so f(g)−1f(g′) = f(g−1g′) = e.Thus f(g) = f(g′) and f(g ker f) = f(g) is a well-defined function whose image isIm f . Moreover

f(g1 ker fg2 ker f) = f(g1g2 ker f) = f(g1g2) = f(g1)f(g2) = f(g1 ker f)f(g2 ker f)

so f is a homomorphism fromG/H to Im f with Im f = Im f . Finally if f(g ker f) =e then f(g) = e i.e. g ∈ ker f . So ker f = {ker f} = {eG/ ker f} and the result followsfrom the special case of the first isomorphism theorem proven in section 1.4. �

Remark. Often a good way to prove that a subset of a group G is a normal subgroupis to show that it is the kernel of some homomorphism from G to another group.

Lecture 16

Example. The homomorphism Z→ Zn that sends a to the remainder after dividing

a by n has kernel nZ and image Zn. Thus it induces an isomorphism Z/nZ '→ Zn.53

Example. Let θ : (R,+) → (C\{0}, ·) be given by θ(r) = e2πir. Then θ(r + s) =e2πi(r+s) = θ(r)θ(s) so θ is a homomorphism. Moreover

Im θ = S1 := {z ∈ C | |z| = 1},the unit circle in C and

ker θ = Zthus we can deduce that R/Z ' S1.

Example. Let θ : D2n → {±1} such that

θ(g) :=

{+1 if g is a rotation−1 if g is a reflection.

Then we can verify that θ is a homomorphism since the product of two reflectionsor two rotations is a rotation and the product of a rotation and a reflection ineither order is a reflection. Moreover Im θ = {±1} and ker θ is the subgroup of allrotations of the regular n-gon. Thus D2n/{rotations in D2n} ' C2.

53The notation Zn as we have defined it is rarely used and instead Z/nZ is used to describeessentially the same thing.

GROUPS 33

Example (Group-theoretic understanding of qth powers mod p). Let p and q bedistinct primes and G = (Zp\{0}, ·p). Define

θ : G→ G;x 7→ xq.

Then for x, y ∈ G, θ(xy) = (xy)q = xqyq = θ(x)θ(y) i.e. θ is a homomorphism.Then

ker θ = {x ∈ G | xq = 1} = {x ∈ G | o(x) = 1 or q}.

We now divide into two cases.First suppose that q is not a factor of p−1. Since |G| = p−1, G has no elements

of order q by Lagrange. Thus ker θ = {1}. It follows that θ induces an isomorphismG ' Im θ. By counting we can conclude that Im θ = G. In particular we see thatevery element of Zp is a qth power when p is not 1 mod q.

Next suppose that q is a factor of p − 1. In this case G does have an elementof order q by Cauchy’s Theorem. Thus | ker θ| > q.54 Since G/ ker θ ' Im θ and|G/ ker θ| = |G|/| ker θ| 6 p−1

q we see that Zp has at most p−1q + 1 qth-powers when

p is 1 mod q.55

Example. If G acts on a set X and K = {g ∈ G | g(x) = x for all x ∈ X} =⋂x∈X StabG(x) then the homomorphism G → S(X) given by the action induces

an isomorphism from G/K to a subgroup of S(X). Thus the action of G on Xinduces a faithful action of G/K on X.56

Example. Suppose that X is a regular tetrahedron in R3. X has six edges and eachedge has four neighbours.57 Thus we can partition the set of edges into three pairswith each pair consisting of non-adjacent edges. Let P denote the set of such pairs.Then the action of Sym(X) on X induces an action on P since if f ∈ Sym(X) and vand w are non-adjacent edges of X then f(v) and f(w) are also non-adjacent edgesof X. Thus by the last example there is a homomorphism θ : Sym(X) → S(P ).It is easy to verify by hand that Im θ = S(P ). Then the isomorphism theorem wecan deduce that Sym(X)/ ker θ ' S(P ). We showed earlier than Sym(X) ' S4 andit is straightforward to see that S(P ) ' S3.58 Thus we can deduce that S4 has anormal subgroup K such that S4/K ' S3.59

Lecture 17

5. Matrix groups

Suppose that throughout this section F denotes either R or C.

54Since an element of order q generates a subgroup of order q contained in the kernel. In factit is not too hard to prove that ker θ has precisely q elements.

55In fact precisely this many.56This means that to understand all actions of a group G it is equivalent to understand all

faithful actions of all quotients of G.57There are two edges sharing each vertex of a given edge.58Or S(P ) ' D6 if you prefer59Can you say which elements of S4 live in K? There must be four of them by Lagrange. If

you find this too hard at this stage then try again when you revise the course having studied the

groups Sn in more detail.

34 SIMON WADSLEY

5.1. The general and special linear groups. Let Mn(F) denote the set of n×nmatrices with entries in F.

Here are some facts proven in Vectors and Matrices.

Facts.(1) Every element A of Mn(F) defines a linear map A : Fn → Fn via A : v 7→ Av.60

Moreover every linear map Fn → Fn arises in this way and A can be recoveredfrom A since the ith column of A is A(ei) where ei denotes the element of Fnwith ith entry 1 and all other entries 0.

(2) AB corresponds to the composite A◦B. Thus associativity of multiplication of(square) matrices follows from associativity of composition of functions Fn →Fn.

(3) The matrix In with 1s down the main diagonal and 0s elsewhere is an identityfor matrix multiplication on Mn(F). Moreover In = idFn .

(4) There is a function det : Mn(F)→ F such that A has an inverse in Mn(F) if andonly if detA 6= 0. Moreover det(AB) = det(A) det(B) for any A,B ∈ Mn(F)and det In = 1.

Definition. The general linear group GLn(F) := {A ∈ Mn(F) | detA 6= 0} is thegroup of invertible n× n matrices with entries in F.

Proposition. GLn(F) is a group under matrix multiplication.

Proof. Since for A,B ∈ Mn(F), detAB = detA detB, if A,B ∈ GLn(F) thendetA 6= 0 and detB 6= 0 so detAB 6= 0. Thus AB ∈ GLn(F) and matrix multipli-cation defines an associative binary operation on GLn(F).

Since det In = 1, In ∈ GLn(F) so GLn(F) has an identity.Finally if A ∈ GLn(F), since detA 6= 0, there is a matrix B such that AB = In =

BA. Then detA detB = detAB = det In = 1. So detB 6= 0 and B ∈ GLn(F). �

Remark. There is a natural action of GLn(F) on Fn via (A, v) 7→ Av. One canshow that the homomorphism GLn(F) → S(Fn) coming from this action inducesan isomorphism GLn(F) with the subgroup of S(Fn) consisting of all invertiblelinear maps Fn → Fn.

Lemma. The function det : GLn(F) → (F\{0}, ·) is a group homomorphism withimage F\{0}.Proof. That it is a homomorphism follows immediately from fact 4 above: forA,B ∈ GLn(F), detAB = detAdetB. To see its image we compute

det

λ 0

. . . 0

0 1. . .

. . .

. . .. . .

. . . 0

0. . . 0 1

= λ.

�

Definition. The special linear group SLn(F) is the kernel of det : GLn(F)→ F\{0}i.e.

SLn(F) := {A ∈Mn(F) | detA = 1}.60Recall that A is linear means that A(λv+µw) = λA(v)+µA(w) for all λ, µ ∈ F and v, w ∈ Fn.

GROUPS 35

Remarks.

(1) The action of GLn(F) on Fn induces an action of SLn(F) on Fn by restrictionand the resulting homomorphism SLn(F) → S(Fn) induces an isomorphismof SLn(F) with the subgroup of S(Fn) consisting of volume preserving linearmaps Fn → Fn.

(2) SLn(F) a normal subgroup of GLn(F and GLn(F)/SLn(F) ' F\{0}.

Examples.

GL2(F) =

{(a bc d

)| ad− bc 6= 0

}and

SL2(F) =

{(a bc d

)| ad− dc = 1

}5.2. Mobius maps as projective linear transformations.

Notation. Given v ∈ C2\{0} let [v] denote the (unique) line {λv | λ ∈ C} through0 and v in C2. The set of all such lines is called the complex projective line typicallywritten P1(C).

The following lemma gives a parameterisation of the elements of P1(C).

Lemma. Every element of P1(C) is either of the form

[(z1

)]with z ∈ C or[(

10

)]. Moreover these lines are all distinct.

Proof. Suppose v ∈ C2\{0}. Then v =

(v1v2

)for some v1, v2 ∈ C not both 0.

If v2 6= 0 then [v] =

[(v1v21

)]since λ

(v1v2

)= v2λ

( v1v21

).

If v2 = 0 then [v] =

[(10

)]since λv = v1λ

(10

).

Finally if λ

(z1

)=

(w1

)then λ = 1 and z = w. And λ

(z1

)=

(10

)is evidently

impossible. �

It follows that we may identify C∞ and P1(C) via z 7→[(z1

)]for z ∈ C and

∞ 7→[(

10

)].

Proposition. GL2(C) acts on P1(C) via (A, [v]) 7→ [Av] for v ∈ C2\{0}.

Proof. First we must show that (A, [v]) 7→ [Av] is a well-defined function

GL2(C)× P1(C)→ P1(C)

i.e. that if A ∈ GL2(C) and v ∈ C2\{0} then Av 6= 0, and if [v] = [w] then [Av] =[Aw]. Now, if Av = 0 then v = A−1Av = A−1(0) = 0 and if [v] = [w] there is somenon-zero µ ∈ C such that v = µw. Then for λ ∈ C, A(λw) = λA(µv) = (λµ)Av so[Aw] ⊂ [Av]. By symmetry we must have [Av] = [Aw].

It remains to observe that [I2v] = [v] and that [A(Bv)] = [(AB)v] for all [v] ∈P1(C) and A,B ∈ GL2(C). �

36 SIMON WADSLEY

We note that under this action of GL2(C) on P1(C)(a bc d

)·[(z1

)]=

[(az + bcz + d

)]=

[(az+bcz+d

1

)]when z 6= −d/c,

(a bc d

)·[(−d/c

1

)]=

[(10

)],

(a bc d

)·[(

10

)]=

[(ac

)]=

[(ac1

)]when c 6= 0, and

(a b0 d

)·[(

10

)]=

[(10

)].

Thus, under the identification of C∞ with P1(C), the homomorphism

θ : GL2(C)→ S(C∞)

corresponding to this action sends the matrix

(a bc d

)to the Mobius map repre-

sented by z 7→ az+bcz+d , and so Im θ =M. Thus M is a subgroup of S(C∞).

Moreover ker θ consists of invertible matrices

(a bc d

)fixing every line through

the origin in C2.Now

StabGL2(C)

([(10

)])=

{(a bc d

)∈ GL2(C) | c = 0

},

StabGL2(C)

([(01

)])=

{(a bc d

)∈ GL2(C | b = 0

}and

StabGL2(C)

([(11

)])=

{(a bc d

)∈ GL2(C) | a+ b = c+ d

}.

Since a Mobius transformation that fixes three distinct points is the identity, ker θis the intersection of these three sets i.e.

ker(GL2(C)→M) =

{(a 00 a

)| a 6= 0

}is the group of non-zero scalar matrices.61

Thus PGL2(C) := GL2(C)/{λI | λ ∈ C 6= 0} ' M. It is not hard to see that asimilar argument shows that PSL2(C) := SL2(C)/{±I} ' M.

We can summarize this discussion with the following theorem.

Theorem. The action of GL2(C) on P1(C) induces an isomorphism from PGL2(C)to M. In particular M is a subgroup of S(C∞).

61We can see this another way: the kernel of θ is certainly contains in the intersection of these

three stabilisers so it would suffice to check that any scalar matrix is in the kernel ie [λI2v] = [v]for all non-zero λ in C. Indeed this is how we showed that a Mobius map that fixes 0, 1 and ∞ is

the identity in §1.5.

GROUPS 37

Lecture 18

5.3. Change of basis. Recall that if A is a linear map Fn → Fn corresponding tothe matrix A and e1, . . . , en is the standard basis for Fn then A(ei) =

∑nj=1Ajiej .

If f1, . . . , fn is another basis for Fn then there is an invertible linear map P suchthat P (ei) = fi for i = 1, . . . , n. i.e. P corresponds to the matrix P whose columnsf1, . . . , fn and fi =

∑nj=1 Pjiej for i = 1, . . . , n. It follows that for j = 1, . . . , n,

n∑k=1

P−1kj fk =

n∑k=1

P−1kj

n∑l=1

Plkel =

n∑l=1

(PP−1)ljel = ej .

Then

A(fi) = AP (ei)

=

n∑j=1

(AP )jiej

=

n∑j=1

(AP )ji(

n∑k=1

P−1kj fk)

=

n∑k=1

(P−1AP )kifk

Thus P−1AP represents A with respect to the basis f1, . . . , fn.

Proposition. GLn(F) acts on Mn(F) by conjugation.

Proof. Consider

· : GLn(F)×Mn(F)→Mn(F); (P,X) 7→ PXP−1.

Then for X ∈ Matn(F),

In ·X = InXI−1n = X

and

P (QXQ−1)P−1 = (PQ)X(PQ)−1

for all P,Q ∈ GLn(F). �

It is now straightforward to see that two distinct matrices in Mn(F) representthe same linear map with respect to different bases if and only if they are in thesame GLn(F)-orbit under this conjugation action.

Example (See Vectors and Matrices). If A : C2 → C2 is a linear map then preciselyone of the following three things is true:

(i) there is a basis for C2 such that A is represented by a matrix of the form(λ 00 µ

)with λ, µ ∈ C distinct — in this case {λ, µ} is determined by A62 but theymay appear in either order in the matrix;

62λ and µ are its eigenvalues and the basis vectors are the corresponding eigenvectors

38 SIMON WADSLEY

(ii) there is a basis for C2 such that A is represented by a matrix of the form(λ 00 λ

)with λ ∈ C — in this case λ is determined by A indeed A = λ idC2 and A isrepresented by this matrix with respect to every basis;

(iii) there is a basis for C2 such that A is represented by a matrix of the form(λ 10 λ

)again λ is determined by A.63

We may interpret this group-theoretically: every GL2(C)-orbit in M2(C) withrespect to the conjugation action is one of the following:

Oλ,µ := OrbGL2(C)

((λ 00 µ

))with λ, µ ∈ C distinct,

O(1)λ := OrbGL2(C)

((λ 00 λ

))with λ ∈ C and

O(2)λ := OrbGL2(C)(

((λ 10 λ

))with λ ∈ C.

These are all disjoint except that Oλ,µ = Oµ,λ. More explicitly,

Oλ,µ = {A ∈M2(C) | det(tI2 −A) = (t− λ)(t− µ) for all t ∈ C},

O(1)λ = {λI2} and

O(2)λ = {A ∈M2(C) | det(tI −A) = (t− λ)2 for all t ∈ C, A 6= λI2}.

We can also compute (λ 00 µ

)(a bc d

)=

(aλ bλcµ dµ

)and (

a bc d

)(λ 00 µ

)=

(aλ bµcλ dµ

)so that for λ 6= µ,

StabGL2(C)

((λ 00 µ

))=

{(a 00 d

)| ad 6= 0

}and StabGL2(C)(λI2) = GL2(C).

Similarly (λ 10 λ

)(a bc d

)=

(aλ+ c bλ+ dcλ dλ

)and (

a bc d

)(λ 10 λ

)=

(aλ a+ bλcλ c+ dλ

)so

StabGL2(C)

((λ 10 λ

))=

{(a b0 a

)| a 6= 0

}.

All other stabilisers are conjugate to these ones. We can easily read off the conju-gacy classes and centralisers in GL2(C) by restricting to the case λ, µ 6= 0.

63it is the unique eigenvalue of A and A 6= λ idC.

GROUPS 39

Exercise. Deduce that in M ' PGL2(C), ccl(z 7→ az) = ccl(z 7→ z 7→ bz) if andonly if b ∈ {a, 1/a} thus provide a description of all the conjugacy classes in Mand compute centralisers of suitable representatives of each class.

Lecture 19

5.4. The orthogonal and special orthogonal groups. Recall that any (square)matrix A has a transpose AT with ATij = Aji and detAT = detA. Moreover if A,B

are square matrices of the same size then (AB)T = BTAT .

Definition. The orthogonal group O(n) := {A ∈ Mn(R) | ATA = In = AAT } ⊂GLn(R) is the group of orthogonal n× n matrices.

Lemma. O(n) is a subgroup of GLn(R).

Proof. In ∈ O(n) and if A,B ∈ O(n) then BT = B−1 so

(AB−1)TAB−1 = (B−1)TATAB−1 = (B−1)T IBT = (BB−1)T = ITn = In

and similarly (AB−1)(AB−1)T = In. �

Recall that Rn comes with an inner product v · w =∑ni=1 viwi which defines a

length function on Rn via |v| = (v ·v)1/2. We also recall the definition of Kronecker’sdelta

δij :=

{1 if i = j;0 if i 6= j.

A basis f1, . . . , fn of Rn is said to be orthonormal if fi · fj = δij .64

Lemma.

(a) If {f1, . . . , fn} ⊂ Rn such that fi ·fj = δij for all 1 6 i, j 6 n, then {f1, . . . , fn}is an orthonormal basis for Rn.

(b) If v, w ∈ Rn then v · w = 14 (|v + w|2 − |v − w|2).

Proof. (a) Suppose∑ni=1 λifi = 0. Then for j = 1, . . . , n,

0 =

(n∑i=1

λifi

)· fj =

n∑i=1

λi(fi · fj) = λj .

Thus {f1, . . . , fn} is linearly independent. Since it has n elements it must spanRn.65

(b) We compute

(v + w) · (v + w) = v · v + v · w + w · v + w · wand

(v − w) · (v − w) = v · v − v · w − w · v − w · wsubtracting and using symmetry of the inner product we see that

|v + w|2 − |v − w|2 = 4v · was required. �

64There is a little apparent notational ambiguity here since we use subscripts to index the

basis vectors as well as to index the coordinates of a vector. Each fi is in Rn so can be written

as∑n

k=1(fi)kek and fi · fj =∑n

k=1(fi)k(fj)k.65In fact if v ∈ Rn, v =

∑ni=1(v · fi)fi.

40 SIMON WADSLEY

Proposition. Suppose that A ∈Mn(R). The following are equivalent:

(i) A ∈ O(n);(ii) Av ·Aw = v · w for all v, w ∈ Rn;

(iii) the columns of A form an orthonormal basis;(iv) |Av| = |v| for all v ∈ Rn.

Proof. Suppose that A ∈ On and v, w ∈ Rn. Then

Av ·Aw =

n∑i=1

(Av)i(Aw)i

=

n∑i=1

n∑j=1

Aijvj

( n∑k=1

Aikwk

)

=

n∑j,k=1

vj(ATA)jkwk

=∑j

vjwj = v · w.

Thus (i) =⇒ (ii).Suppose now that Av ·Aw = v ·w for all v, w ∈ Rn. Then in particular Aei ·Aej =

ei · ej = δij for each 1 6 i, j 6 n and Ae1, . . . , Aen66 is an orthonormal basis for Rn

by part (a) of the last lemma. Thus (ii) =⇒ (iii).Next, if Ae1, . . . , Aen form an orthonormal basis then for 1 6 i, j 6 n

δij = Aei ·Aej

=

n∑k=1

AkiAkj = (ATA)ij

and ATA = In. Thus (iii) =⇒ (i).(ii) =⇒ (iv) is immediate from the definitions. To see that (iv) =⇒ (ii) we use

(b) of the lemma:

Av ·Aw =1

4(|A(v + w)|2 − |A(v − w)|2

and

v · w =1

4(|v + w|2 − |v − w|2).

If (iv) holds then RHSs of these equations to be equal for all v, w and thus theLHSs are equal for all v, w and (ii) also holds. �

Thus O(n) is isomorphic to the subgroup of S(Fn) consisting of linear maps thatpreserve the scalar product or equivalently to the subgroup of S(Fn) consisting oflinear maps that preserve length.

The conjugation action GLn(R) on Mn(R) restricts to an action of O(n) onMn(R). The equivalence of (i) and (iii) in the propostion shows that two distinctmatrices in Mn(R) are in the same O(n)-orbit if and only if they represent the samelinear map with respect to two different orthonormal bases (see the last lecture).

Proposition. det : O(n)→ (R\{0}, ·) has image {±1}.66i.e. the set of columns of A

GROUPS 41

Proof. If A ∈ O(n) then detA = detAT so 1 = det In = detAAT = (detA)2 anddetA = ±1. Since

−1 00 1 0

. . .

0 1

∈ O(n)

has determinant −1 both 1 and −1 are in the image. �

Definition. The special orthogonal group

SO(n) := O(n) ∩ SLn(R) = ker(det : O(n)→ {±1}).SO(n) is isomorphic to the subgroup of S(Rn) consisting of linear maps that

preserve the scalar product and orientation.67 It is a normal subgroup of O(n) andO(n)/SO(n) ' C2.

There are complex versions of the orthogonal group and the special orthogonalgroup called the unitary group and the special unitary group. We won’t have timeto discuss them but they do appear on Example Sheet 4.

Lecture 20

5.5. Reflections.

Definition. Suppose that n ∈ Rm has length 1 then the reflection in the planenormal to n is the function Rn : Rm → Rm given by

Rn(x) = x− 2(x · n)n.

Note that if y · n = 0 then Rn(y) = y, and Rn(n) = n− 2n = −n.

Lemma. Suppose n ∈ Rm has length 1 then

(a) Rn ∈ O(m);(b) Rm has a basis with respect to which Rn is represented by a diagonal matrix D

such that D11 = −1, Dii = 1 for 2 6 i 6 m;(c) (Rn)2 = idRm and;(d) detRn = −1.

Proof. (a) First we show that Rn is linear: if x, y ∈ Rm and λ, µ ∈ R then

Rn(λx+ µy) = (λx+ µy)− 2((λx+ µy) · n)n

= λ(x− (2x · n)n) + µ(y − (2y · n)n)

= λRn(x) + µRn(y).

Now we show that Rn preserves the inner product: if x, y ∈ Rm then

Rn(x) ·Rn(y) = (x− 2(x · n)n) · (y − 2(y · n)n)

= x · y − 2(x · n)(n · y)− 2(y · n)(x · n) + 4(x · n)(y · n)(n · n)

= x · y.(b) Notice that U := {y ∈ Rm | y · n = 0} = ker (Rm → R; y 7→ y · n) is a

subspace of Rm of dimension m− 1 by the rank-nullity theorem. Moreover n 6∈ U .

67I have not defined an orientation of Rn. One way would be as an SO(n)-orbit of orthonormal

bases for Rn which would make this completely tautological. There are more sophisticated waysthat make it less so. With this definition the next sentence gives that there are exactly two

orientations of Rn.

42 SIMON WADSLEY

So we may find a basis f1, . . . , fm for Rm such that f1 = n and f2, . . . , fm ∈ U .Then Rn(f1) = −f1 and Rn(fi) = fi for 2 6 i 6 m as required.

(c), (d) If P is the change of basis matrix so that P−1RnP = D then P−1(Rn)2P =(P−1RnP )2 = Im. Thus (Rn)2 = PImP

−1 = Im. Moreover

−1 = detD = (detP )−1 detRn detP = detRn

as required. �

Proposition. If x, y ∈ Rm with x 6= y but x ·x = y · y then there is n ∈ Rm of unitlength such that Rn(x) = y. Moreover n may be chosen to be parallel to x− y.

Proof. Let n = (x−y)((x−y)·(x−y))1/2 so that n · n = 1.68 Then Rn(x) = x − 2(x · n)n.

But

2x · (x− y) = 2x · x− 2x · y = (x− y) · (x− y)

since x · x = y · y. Thus 2(x · n)n = x− y as required. �

Theorem. Every element of O(3) is a product of at most three reflections of theform Rn with n ∈ R3 of length 1. 69

Proof. Let A ∈ O(3). For this proof only we’ll write R0 to denote idR3 .Consider A(e1). If A(e1) = e1 then let n1 = 0. Otherwise use the proposition to

find n1 of unit length such that Rn1A(e1) = e1. In either case, Rn1

A(e1) = e1.Next consider Rn1

A(e2). If Rn1A(e2) = e2 let n2 = 0. Otherwise

Rn1A(e2) · e1 = e2 · e1 = 0

since Rn1A is orthogonal and fixes e1. Thus we may use the proposition to findn2 of unit length parallel to Rn1A(e2)− e2 such that Rn2Rn1A(e2) = e2. Morevern2 · e1 = 0 since n2 is parallel to Rn1

A(e2) − e2 and Rn1A(e2) and e2 are both

orthogonal to e1. Thus Rn2Rn1

A fixes e1 and e2 in every case.Now Rn2

Rn1A(e3) has length 1 and is orthogonal to both e1 and e2 since e3

is orthogonal to both e1 and e2, and Rn2Rn1A is orthogonal and fixes e1 and e2.Thus Rn2Rn1A(e3) = ±e3. If Rn2Rn1A(e3) = e3 let n3 = 0, otherwise let n3 = e3.In either case we can compute that Rn3

Rn2Rn1

A fixes e1, e2 and e3. Thus as it isa linear map it must be the identity. We conclude A = Rn1

Rn2Rn3

since each Rnihas order 1 or 2 according as ni = 0 or ni 6= 0. In particular A is a product of atmost three reflections. �

Proposition. If A ∈ O(2) then either

(i) A = SO(2) and there is some 0 6 θ < 2π such that

A =

(cos θ sin θ− sin θ cos θ

)70 or

(ii) A 6∈ SO(2) and A = Rn for some n ∈ R2 of unit length.

68x 6= y means (x− y) · (x− y) > 0.69There is nothing special about three here. In general every element of O(m) is a product of

at most m reflections of the form Rn. The proof is exactly similar to this one.70i.e. A is a rotation

GROUPS 43

Proof. In either case we know

A =

(a bc d

)for some a, b, c, d ∈ R such that(

a bc d

)(a cb d

)= I2.

Thus a2 + b2 = c2 + d2 = 1, ac+ bd = 0 and ad− bc = ±1.Let a = cos θ, b = sin θ, c = sinφ and d = cosφ with 0 6 θ, φ < 2π. Then

0 = cos θ sinφ+ cosφ sin θ = sin(θ + φ)

and

ad− bc = cos θ cosφ− sin θ sinφ = cos(θ + φ).

In case (i) ad− bc = 1, θ + φ is a multiple of 2π and cos(φ) = cos(−θ) = cos(θ)and sin(φ) = sin(−θ) = − sin θ as required.

In case (ii) ad−bc = −1, and θ+φ is π or 3π and cos(φ) = cos(π−θ) = − cos(θ).Then (

cos θ sin θsin θ − cos θ

)(sin θ/2− cos θ/2

)=

(sin−θ/2cos θ/2

)= −

(sin θ/2− cos θ/2

)and (

cos θ sin θsin θ − cos θ

)(cos θ/2sin θ/2

)=

(cos θ/2sin θ/2

).

Since (sin θ/2− cos θ/2

)·(

cos θ/2sin θ/2

)= 0

we see that A = Rn for n =

(sin θ/2− cos θ/2

).71 �

Theorem. If A ∈ SO(3) then there is some non-zero v ∈ R3 such that Av = v.72

Proof. By the last theorem A is a product of at most three reflections Rn. SincedetA = 1 and each Rn has determinant −1, A must be a product of either 0 or 2reflections. If 0 then A = idR3 and any v will work. If A = Rn1Rn2 with n1, n2 ofunit length. Then consider the linear map

R3 → R2;x 7→(x · n1x · n2

).

By the rank-nullity theorem it has non-trivial kernel. i.e. there is v ∈ R3 such thatv · n1 = v · n2 = 0. Then Rn1

Rn2(v) = Rn1

(v) = v. �

71To handle case (ii) we could instead appeal to the result that any element of O(2) is a product

of at most 2 reflections and the product of 0 or 2 reflections is in SO(2).72That is every rotation in R3 has an axis.

44 SIMON WADSLEY

Lecture 21

Corollary. Every A in SO(3) is conjugate in SO(3) to a matrix of the form1 0 00 cos θ sin θ0 − sin θ cos θ

.

Proof. Suppose A ∈ SO(3) and v1 ∈ R3 non-zero such that Av1 = v1. By replacingv1 by v1/|v1| we may assume |v1| = 1. Let U := ker(R3 → R;x 7→ x · v1). Bythe rank-nullity theorem dimU = 2. We can choose an orthonormal basis v2, v3for U . Then v1, v2, v3 is an orthonormal basis for R3 and so the matrix P withcolumns v1, v2 and v3 is in O(3). If detP = −1 then we may swap v2 and v3 sothat P ∈ SO(3). We claim that B := P−1AP is of the required form.

Certainly B ∈ SO(3) and B(e1) = P−1Av1 = P−1(v1) = e1. Moreover

e1 ·Be2 = Be1 ·Be2 = e1 · e2 = 0

and similarly e1 ·Be3 = 0. So

B =

1 0 00 C11 C12

0 C21 C22

for some C ∈M2(R).

Then BTB = I3 gives CTC = I2.73 Since detB = 1, detC = 1 and C ∈ SO(2).The result follows from the last proposition about O(2). �

6. Permutations

Recall that a permutation of a set X is an element of the group S(X); that isan invertible function X → X. In this chapter we will study permutations of finitesets. More particularly we will study permutations of [n] := {1, 2, . . . , n}. Sincethere is a 1-1 correspondence (i.e. invertible function) between any finite set and[n] for some value of n this amounts to the same thing.

6.1. Permutations as products of disjoint cycles.

Definition. We say that a permutation σ : [n]→ [n] is a cycle if the natural actionof the cyclic subgroup of Sn generated by σ has precisely one orbit of size greaterthan one.

Example. If σ : [5] → [5] such that σ(1) = 3, σ(2) = 2, σ(3) = 5, σ(4) = 4 andσ(5) = 1 then σ ∈ S5. We can draw σ as follows:

1 442 qq 3 444-- 5yy

.

We can compute σk(2) = 2 and σk(4) = 4 for all k ∈ Z. We can also computeσ2(1) = σ(3) = 5, σ2(3) = σ(5) = 1 and σ2(5) = σ(1) = 3. Finally σ3(1) = σ(5) =1, σ3(3) = σ(1) = 3 and σ3(5) = σ(3) = 5. So σ3 = id, the group generated by σis {id, σ, σ2} and the orbits are {1, 3, 5}, {2} and {4}. Thus σ is a cycle.

73Alternatively we could’ve shown that B acts on the span of e2 and e3 by a length preservingtransformation C and so C ∈ O(2).

GROUPS 45

Suppose that σ is a cycle of order k. Then σ generates the subgroup

〈σ〉 := {id, σ, . . . , σk−1}.

For any b ∈ [n] in an orbit of size 1

σi(b) = b for all i ∈ Z

and if a ∈ [n] is in the orbit of size greater than one then for c = σj(a) ∈ Orb〈σ〉(a),

σi(c) = σi+j(a) = σj(σi(a)).

Thus σi(c) = c whenever σi(a) = a. i.e. σi ∈ Stab〈σ〉(a) only if σi = id. ThusStab〈σ〉(a) = {e} and |Orb〈σ〉(a)| = k.

Notation. If σ is a cycle of order k such that the orbit of size greater than onecontains the element a ∈ [n] then we write

σ = (aσ(a)σ2(a) · · ·σk−1(a)).

The discussion above shows that the elements a, σ(a), . . . , σk−1(a) are all distinctand exhaust the orbit of a under 〈σ〉. Thus (aσ(a) · · ·σk−1(a)) uniquely determinesσ since σ(b) = b for b 6∈ {a, σ(a), . . . , σk−1(a)} and σ(σi(a)) = σi+k1(a).

Example. If σ : [5]→ [5] is as in the example above then σ = (135) = (351) = (513).

Definition. We say that (a1, . . . , ak) and (b1, . . . , bl) are disjoint cycles if

{a1, . . . , ak} ∩ {b1, . . . , bl} = ∅.

Lemma.(a) For a1, . . . , am ∈ [n] distinct

(a1a2 · · · am) = (a2a3 · · · ama1) = (a3a4 · · · ama1a2) = · · ·

i.e. cycles can be cycled.(b) If σ and τ are disjoint cycles then στ = τσ i.e. disjoint cycles commute.

Proof. (a) comes from choosing different elements of the non-trivial orbit to startthe cycle.

(b) Suppose that k ∈ [n]. Since σ and τ are disjoint either σ fixes k or τ fixes k(or possibly both).

Without loss of generality we may assume σ(k) = k. Then τσ(k) = τ(k). Thusto show τσ(k) = στ(k) it suffices to show that σ fixes τ(k).

If τ(k) = k then στ(k) = σ(k) = k = τ(k).If τ(k) 6= k then Orb〈τ〉(τ(k)) = Orb〈τ〉(k) is the non-trivial 〈τ〉-orbit so by

disjointness of σ and τ any element of it, in particular τ(k), is fixed by σ as required.�

Theorem (Disjoint cycle decomposition). Every π ∈ Sn may be written as a (possi-bly empty) product of disjoint cycles. Moreover the representation of π as a productof disjoint cycles is unique up to reordering.

Example. Consider (135) and (145) in S5. How is (135)(145) expressed as a productof disjoint cycles? We can chase elements one at a time. (145) sends 1 to 4 and(135) fixes 4. (145) sends 4 to 5 and (135) sends 5 to 1. Thus (14) is one of thecycles in the disjont cycle decompositon of (135)(145). 2 is fixed by both (145) and(135) so we can ignore it. (145) fixes 3 and (135) sends 3 to 5. (145) sends 5 to 1

46 SIMON WADSLEY

and (135) sends 1 to 3. So (35) is another cycle in the decomposition. It followsthat (135)(145) = (14)(35). Pictorally

1

))

2

��

3

��

4

��

5

ss

1

$$

2

��

3

��

4

zz

5

��

1

''

2

��

3

''

4

��

5

ss

=

1 2 3 4 5 1 2 3 4 5

whereas

1

''

2

��

3

''

4

��

5

ss

1

��

2

��

3

��

4

��

5

��

1

))

2

��

3

��

4

��

5

ss

=

1 2 3 4 5 1 2 3 4 5

i.e. (145)(135) = (13)(45).

Lecture 22

Proof of disjoint cycle decomposition theorem. Suppose that π ∈ Sn. The 〈π〉-orbits partition [n],

[n] =

m⋃i=1

Oi

say. By re-ordering the orbits we may assume that ki := |Oi| > 1 for i = 1, . . . land |Oi| = 1 for i = l + 1, . . .m.74.

For 1 6 i 6 l pick ai ∈ Oi and define

σi := (aiπ(ai) · · ·πki−1(ai)).

Since the Oi are disjoint, the σi are disjoint cycles. We claim that π =∏li=1 σi.

75

To prove the claim suppose that a ∈ [n]. Since the Oi partition [n], a ∈ Oia forprecisely one ia ∈ [m].

If ia > l then π(a) = a and σi(a) = a for all i ∈ [l]. Thus

π(a) = a =

(l∏i=1

σi

)(a).

If ia 6 l then σi(a) = a for i ∈ [l]\{ia} and σia(a) = π(a) by definition. Thus(l∏i=1

σi

)(a) = σia

l∏i=1

i 6=ia

σi(a) = σia(a) = π(a)

as required.

74we allow l = 0 or l = m in which cases one of these lists is empty75Since the σi are disjoint the order in the product does not matter by part (b) of the last

lemma.

GROUPS 47

Uniqueness follows easily from the construction above — a cycle σ will appearin the product if and only if its non-trivial orbit O is a non-trivial orbit of 〈π〉 andσ(a) = π(a) for all a ∈ O. �

Lemma. If π is a product of disjoint cycles of order n1, n2, . . . , nk then

o(π) = lcm(n1, . . . , nk).

Proof. Let π =∏ki=1 σi with σi pairwise disjoint and o(σi) = ni. Then πn =∏k

i=1 σni since the σi commute. Moreover πn = id if and only if σni = id for each

i.76 Thus πn = id if and only if ni divides n for each i = 1, . . . k i.e. if and only ifn is a multiple of the lcm of the ni. �

6.2. Permuations as products of transpositions.

Definition. We call a cycle of order 2 a transposition.

Lemma. Every π ∈ Sn is a product of transpositions.

Proof. By the decomposition into disjoint cycles theorem it suffices to show thatevery cycle is a product of transpositions. One may easily check that

(a1a2 · · · ak) = (a1a2)(a2a3) · · · (ak−1ak).

�

Remark. The representation of π as a product of transpositions is not unique. Forexample

(12)(23)(34) = (1234) = (14)(13)(12).

Despite the remark it is true that π ∈ Sn cannot be written both as a prod-uct of an even number of transpositions and as a product of an odd number oftranspositions.

Theorem. Given π ∈ Sn let l(π) be the number of orbits of 〈π〉 in [n]. For anyπ ∈ Sn and any transposition (ab) ∈ Sn,

l(π(ab)) = l(π)± 1.

Proof. Suppose first that a and b are in the same orbit in π. i.e. when we write πas a product of disjoint cycles π =

∏mi=1 σi one of the σj = (ax2 · · ·xkbxk+2 · · ·xm).

Then we can compute

σj(ab) = (axk+2 · · ·xm)(bx2 · · ·xk).

Since a and b don’t appear in any of the other cycles we see that l(π(ab)) = l(π)+1in this case.77

Suppose instead that a and b lie in different orbits in π. Then some σi =(ax2 · · ·xk) and some σj = (by2 · · · yl).78 Then σiσj(ab) = (ay2 · · · ylbx2 · · ·xk).Since a and b don’t appear in any of the other cycles we see that l(π(ab)) = l(π)−1in this case. 79

Thus in every case l(π(ab)) = l(π)± 1. �

76This can be seen by considering what πn does to the non-trivial orbit of each σi.77The orbits are the same except that the orbit containing a and b splits into two.78If a or b is fixed by π this isn’t quite true but one can check that the argument can be

adapted is a straightforward manner by allowing ’cycles of length 1’ in the decomposition.79This time the orbits are the same except that the orbits containing a and b are joined

together.

48 SIMON WADSLEY

Corollary. There is a well-defined group homomorphism

ε : Sn → ({±1}, ·)

such that ε(π) = 1 if π is a product of an even number of transpositions andε(π) = −1 if π is a product of an odd number of transpositions. Moreover, forn > 2, Im ε = {±1}.

Proof. Suppose that π =∏mi=1 τi with each τi a transposition. By induction on m

we see that l(π) ≡ l(id)+m mod 2. Thus if also π =∏m′

i=1 σi with each σi a trans-position then m ≡ m′ mod 2 and the function ε as defined in the statement makessense.80 It is now easy to verify that ε is a homomorphism and, since ε((12)) = −1,that Im ε = {±1} for n > 2. �

Definition. Given π ∈ Sn we say that π is even if ε(π) = 1 and that π is odd ifε(π) = −1.

Remark. Notice that a cycle of odd order is even and a cycle of even order is odd.81

Definition. The alternating group on [n], An := ker(ε : Sn → {±1}) is the normalsubgroup of Sn consisting of all even permutations.

Since |Sn| = n! it follows easily from the isomorphism theorem that, for n > 2,|An| = n!

2 .

Lecture 23

6.3. Conjugacy in Sn and in An. We now try to understand the conjugacyclasses in Sn and in An. In Sn they have a remarkably simple description.

Lemma. If σ ∈ Sn and (a1 · · · am) is a cycle then

σ(a1 · · · am)σ−1 = (σ(a1) · · ·σ(am)).

Proof. If a ∈ [n] then consider

σ(a1 · · · am)σ−1(σ(a)) =

{σ(ai+m1) if a = aiσ(a) if a 6∈ {a1, . . . , am}.

Since every b ∈ [n] is uniquely σ(a) for some a ∈ [n] we deduce that

σ(a1 · · · am)σ−1(b) = (σ(a1) · · ·σ(am))(b)

for all b ∈ [n] as claimed. �

Theorem (Conjugacy classes in Sn). Two elements of Sn are conjugate if and onlyif they are the product of the same number of disjoint cycles of each length.82

80This is really the content of the corollary. Without the Theorem, or something like it, this

is not at all clear.81This is just another of those frustrating facts of life.82We sometimes say that they have the same cycle type.

GROUPS 49

Proof. The lemma tells us that if π =∏ki=1(ai1ai2 · · · aim(i)) is a decomposition of

π into a product of disjoint cycles then

σπσ−1 =

m∏i=1

(σ(ai1ai2 · · · aim(i))σ−1)

=

m∏i=1

(σ(ai1)σ(ai2) · · ·σ(aim(i)).

Since the right-hand-side is a product of disjoint cycles we see that anything con-jugate to π must be a product of the same number of disjoint cycles of the samelength.

Suppose now that τ =∏ki=1(bi1bi2 · · · bim(i)) is a decomposition of τ into a

product of disjoint cycles (with the same number of each length as for π). Then let

{a01, . . . , a0m(0)} = [n]\{aij | 1 6 i 6 k, 1 6 j 6 m(i)}83

and

{b01, . . . b0m(0)} = [n]\{bij | 1 6 i 6 k, 1 6 j 6 m(i)}.84

Then we define σ(aij) = bij for 0 6 i 6 k and 1 6 j 6 m(i). Thus σ ∈ Sn andτ = σπσ−1 by the lemma. �

Example. Conjugacy classes in S4

representative element e (12) (12)(34) (123) (1234)cycle type 14 2.12 22 3.1 4number of elements in class 1 6 3 8 6size of centraliser 24 4 8 3 4sign 1 −1 1 1 −1

Corollary (Conjugacy classes in An). If π ∈ An then its conjugacy class in An isequal to its conjugacy class in Sn if and only if CSn(π) contains an odd element.Moreover if CSn(π) ⊂ An then the conjugacy class of π in Sn is a union of twoconjugacy classes in An of equal size.

Proof. Under the conjugation action ofAn and Sn respectively OrbAn(π) ⊂ OrbSn(π).It follows that it suffices to show that

|OrbAn(π)| ={|OrbSn(π)| if CSn(π) 6⊂ An12 |OrbSn(π)| if CSn(π) ⊂ An.

Moreover by the orbit-stabiliser theorem

|OrbAn(π)| = |An||CAn(π)|

=|Sn|

2|(CAn(π))|= |OrbSn(π)| |CSn(π)|

2|CSn(π) ∩An|.

Thus if CSn(π) ⊂ An then we see that |OrbAn(π)| = 12 |OrbSn(π)|.

Otherwise consider ε|CSn (π) : CSn(π) → {±1}. Since ε(CSn(π)) 6= {1} we seethat Im ε|CSn (π) = {±1} and

CSn(π)/CAn(π) = CSn(π)/ ker ε|CSn(π)' {±1}.

It follows by Lagrange that |CSn(π)| = 2|CAn(π)| and OrbAn(π) = OrbSn(π). �

83So∑n

i=0m(i) = n and every element of [n] is equal to precisely one aij .84So every element of [n] is also equal to precisely one bij .

50 SIMON WADSLEY

Example (Conjugacy classes in A4).The even cycle types in S4 are 14, 22 and 3.1. Now (12) ∈ CS4(e) and (12) ∈

CS4((12)(34)) so the centralisers of elements of conjugacy classes of cycle type 14

and 22 contain elements of odd order. Thus these classes are the same in A4 andS4.

Since CS4((123)) has order 3 it must be generated by (123) and so it is equal to

CA4((123)). Thus the conjugacy class with cycle type 3.1 splits into two parts ofequal size.

representative element e (12)(34) (123) (132)cycle type 14 22 3.1 3.1number of elements in class 1 3 4 4size of centraliser 12 4 3 3

Corollary. A4 has no subgroup of index 2.

Proof. If H 6 A4 were index 2 then it would be normal (Example Sheet 3 Q1) andso be a union of conjugacy classes (Example Sheet 3 Q3). But |A4|/2 = 12/2 = 6and H must contain id. No set of conjugacy classes including {e} has a total of 6elements. �

Lecture 24

6.4. Simple groups.

Definition. We say a non-trivial group G is simple if G has no normal subgroupsexcept itself and its trivial subgroup.

Since if N is a normal subgroup of G one can view G as ‘built out of’ N andG/N , one way to try to understand all groups is to first understand all simplegroups and then how they can fit together.

Example. If p is prime then Cp is simple since Cp has no non-trivial proper sub-groups. These are the only abelian simple groups.

Theorem. A5 is simple.

Proof. The cycle types of even elements of S5 are 15, 22.1, 3.12 and 5. Thesehave 1, 5 × 3 = 15,

(52

)× 2 = 20 and 4! = 24 elements respectively — note that

1 + 15 + 20 + 24 = 60 = |A5| so we have got them all.The element (12) is in the centralisers of id, (12)(34) and (345) so the classes

with cycle type 15, 22.1, and 3.12 have centralisers containing odd elements and sothese classes do not split as conjugacy classes in A5.

Since there are 4! 5-cycles in S5, |CS5((12345))| = 5!/4! = 5. But all powers of

(12345) commute with (12345) so CS5((12345)) = 〈(12345)〉 ⊂ A5. Thus the classwith cycle type 5 in S5 splits into two classes of order 12 in A5.

In summary,

representative element e (12)(34) (123) (12345) (12345)2

cycle type 15 22.1 3.12 5 5number of elements in class 1 15 20 12 12size of centraliser 60 4 3 5 5

Now if N is a non-trivial proper normal subgroup of A5 then it is a union ofconjugacy classes including the class 15. Moreover |N | is a factor of 60 by Lagrange.

GROUPS 51

If we take the identity and one of the smallest non-trivial classes (of size 12)we already have 13 elements so such an N must have order 15, 20 or 30. It isstraightforward to check that no union of conjugacy classes including {e} can havethese orders. �

The remainder of the course is non-examinable.In fact we can prove a stronger result.

Theorem. An is simple for all n > 5.

Remark. A4 is not simple since it has a normal subgroup of order 4 namely V :={id, (12)(34), (13)(24), (14)(23)}. A3 ' C3 is simple, A2 is trivial so not simple.

Proof. First we show that all 3-cycles are conjugate in An (for n > 5). Usingour theorem about conjugacy classes in An this follows from the fact that (45) ∈CAn(123) is odd.

Next we show that An is generated by 3-cycles. We know that every element ofAn is a product of an even number of transpositions. So we must show that everyproduct of two transpositions is also a product of 3-cycles. Suppose a, b, c and dare distinct. Then (ab)(ab) = id, (ab)(bc) = (abc) and (ab)(cd) = (abc)(bcd). ThusAn is indeed generated by 3-cycles.

Now it suffices to show that every non-trivial normal subgroup N of An containsa 3-cycle — if N contains one it must contain all and so it generates An and so itis An. To this end we pick π ∈ N\{e} with the most fixed points. We show that πis a 3-cycle.

If π has two transpositions in its decomposition as a product of disjoint cycles sayπ swaps a and b and also swaps c and d. Let σ = (cde) for some e 6∈ {a, b, c, d}. Thenτ := π−1(σπσ−1) ∈ N since N is a normal subgroup of An. Moreover τ(a) = a,τ(b) = b, τ(e) = c 6= e, and if π(i) = i for i 6= e then τ(i) = i. Thus τ 6= id and τhas more fixed points than π which contradicts our definition of π.

Now we suppose that 〈π〉 has an orbit of size at least 3 and is not a 3-cycle.Suppose a, π(a), π2(a), b, c are all distinct and not fixed by π. 85 Let σ = (π2(a)bc)and τ = (σπσ−1)π−1 ∈ N since N is a normal subgroup of An this time we computeτ(π(a)) = π(a), τ(π2(a)) = b 6= π2(a), and if i is fixed by π then i is also fixed byτ . Thus τ 6= id has more fixed points than π. �

A triumph of late 20th century mathematics was the classification of all finitesimple groups. Roughly speaking this says that every finite simple group is either

• cyclic of prime order;• an alternating group;• a matrix group over a field of finite order (for example PSLn(Z/pZ));• one of 26 so-called sporadic simple groups the largest of which is known as

‘the monster’ and has approximately 8× 1053 elements.

One first important step in the proof was a result by Feit and Thompson that thereis no non-abelian simple group of odd order that first appeared as a circa 250 pagepaper in 1963. The first proof of the whole classification theorem was annouced inthe early 1980s. It ran to over ten of thousand pages spread across a large numberof journal articles by over 100 authors. It turned out not to be quite complete. In2004 it appeared to experts to be complete.

85if there were precisely 4 points that π does not fix then π would be a 4-cycle which wouldbe odd or a product of disjoint transpositions which we have already ruled out.

52 SIMON WADSLEY

In this course we have seen a little of how symmetry can be understood usingthe language of groups. But even when considering only finite groups there is muchmore to learn.