GROUPS Contents 1. Examples of Groups 2 1.1. a Motivating Example

GROUPS

SIMON WADSLEY

Contents 1. Examples of groups2 1.1. A motivating example2 1.2. Some initial definitions4 1.3. Further geometric examples7 1.4. Subgroups and homomorphisms9 1.5. The MöbiusGroup 13 2. Lagrange’s Theorem 17 2.1. Cosets 17 2.2. Lagrange’s Theorem 18 2.3. Groups of order at most 8 19 2.4. The Quaternions 22 2.5. Fermat–Euler theorem 22 3. Group Actions 23 3.1. Definitions and examples 23 3.2. Orbits and Stabilisers 24 3.3. Conjugacy classes 27 3.4. Cayley’s Theorem 28 3.5. Cauchy’s Theorem 29 4. Quotient Groups 29 4.1. Normal subgroups 29 4.2. The isomorphism theorem 32 5. Matrix groups 33 5.1. The general and special linear groups 34 5.2. Möbiusmaps as projective linear transformations 35 5.3. Change of basis 37 5.4. The orthogonal and special orthogonal groups 39 5.5. Reflections 41 6. Permutations 44 6.1. Permutations as products of disjoint cycles 44 6.2. Permuations as products of transpositions 47 6.3. Conjugacy in Sn and in An 48 6.4. Simple groups 50

1 2 SIMON WADSLEY

Purpose of notes. Please note that these are not notes of the lectures but notes made by the lecturer in preparation for the lectures. This means they may not exactly 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 algebraic tool designed to abstract the notion of symmetry. Symmetry arises all over mathematics; which is to say that groups arise all over mathematics. Roughly speaking a symmetry is a transformation of an object that preserves certain properties of the object. As I understand it, the purpose of this course is two-fold. First to introduce groups so that those who follow the course will be familiar with them and be better equipped to study symmetry in any mathematical context that they encounter. Second as an introduction to abstraction in mathematics, and to proving things about abstract mathematical objects. It is perfectly possible to study groups in a purely abstract manner without geometric motivation. But this seems to both miss the point of why groups are interesting 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 do we mean by that?

1.1. A motivating example. Question. What are the distance preserving functions from the integers to the integers? 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; they describe all the ways of ‘rearranging’ the integers that preserve the distance between any 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 integers then

(ta ◦ tb)(n) = ta(b + n) = a + b + n = ta+b(n) 1 for every integer n, that is that ta+b = ta ◦ tb . Moreover t0 is the identity or ‘do nothing’ function id: Z → Z that maps every integer n to itself. Thus for every a ∈ 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 if they 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 the function 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. What does this tell us about g? For example, what does it tell us about g(1)? Since g is a 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 g is 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 of the form ta ◦ s (with a ∈ Z in either case). That is all symmtries of Z are of the form 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 argument above 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 itself a symmetry of the integers. In fact, we’ve only used this for some special cases 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 symmetry f there is a symmetry g such that g ◦ f = id = f ◦ g (in fact we’ve only used this for f = ta and f = s and only that there is a g such that g ◦ 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 for symmetries 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 ordered pair of elements of S and uses them to produce an element of S. If x ◦ y = y ◦ x then we say that x and y commute (with respect to ◦). We say ◦ is commutative if 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 on Z 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 ∈ S i.e. it doesn’t matter which of the two ◦ we use first. It will be instructive to convince yourself that if ◦ is an associative binary operation on S and w, x, y, z ∈ S then w ◦ (x ◦ y ◦ z) = (w ◦ x) ◦ (y ◦ z) = (w ◦ x ◦ y) ◦ z. Having done this you should also convince yourself that there is nothing special about four and the obvious generalisation holds for any (finite) number of elements of S whenever ◦ is associative. This means that whenever ◦ is an associative binary operation we may (and will!) omit brackets, writing for example w◦x◦y◦z without ambiguity. 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 operations on Z or N. Subtraction is not associative on Z since ((0−1)−2) = −3 but 0 − (1 − 2) = 1 6= −3. 2 (2) Exponentiation (a, b) 7→ ba is not associative on N since 23 = 29 but (23)2 = 26 6= 29. (3) Composition is always an associative operation on the set of functions from X 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 element e ∈ 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 is an identity for multiplication on both these sets. Subtraction on Z does not have an identity since if e − x = x for all x ∈ Z then e = 2x for all x ∈ Z and this is absurd. Note however that x − 0 = x for all x ∈ Z. We sometimes 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 obvious sense. (3) If X is any set then the identity function id: X → X; s 7→ s is an identity for 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 e0 are both identities on S. Then e ◦ e0 = e0 = e0 ◦ e since e is an identity. But also e0 ◦ e = e = e ◦ e0 since e0 is an identity. Thus e = e0 as required.

Lecture 2 Definition. If a binary operation ◦ on a set S has an identity e then we say that it 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 Q with an identity (1) but it still does not have inverses. Although for every non-zero rational q, 1/q is also rational and q · 1/q = 1 = 1/q · q, 0 is also rational and there is no rational r such that r·0 = 1. However multiplication does have inverses on the set Q\{0}. (3) In general composition on the set of functions X → X does not have inverses. For example the function f : Z → Z; n 7→ 0 has no inverse since if g : Z → Z were an inverse then we’d have f(g(n)) = n for all n ∈ Z but in fact however g is defined f(g(1)) = 0. This idea can be adapted to show that 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

2 (4)( Z, −) is not a group since − is not associative. (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 functions have 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 nearly always call the identity of a group e (or eG if we want to be clear which group it is the 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 a set. 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 remainder 3 after dividing x + y by n . It is straightforward to see that Zn is an abelian group of 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 such that 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 must show that there is at most one. Suppose that h, k ∈ G such that gh = e = hg and gk = 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 −1 −1 and similarly (h g )(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 and gn := (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 is a natural number n such that gn = e. If g has finite order, we call the smallest natural 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. 3 Z12 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 set D2n of all symmetries of a regular polygon P with n vertices (for n > 3) living in the complex plane C. By symmetry of P we will mean a distance preserving transformation of the plane that maps P to itself. We might as well assume that the 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 defines a binary operation on D2n. Since composition of functions is always associative this operation on D2n is associative. Moreover the identity function id is obviously a symmetry of P . Thus to see that D2n is a group it remains to show that every element of D2n is invertible. We could do this directly but instead we’ll use a nailing to 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. Then r does preserve distances in C as |r(z) − r(w)| = |e2πi/n||z − w| = |z − w| m n and r(P ) = P i.e. r ∈ D2n. Moreover r ∈ D2n for all m ∈ N and r = id; i.e. r−1 = rn−1. Similarly let s: C → C; z 7→ z¯ denote complex conjugation; i.e. reflection in the real 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 will 5 preserve V ; that is to say if g ∈ D2n then g(V ) = V . 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 of C 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 transformation of C fixing 0, 1 and e2πi/n. Thus srn−kg = id and g = rks. So we have computed that k k D2n = {r , r s|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 denote reflection 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 geometric meaning 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 of R3 such that ρ(X) = X. These form a group. We will consider the cases X a tetrahedron 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 invertible functions from X to X forms a group under composition; that is the set of functions 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 is a binary operation on this set; that is that the composition of two invertible functions is invertible. Some people would say that we need to check that the binary operation 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 is invertible.

Proof. There are functions g1, g2 : X → X such that f1 ◦ g1 = id = g1 ◦ f1 and f2 ◦ 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 a group 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} we write 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 symmetries of an object we want to restrict ourselves to considering symmetries that preserve 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 of symmetries under composition. The groups D2n and Sym(X) for X a regular solid are deﬁned to consist of those symmetries that preserve a certain subset of the whole space. Similary, instead of considering D2n, the group of all symmetries of a regular n-gon we might want to restrict only to those symmetries that preserve orientation, 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 to 7 a binary operation on H and (H, ◦) is a group. We write H 6 G to denote that H 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 if the 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 operation is associative follows immediately from the associativity of its extension to G. Condition (ii) tells us that it has an identity8 and condition (iii) tells us H has inverses9. 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 is the identity in H then eH e = eH = eeH since e is the identity in G. But also eH eH = eH since eH is the identity in H. Thus eH e = eH eH . But eH has an −1 −1 inverse in G and we see that eH eH e = eH eH eH ie e = eH ∈ H i.e. condition (ii) holds. That condition (iii) holds follows from uniqueness of inverses in G since for h ∈ 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 noted in the proof it is immediate from the associativity of the operation on G.

7 precisely h1 ◦ h2 ∈ H for all h1, h2 ∈ H 8the same identity as G since if h ∈ H then h ∈ G so eh = h = he 9the 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. −1 Corollary. A subset H of G is a subgroup precisely if it is non-empty and h1 h2 ∈ H for all h1, h2 ∈ H. Proof. Suppose that conditions (i)-(iii) of the last lemma hold. Condition (ii) tells −1 us that H is non-empty and conditions (iii) and (i) combined give that h1 h2 ∈ H for 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 assumption −1 −1 i.e. condition (iii) holds. Finally, for h1, h2 ∈ H, h1h2 = (h1 ) h2 ∈ H. Example. The set H = {f ∈ Isom(Z)|f(0) = 0} is a subgroup of Isom(Z). We can see this using the corollary. Certainly id(0) = 0 so H 6= ∅. Moreover if h1, h2 ∈ H then −1 −1 −1 h1 h2(0) = h1 (0) = h1 h1(0) = id(0) = 0. Note that this argument isn’t much simpler than verifying conditions (i)-(iii) of the lemma in practice. We will also be interested in maps between groups. However we won’t typically be interested an arbitary functions between two groups but only those that respect the 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 homomorphism (or just homomorphism) precisely if θ(h1 ∗ h2) = θ(h1) ◦ θ(h2) for all h1, h2 ∈ H. Definition. A group homomorphism θ : H → G is an isomorphism if θ is invertible −1 −1 as a function; ie if there is a function θ : G → H such that θ ◦ θ = idG and −1 θ ◦ 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, +) to G arises in this way. + 10 (4) θ : Z → Isom (Z); n 7→ tn is an isomorphism. (5) The exponential function defines an isomorphism a exp: (R, +) → ({r ∈ R | r > 0}, ·); a 7→ e .

The inverse map is given by log = loge. If you are alert you will be asking why we don’t require homomorphisms θ : H → −1 −1 G to satisfy θ(eH ) = eG and θ(h ) = θ(h) for all h ∈ H. The following lemma shows 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.

10 recall tn denotes translation by n GROUPS 11

Proof. (i) Since θ is a homomorphism θ(eH ) = θ(eH eH ) = θ(eH )θ(eH ). Thus −1 −1 eG = θ(eH ) θ(eH ) = θ(eH ) θ(eH )θ(eH ) = θ(eH ) as required. −1 −1 (ii) Pick h ∈ H. Then θ(h)θ(h ) = θ(hh ) = θ(eH ) = eG (by (i)). Similarly −1 −1 −1 −1 θ(h )θ(h) = θ(h h) = θ(eH ) = eG. Thus θ(h ) = θ(h) as required. Definition. If θ : H → G is a group homomorphism then the kernel of θ is defined by 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 and Im 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. −1 −1 −1 −1 k1k2 ∈ ker θ. Similarly θ(k1 ) = θ(k1) = eG = eG, i.e. k1 ∈ ker θ . So ker θ 6 H. Suppose now that θ(h1), θ(h2) ∈ Im θ. Then θ(h1)θ(h2) = θ(h1h2) so θ(h1)θ(h2) ∈ −1 −1 −1 Im θ. Similarly θ(h1) = θ(h1 ), so θ(h1) ∈ Im θ. So Im θ 6 G.

Lecture 5 Theorem (Special case of the isomorphism theorem). A group homomorphism θ : H → G is an isomorphism if and only if ker θ = {eH } and Im θ = G. In this case, θ−1 : G → H is a group homomorphism (and so also an isomorphism).

−1 Proof. Suppose that θ has an inverse. Certainly eH ∈ ker θ and so θ (eG) = −1 −1 −1 θ θ(eH ) = eH . Thus if k ∈ ker θ then k = θ θ(k) = θ (eG) = eH and so −1 −1 ker θ = {eH }. Moreover for each g ∈ G, θ (g) ∈ H and g = θ(θ (g)). Thus G = Im θ. Conversely, suppose that ker θ = eH and Im θ = G. By the latter property for each g ∈ G we may choose an element hg ∈ H such that θ(hg) = g. Now if s: G → H is the function g 7→ hg then θs(g) = θ(hg) = g for all g ∈ G. It follows that for all h ∈ H, −1 −1 −1 −1 θ(s(θ(h))h ) = θ(s(θ(h)))θ(h ) = θ(h)θ(h ) = θ(hh ) = θ(eH ) = eG, −1 i.e. s(θ(h))h ∈ ker θ = {eH }. It follows that sθ(h) = h for all h ∈ H and so s is an inverse of θ. Finally, suppose that θ is an isomorphism and g1, g2 ∈ G. Then −1 θ(θ (g1g2)) = g1g2 and −1 −1 −1 −1 θ(θ (g1)θ (g2)) = θ(θ (g1))θ(θ (g2)) = g1g2 since θ is a homomorphism. Thus −1 −1 −1 −1 −1 −1 −1 −1 θ (g1g2) = θ (θ(θ (g1g2))) = θ (θ(θ (g1)θ(θ (g2)))) = θ (g1)θ (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 ∈ K then θ1(θ2(k1k2)) = θ1(θ2(k1)θ2(k2)) = θ1θ2(k1)θ1θ2(k2) ie θ1θ2 is a homomor- −1 −1 phism. Finally, if θ1 and θ2 are invertible then θ2 θ1 is an inverse of θ1θ2.

Deﬁnition. We say that a group G is cyclic if there is a homomorphism f : Z → G such 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 the form gi with i ∈ Z. More generally we say that a subset S of G generates G if every element of G is a product of elements of S and their inverses — that is if G the 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 1 and −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} 12 generates Zn if and only if it is coprime to n . The ‘if’ part is a conseqeunce from Euclid’s algorithm; the only if part is elementary.

Lemma. Suppose that G is a group containing an element g with gn = e. There is a unique group homomorphism f : Zn → G such that f(1) = g. In particular every group 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 for a = 0, 1, . . . n − 1, f(a +n 1) = f(a)f(1) = f(a)g. Thus we can see inductively that a f(a) = g for all a ∈ Zn. Thus if f exists then it is unique. a We now see how to construct f. We deﬁne f(a) = g for all a ∈ Zn and we must prove that this deﬁnes a homomorphism. That is we must show that ga+nb = a+b f(a+n b) and g = f(a)f(b) are equal for all a, b ∈ Zn. Since a+b−(a+n b) = kn for 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 a there is a homomorphism f : Zn → G such that f(a) = g for each a ∈ Zn. Suppose a−nb that f(a) = f(b) for a, b ∈ Zn. Then f(b −n a) = g = e thus a = b else g would have order strictly smaller than n. It follows that ker f = {e} and | Im f| has n elements 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 veriﬁed that 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 good 12Recall that non-negative integers a, b are coprime if and only if their only common factor is 1. GROUPS 13

i i Recall that we showed that D2n = {r , r s | i = 0, 1, . . . , n − 1} where r denotes a 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 such that gn = e, h2 = e and hgh−1 = g−1. There is a unique group homomorphism f : D2n → G such that f(r) = g and f(s) = h. Moreover if o(g) = n and |G| = 2n then f is an isomorphism.

Proof. This is similar to the last proof. Any homomorphsim f : D2n → G such that f(r) = g and f(s) = h must satisfy f(ri) = gi and f(ris) = gih. So we must show that this does deﬁne a homomorphism i.e. show that f(xy) = f(x)f(y) when i i x, y ∈ {r , r s} = D2n. We’ll do the most diﬃcult case and leave the rest as an exercise. 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 then gih = e so h = gi. Then g−1 = hgh−1 = gigg−i = g and so g2 = e contradicting o(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 the pigeonhole principle.

Lecture 6 1.5. The MöbiusGroup. Informally, a Möbiustransformation is a function f : C → C of the form az + b f : z 7→ cz + d with a, b, c, d ∈ C and ad − bc 6= 0. The reason for the condition ad − bc 6= 0 is that for such a function if z, w ∈ C then az + b aw + b (z − w) f(z) − f(w) = − = (ad − bc) cz + d cw + d (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 not divide by zero in the complex numbers. This makes composition of Möbiustrans- formations problematic since the image of one Möbiustransformation may not coincide with the domain of defintion of another. We will fix this by adjoining an 14 additional point to C called ∞.

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

13This is bad because we’re really interested in invertible functions 14And pronounced ‘inﬁnity’. 14 SIMON WADSLEY

Deﬁnition. Given (a, b, c, d) ∈ C4 such that ad − bc 6= 0 we can deﬁne a function f : C∞ → C∞ as follows: if c 6= 0 then  az+b if z ∈ \{−d/c};  cz+d C f(z) := ∞ if z = −d/c;  a/c if z = ∞; if c = 0 then az+b if z ∈ f(z) := cz+d C ∞ if z = ∞.

We call all functions from C∞ to C∞ that arise in this way Möbiustransformations and let M := {f : C∞ → C∞ | f is a Möbiustransformation}. We’ll see another way to interpret this definition later in the course involving projective geometry. But for now we’ll work with it as it stands and also take for granted 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(∞) = ∞. Then f = id. az+b Proof. If f = cz+d then f(∞) = ∞ forces c = 0 and f(0) = 0 forces b = 0 and then az+0 f(1) = 1 forces a/d = 1 and so f : z 7→ 0z+a = z for all z ∈ C as required.

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

Proof. First we consider the case w1 = 0, w2 = 1 and w3 = ∞. (z2−z3)(z−z1) If none of z1, z2, z3 are ∞ then f(z) = does the job. (z2−z1)(z−z3) 1 If some zi = ∞ then choose z4 ∈ ∞\{z1, z2, z3} and let s(z) = then C z−z4 (s(z1), s(z2), s(z3)) does not contain ∞ so by the above we may find g ∈ M such that 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, −1 g(z3) = ∞ and h(w1) = 0, h(w2) = 1 and h(w3) = ∞. Then f = h g(zi) = wi for i = 1, 2, 3 and we’ve shown existence in general. Moreover if k is another such map then hkg−1 fixes 0, 1 and ∞ so by the lemma is the identity. Thus k = h−1g = f and we’ve shown uniqueness.

Deﬁnition. Given distinct points z1, z2, z3, z4 ∈ C∞ the cross-ratio of z1, z2, z3, z4 written [z1, z2, z3, z4] := f(z4) where f is the unique M¨obius transformation such 16 that f(z1) = 0, f(z2) = 1 and f(z3) = ∞.

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

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) is the unique M¨obius (z2−z1)(z−z3) 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]. −1 Proof. Suppose that f(z1) = 0, f(z2) = 1 and f(z3) = ∞. Then (fg )(g(z1)) = 0, −1 −1 (fg )(g(z2)) = 1 and (fg )(g(z3)) = ∞. Thus −1 [g(z1), g(z2), g(z3), g(z4)] = fg (g(z4)) = f(z4) = [z1, z2, z3, z4] as required. Proposition. Every element of M is a composite of M¨obiustransformations of the following forms. az+0 (a) Da : z 7→ az = 0z+1 with a ∈ C\{0} (rotation/dilations); 1z+b (b) Tb : z 7→ z + b = 0z+1 with b ∈ C (translations); 0z+1 (c) S : z 7→ 1/z = 1z+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 18 f by ST−f(∞)f if necessary, without loss of generality f(∞) = ∞ . 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’re done. az+b a (bc−ad) Alternatively if c 6= 0 then cz+d = c + 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} 19 for some a ∈ C and r > 0 or of the form {z ∈ C | aRe(z) + bIm(z) = c} ∪ {∞} 20 for some a, b, c ∈ R with (a, b) 6= (0, 0)

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 21 with A, C ∈ R, B ∈ C and AC < |B|2.

18 Since 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(∞)S 19 i.e. a usual circle in C 20 i.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. 2 The latter deﬁnes a line in C if and only if B 6= 0 ie |B| > 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 of the 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 to ¯ Cww¯ + Bw + Bw¯ + A.

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

Proof. Using triple transitivity we can find f ∈ M such that f(z1) = 0, f(z2) = 1 and f(z3) = ∞. By preservation of circles z1, z2, z3 and z4 lie on a circle if and only if 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 argument to deduce that Möbius transformations preserve of circles from it. Definition. Given two elements x, y of a group G we say y is conjugate to x if there 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 conjugate is symmetric in x and y. Morever if also z = hyh−1 then z = (gh)x(gh)−1 so if z is conjugate to y and y is conjugate to x then z is conjugate to x. Proposition. Every Möbius transformation f except the identity has precisely one or two fixed points. If f has precisely one fixed point it is conjugate to the translation z 7→ z + 1. If f has precisely two fixed points it is conjugate to a map of the form z 7→ az with a ∈ C\{0}. az+b Proof. Let f be the map extending z 7→ cz+d and assume that f 6= id. Suppose first that c = 0. In this case ∞ is a fixed point. Moreover for z ∈ C, f fixes 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 only if cz2 + dz − az − b = 0. This is a quadratic equation and the coefficient of z2 is non-zero so it has one or two solutions (two unless there is a repeated root). Thus again 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 not fixed by f. Then f(x) 6∈ {x, w}.22 Let g be the unique Möbiustransformation such −1 that g(w) = ∞, g(x) = 0 and g(f(x)) = 1. Now for z ∈ C∞, gfg (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 only fixed point of f, ∞ = g(w) is the only fixed point of gfg−1. Thus by the discussion −1 az+b above gfg extends z 7→ 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 Möbiustransformation such that g(z1) = 0 and g(z2) = ∞. Then −1 gfg (0) = gf(z1) = g(z1) = 0 and −1 gfg (∞) = gf(z2) = g(z2) = ∞. −1 Thus gfg (z) = az for some a ∈ C∞.

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

n −1 −1 so f (z) = g (g(z)+n) for z ∈ C∞ not ﬁxed by f. Similarly if gfg : z 7→ az then n −1 n for n > 0, f (z) = g (a g(z)) for z not ﬁxed by f. Thus we can use conjugation to 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 of H in G is a set of the form g ◦ H := {g ◦ h | h ∈ H} for some g ∈ G. Similarly a right 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 the set 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, +). Then 0 + nZ = nZ = n + nZ. 1 + nZ = {1 + an | a ∈ Z} = (1 − n) + nZ. More 24 generally, b + nZ is the set of integers x such that x − b is a multiple of n. 2 2 (2) Suppose that G = D6 = {e, r, r , s, rs, r s} 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? 24 Note 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. 2 2 2 However if K = {e, r, r } 6 D6 then K = eK = rK = r K = Ke = Kr = Kr and {s, rs, r2s} = sK = rsK = r2sK = Ks = Krs = Kr2s. So in this case the left and right cosets are the same. (3) Suppose that M is the Möbiusgroup and H = {f ∈ M | f(0) = 0}. Then, for g ∈ 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 subgroup of 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) that every element of G lives inside some left coset; and (ii) if two left cosets have a common 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 −1 −1 −1 g = g1h1 = g2h2. Then for all h ∈ H, g1h = gh1 h = g2h2h1 h. Since h2h1 h ∈ H, we see that g1H ⊂ g2H. By symmetry we may conclude that g2H ⊂ g1H and so g1H = g2H as required. Now for the last part we observe that when G is finite, H must also be finite P since 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 function −1 −1 lg is given by gh 7→ g gh = 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 ∈ Z n there are integers q, r with 0 6 r < o(g) and n = qo(g) + r and then g = (go(g))qgr = gr. Moreover the elements e, g, . . . , go(g)−1 are distinct since if gi = gj j−i with 0 6 i < j < o(g) then g = e contradicting the definition of o(g). Proposition. Suppose that p is prime. Then every group of order p is isomorphic to 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 an earlier 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 of order at most 8 under the perspective that two groups are the same precisely if they are isomorphic. We’ve already seen that every group of order 2, 3, 5 or 7 is isomorphic to the cyclic group of the same order. It is evident that the trivial group is the only group of order 1 up to isomorphism. Before we begin this we’ll need the following construction that enables us to build new groups from old ones. Example. Suppose that G and H are groups. We can deﬁne a binary operation on G × H via (g1, h1)(g2, h2) = (g1g2, h1h2) for g1, g2 ∈ G and h1, h2 ∈ H. We claim that 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. −1 −1 −1 −1 Finally since (g , h )(g, h) = (eG, eH ) = (g, h)(g , h ) the operation on G × H has inverses.

Exercise. Show that if G1,G2 and G3 are groups then G1 × G2 is isomorphic to G2 × 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 and h2, 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 −1 (h1, h2) ∈ ker f then h1h2 = e so h1 = h2 ∈ H1 ∩ H2 = {e} (by (i)). Thus ker f = {e}. Finally Im f = {h1h2 | h1 ∈ H1, h2 ∈ H2} = G (by (iii)) and so f is the 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 then G 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 socks lemma it follows that y−1x−1 = xy. Since x−1 = x and y−1 = y, yx = xy as required. We also recall that every a group of order n with an element of order n is isomorphism to Cn and that every group of order 2n that has an element g of order −1 n and an element h of order 2 such that hg = g h is isomorphic to D2n.

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

Proof. First we’ll show that C4 and C2 × C2 are not isomorphic. Suppose (g, h) ∈ 2 C2 × C2. Then (g, h) = (e, e). So every element of C2 × C2 has order 2 but C4 has an 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 4 then it is isomorphic to C4 so suppose not. By Lagrange every non-identity element of G has order 2. Let g, h be two distinct such elements. Let H1 = {e, g}' C2 and H2 = {e, h}' C2. It suﬃces to show that the three conditions of the direct product theorem hold. Condition (i) is immediate since g 6= h. Condition (ii) follows from the 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 abelian 27 and D6 is not . Now suppose that G is any group of order 6. If G contains an element of order 6 then it is isomorphic to C6 so suppose not. By Lagrange every non-identity element of G has order 2 or 3. If there were no element of order 3 then any two non-identity elements would generate a subgroup of order 4 contradicting Lagrange. Thus G contains an element g of order 3. Let K = {e, g, g2} be the subgroup of G generated by g which is isomorphic to C3. By Lagrange for any h 6∈ K, G is partitioned by K 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 that G has no elements of order 6. Thus h2 = e. By the right coset version of Lagrange, G is also partitioned by K and Kh thus Kh = 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. −1 Thus gh = hg And we can deduce that G ' D6.

Example. The following set of matrices form an non-abelian group Q8 of order 8 1 0 i 0 0 1 0 i ± , ± , ± , ± . 0 1 0 −i −1 0 i 0

26 2 If f : Z4 → C2 ×C2 were an isomorphism then f(2) = f(1) = e giving f a non-trivial kernel. 27 If 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 = k 29had 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 0 i 0 0 1 0 i 1 = , i = , j = and k = . 0 1 0 −i −1 0 i 0

Then Q8 = {±1, ±i, ±j, ±k}. Then we can compute that 1 is an identity, −1 has order 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 necessarily 30 isomorphic to C2 × C2. 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 the 31 argument above for C2 × C2. So we’re reduced to classifying groups G of order 8 with no element of order 8 and at least one element g of order 4. In this setting the set K = {e, g, g2, g3} is a subgroup of G isomorphic to C4. Let h ∈ G\K. Then G = K∪hK by Lagrange so G = {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 G has no such elements. So there are two cases remaining: case A where h2 = e and case B where h2 = g2. Suppose that we’re in case A and h2 = e. Then consider the product gh. By Lagrange again, gh ∈ hK = {h, hg, hg2, hg3}. If gh = hgi then h−1gh = gi and

2 g = h−2gh2 = h−1(h−1gh)h = h−1gih = (h−1gh)i = (gi)i = gi .32 Thus i = 1 or 3. If i = 1 then G is abelian and we can use the direct product theorem to show that G ' K × {e, h}' C4 × C2. If instead i = 3 then G has an element g of order 4 and an element h of order 2 such that gh = hg−1 so as G has order 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 = hg or 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.

30 the 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 Q14 32since 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 are isomorphic. 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 0 i 0 0 1 0 i 1 = , i = , j = and k = . 0 1 0 −i −1 0 i 0

The sum or product of two elements of H lives in H and + and · obey the same associativity and distributivity laws as Q, R and C with identities 0 and 1 respectively. 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 the latter we can deﬁne ‘quaternionic conjugation’ by (a1 + bi + cj + dk)∗ = a1 − bi − cj − dk and then verifying that if x = a1 + bi + cj + dk then xx∗ = x∗x = (a2 + b2 + c2 + d2)1 −1 1 ∗ so x = a2+b2+c2+d2 x for x 6= 0. 2.5. Fermat–Euler theorem. We can deﬁne a multiplication operation on the set Zn = {0, 1, . . . , n − 1} by setting a ·n b to be the remainder after dividing ab by n.

Deﬁnition. Let Un := {a ∈ Zn | ∃b ∈ Zn s.t. a ·n b = 1} be the set of invertible elements 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. Indeed Un = {a ∈ Zn | (a, n) = 1}.

Lemma. (Un, ·n) is an abelian group.

Proof. ·n deﬁnes an associative and commutative operation on Zn with an identity 34 1. 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 and −1 if a ∈ Un then there is b ∈ Zn with a ·n b = b ·n a = 1. Then b ∈ Un and b = a . So Un has inverses. Theorem (Fermat–Euler Theorem). If (a, n) = 1 then aϕ(n) ≡ 1 mod n.

33 We 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. 34 for 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 theorem that every element of (Un, ·n) has order dividing ϕ(n). The result follows immediately.

Lecture 11 3. Group Actions We started the course by saying that groups are fundamentally about symmetry but the connection has been opaque for the last three lectures. In this chapter we will 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 · x such 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 Möbiusgroup M 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 a function. 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 of vertices 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 sets of 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 = gkH for g, k ∈ G. To see this we need to check that if kH = k0H then gkH = gk0H. But if kH = k0H then k0 = kh for some h ∈ H so gk0 = gkh ∈ gk0H ∩ gkH and gkH = gk0H by Lagrange. Given this we see that for all k ∈ G, ekH = kH and 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 all g ∈ 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 formula θ(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) it suffices 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 this in precisely one way. So suppose that θ is any such homomorphism, θ corresponds to an action ·: G × X → X precisely if g · x = θ(g)(x) defines an action. So it remains to check that e · x = x and g · (h · x) = (gh) · x for all x ∈ X and g, h ∈ G. But 35 e · x = θ(e)(x) = idX (x) = x and g · (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 corresponding 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 x under 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. T 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) = Z and

StabIsom(Z) = {id, m 7→ 2n − m} (2) Under the natural action of M on C∞, for all z ∈ C∞ OrbM(z) = C∞ and az + b Stab (∞) = z 7→ | ad 6= 0 . M 0c + d

(3) Under the action of D2n on the set of points of a regular n-gon the orbit of a vertex of the n-gon is the set of all vertices of the n-gon and the stabliser of a vertex consists of the identity and reﬂection in the line through the centre of the 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 and StabG(eH) = {g ∈ G | gH = eH} = H. More generally −1 −1 StabG(kH) = {g ∈ G | gkH = kH} = {g ∈ G | k gkH = H} = kHk . (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 m X |X| = |Oi| i=1

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 · x −1 37 so g ∈ StabG(x). Thus StabG(x) 6 G as required. (ii) Since for any x ∈ X, x = e · x so x ∈ OrbG(x), it suﬃces to prove that for any x, y ∈ X either OrbG(x) = OrbG(y) or OrbG(x) ∩ OrbG(y) = ∅. Suppose that z ∈ OrbG(x) ∩ OrbG(y). Then there are some elements g, h ∈ G such that −1 g · x = z = h · y. Thus x = (g h) · y. Now if w ∈ OrbG(x) then there is some f ∈ G such that w = f · x. Whence we can compute w = (fg−1h) · y and so OrbG(x) ⊂ OrbG(y). By symmetry OrbG(y) ⊂ OrbG(x). Now we can see that 38 OrbG(x) = OrbG(y) as required. Deﬁnition. We say that an action of G on X is transitive if there is only one orbit i.e. if X = OrbG(x) for any x ∈ X. Theorem (Orbit-Stabiliser Theorem). Suppose a group G acts on a set X and x ∈ 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 Stab (0) Isom( ). Isom(Z) 6 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 since it is non-empty and if g · x = y then for h ∈ G h · x = y ⇐⇒ g−1 · (h · x) = x −1 ⇐⇒ g h ∈ StabG(x)

⇐⇒ h ∈ g StabG(x).

Thus there is a function G/ StabG(x) → OrbG(x) given by h StabG(x) 7→ h · x with 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- 39 grange. 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 when we 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 we

see 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| = n StabSn (n). But StabSn (n) is isomorphic to Sn−1 by restricting f ∈ Sn that ﬁxes n to a permutation of {1, . . . , n − 1}. Thus |Sn| = n|Sn−1|. Since 40 |S1| = 1 we deduce that |Sn| = n!.

Lecture 13 Fact. If f : R3 → R3 is an isometry that fixes 4 non-coplanar points then f is the identity. (4) Let X be a regular tetrahedron. Then Sym(X) acts transitively on the set of 4 vertices of X and the stabiliser of a vertex v ∈ X consists of three rotations and three reflections. Thus | Sym(X)| = 6 · 4 = 24. This calculation enables us to prove that Sym(X) ' S4: if we label the vertices by the numbers 1, 2, 3, 4 then the action of Sym(X) on the vertices 3 defines a homomorphism θ : Sym(X) → S4. Since any isometry of R fixing all four vertices is the identity we can conclude that ker θ = {id}. By counting we can deduce Im θ = S4. (5) Let X be a cube. Then Sym(X) acts transitively on the set of 6 faces of X and the 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 diagonally opposite vertex in F then 42 StabH (v) = {e, reflection in plane containing v, w and the centre of X} .

39You might like to think about whether you can do this last part without recourse to Lagrange. 40 Or if you prefer |S0| = 1 41This can be seen by considering rotations about an axis through the centre of F and the centre of its opposite face. 42 3 Since if an isometry of R 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 the conjugacy 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 be denoted 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 any a, b ∈ Z, −1 tbtatb = ta and for n ∈ Z, −1 sbtasb (n) = sbta(b − n) = sb(a + b − n) = n − a ie −1 sbtasb = t−a so for a 6= 0 44 CG(ta) = H and ccl(ta) = {ta, t−a} . Similarly for n ∈ Z −1 tbsatb (n) = tbsa(n − b) = tb(a − (n − b)) = (2b + a − n) = s2b+a(n) and −1 sbsasb (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}. 45 That is there are two conjugacy classes of reﬂections ccl(s0) and ccl(s1).

43Since conjugacy classes are orbits of an action they partition G; that is every element of G lies in precisely one conjugacy class. 44 Of 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 Möbiusgroup M the conjugacy class of z 7→ z + 1 consists of all Möbiustransformations with precisely one fixed point i.e. ccl(z 7→ z + 1) = {f ∈ M | f has precisely one fixed point} and that every Möbius transformation with precisely two fixed points is in the same conjugacy class as a Möbiustransformation of the form z 7→ az. We will return later to the question of when ccl(z 7→ az) = ccl(z 7→ bz) and what the 46 centralisers of these elements are. Of course ccl(id) = {id} and CM(id) = M. Definition. The kernel of the homomorphism G → S(G) given by the conjugation action 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}. T (b) Z(G) = {g ∈ G | gx = xg for all x ∈ G} = x∈G CG(x). (c) Z(G) = {g ∈ G | |ccl(g)| = 1}.

−1 Proof. (a) For g, x ∈ G, g ∈ CG(x) if and only if gxg = x that is if and only if gx = xg. (b) For g ∈ G, g ∈ Z(G) if and only if gxg−1 = x for all x ∈ G i.e. if and only if gx = 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 ∈ G −1 i.e. if and only if xgx = g for all x ∈ G. 3.4. Cayley’s Theorem. Cayley’s Theorem will tell us that every group is isomorphic to a subgroup of a symmetric group. Deﬁnition. If G is a group then the left regular action of G on itself is given by the function ·: G × G → G; g · x = gx.

Example. The left regular action of Z on itself is by translations. i.e. the corre- 47 sponding homomorphism Z → S(Z) is given by n 7→ tn. Lemma. The left regular action of G on G is an action that is both transitive and faithful. Proof. First we should check that the left regular action is indeed an action: for all g, 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 subgroup of S(G).

46 Spoiler: 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 prove these 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(∞)? 47 recall tn denotes translation by n GROUPS 29

Proof. The left-regular action defines a homomorphism θ : G → S(G). Since Im θ is a subgroup of G we may view this as a homomorphism G → Im θ whose image is still Im θ. Since the action is faithful, ker θ = {e} and we see that G is isomorphic to Im θ. It perhaps should be said that this theorem is simultaneously deep and almost useless. Deep because it tells us that anything satisfying our abstract definition of a group can be viewed as symmetries of something. Almost useless because knowing this 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 group whose order is a multiple of p. Then G contains an element of order p. p Proof. Consider the action of Zp on G 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. p p p−1 Let X = {(g0, . . . , gp−1) ∈ G | g0g1 ··· gp−1 = e} ⊂ G . Then |X| = |G| since however we choose g0, . . . , gp−2 ∈ G there is precisely one choice of gp−1 ∈ G 48 such g0g1 ··· gp−1 = e . Moreover the ‘constant tuple’ (g, g, . . . , g) is in X if and only gp = e — and if g 6= e this is equivalent to o(g) = p. p Now the cyclic permutation action of Zp on G restricts to an action on X since if g0g1 ··· gp−1 = e and i ∈ Zp then −1 gig1+pi ··· g(p−1)+pi = (g0 ··· gi−1) (g0 ··· gp−1)(g0 ··· gi−1) = e.

Since Zp has prime order the orbit-stabiliser theorem tells us that every orbit OrbZp (x) has order 1 or p for x ∈ X. Since the orbits partition X and p is a factor of |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 be at least p − 1 other such orbits. If (g0, . . . , gp−1) is one such orbit then, since i · (g0, . . . , gp−1) = (g0, . . . , gp−1) for all i ∈ Zp, we can conclude that gi = g0 for all p i ∈ Zp and so g0 = e. 4. Quotient Groups 4.1. Normal subgroups. Suppose that G is a group. Let P(G) denote the set of subsets 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.

48 −1 That is gp−1 = (g0g1 ··· gp−2) . 30 SIMON WADSLEY

Lemma. This operation on P(G) is associative and has an identity but does not have 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∅ = ∅, 49 so ∅ has no inverse. 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 of left 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. However in general things are not so straightforward.

2 2 Example. Consider G = D6 = {e, r, r , s, rs, r s} where r denotes a non-trivial rotation 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 and sH. 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 always a left coset of H in G if and only if gHg−1 = H51 for all g ∈ G. In this case g1Hg2H = g1g2H for all g1, g2 ∈ G.

Lecture 15

Proof. For g1, g2 ∈ G we compute −1 g1Hg2H = {g1h1g2h2 | h1, h2 ∈ H} = {g1g2g2 h1g2h2 | h1, h2 ∈ H}. −1 Thus if g2 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 later 51Here gHg−1 means {g}H{g−1} GROUPS 31

−1 −1 we need g2 h1g2 ∈ H for all h1 ∈ H since (g1g2)g2 h1g2e ∈ g1Hg2H. So if the product 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 all g ∈ 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 for all g ∈ G.

Warning. To show that a subset of G is a normal subgroup we must show that 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 a right coset.52 Proof. Suppose that gHg−1 = H. Then gH = gHg−1g = Hg so the left coset gH is a right coset. Conversely suppose that every left coset is a right coset and g ∈ G. Then gH = Hk for some k ∈ G. So g = hk for some h ∈ H. Thus gH = Hk = Hkg−1g = Hh−1g = Hg

−1 and gHg = H as required. Proposition. If H is a normal subgroup of G then the restriction of the binary operation 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) restricts to an associative binary operation on G/H when H is normal and moreover that g1Hg2H = g1g2H. Suppose that gH ∈ G/H. Then eHgH = gH = gHeH so eH = H is an identity −1 −1 in G/H. Finally gHg H = eH = g HgH 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 group homomorphism. Then ker f is a normal subgroup of G, Im f is a subgroup of H and 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 show that 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 see that ker f is normal. Now if g ker f = g0 ker f then g−1g0 ∈ ker f and so f(g)−1f(g0) = f(g−1g0) = e. Thus f(g) = f(g0) and f(g ker f) = f(g) is a well-defined function whose image is Im 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 from G/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 follows from 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 subgroup is 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 ' 53 a by n has kernel nZ and image Zn. Thus it induces an isomorphism Z/nZ → Zn. 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 1 Im θ = S := {z ∈ C | |z| = 1}, the unit circle in C and ker θ = Z thus we can deduce that R/Z ' S1.

Example. Let θ : D2n → {±1} such that +1 if g is a rotation θ(g) := −1 if g is a reflection. Then we can verify that θ is a homomorphism since the product of two reflections or two rotations is a rotation and the product of a rotation and a reflection in either order is a reflection. Moreover Im θ = {±1} and ker θ is the subgroup of all rotations of the regular n-gon. Thus D2n/{rotations in D2n}' C2.

53 The notation Zn as we have deﬁned it is rarely used and instead Z/nZ is used to describe essentially the same thing. GROUPS 33

Example (Group-theoretic understanding of qth powers mod p). Let p and q be distinct 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 isomorphism G ' Im θ. By counting we can conclude that Im θ = G. In particular we see that every 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 element 54 of order q by Cauchy’s Theorem. Thus | ker θ| > q. Since G/ ker θ ' Im θ and p−1 p−1 |G/ ker θ| = |G|/| ker θ| 6 q we see that Zp has at most q + 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} = T 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 X induces a faithful action of G/K on X.56

Example. Suppose that X is a regular tetrahedron in R3. X has six edges and each edge has four neighbours.57 Thus we can partition the set of edges into three pairs with 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 v and w are non-adjacent edges of X then f(v) and f(w) are also non-adjacent edges of 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 we can deduce that Sym(X)/ ker θ ' S(P ). We showed earlier than Sym(X) ' S4 and 58 it is straightforward to see that S(P ) ' S3. Thus we can deduce that S4 has a 59 normal subgroup K such that S4/K ' S3.

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 fact it 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. 58 Or S(P ) ' D6 if you prefer 59 Can 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×n matrices with entries in F. Here are some facts proven in Vectors and Matrices. Facts. n n 60 (1) Every element A of Mn(F) deﬁnes a linear map A: F → F via A: v 7→ Av. Moreover every linear map Fn → Fn arises in this way and A can be recovered n from A since the ith column of A is A(ei) where ei denotes the element of F with 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 identity n for matrix multiplication on Mn(F). Moreover In = idF . (4) There is a function det: Mn(F) → F such that A has an inverse in Mn(F) if and only if det A 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) | det A 6= 0} is the group 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), det AB = det A det B, if A, B ∈ GLn(F) then det A 6= 0 and det B 6= 0 so det AB 6= 0. Thus AB ∈ GLn(F) and matrix multiplication 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 det A 6= 0, there is a matrix B such that AB = In = BA. Then det A det B = det AB = det In = 1. So det B 6= 0 and B ∈ GLn(F). n Remark. There is a natural action of GLn(F) on F via (A, v) 7→ Av. One can n show that the homomorphism GLn(F) → S(F ) coming from this action induces n an isomorphism GLn(F) with the subgroup of S(F ) consisting of all invertible linear maps Fn → Fn.

Lemma. The function det: GLn(F) → (F\{0}, ·) is a group homomorphism with image F\{0}. Proof. That it is a homomorphism follows immediately from fact 4 above: for A, B ∈ GLn(F), det AB = det A det B. To see its image we compute  .  λ 0 .. 0    .. ..   0 1 . .  det   = λ. ......   . . . 0   .  0 .. 0 1

Deﬁnition. The special linear group SLn(F) is the kernel of det: GLn(F) → F\{0} i.e. SLn(F) := {A ∈ Mn(F) | det A = 1}.

60 n Recall that A is linear means that A(λv+µw) = λA(v)+µA(w) for all λ, µ ∈ F and v, w ∈ F . GROUPS 35

Remarks. n n (1) The action of GLn(F) on F induces an action of SLn(F) on F by restriction n and the resulting homomorphism SLn(F) → S(F ) induces an isomorphism n of SLn(F) with the subgroup of S(F ) consisting of volume preserving linear maps Fn → Fn. (2) SLn(F) a normal subgroup of GLn(F and GLn(F)/SLn(F) ' F\{0}. Examples. a b GL ( ) = | ad − bc 6= 0 and 2 F c d a b SL ( ) = | ad − dc = 1 2 F c d 5.2. M¨obiusmaps as projective linear transformations.

Notation. Given v ∈ C2\{0} let [v] denote the (unique) line {λv | λ ∈ C} through 0 and v in C2. The set of all such lines is called the complex projective line typically written P1(C). The following lemma gives a parameterisation of the elements of P1(C). z Lemma. Every element of 1( ) is either of the form with z ∈ or P C 1 C 1 . Moreover these lines are all distinct. 0 2 v1 Proof. Suppose v ∈ C \{0}. Then v = for some v1, v2 ∈ C not both 0. v2 v1 v1 v v1 v If v2 6= 0 then [v] = 2 since λ = v2λ 2 . 1 v2 1 1 1 If v = 0 then [v] = since λv = v λ . 2 0 1 0 z w z 1 Finally if λ = then λ = 1 and z = w. And λ = is evidently 1 1 1 0 impossible. z It follows that we may identify and 1( ) via z 7→ for z ∈ and C∞ P C 1 C 1 ∞ 7→ . 0

1 2 Proposition. GL2(C) acts on P (C) via (A, [v]) 7→ [Av] for v ∈ C \{0}. Proof. First we must show that (A, [v]) 7→ [Av] is a well-deﬁned function 1 1 GL2(C) × P (C) → P (C) 2 i.e. that if A ∈ GL2(C) and v ∈ C \{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 some non-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] ∈ 1 P (C) and A, B ∈ GL2(C). 36 SIMON WADSLEY

1 We note that under this action of GL2(C) on P (C)

a b z az + b az+b · = = cz+d when z 6= −d/c, c d 1 cz + d 1

a b −d/c 1 · = , c d 1 0

a b 1 a a · = = c when c 6= 0, and c d 0 c 1

a b 1 1 · = . 0 d 0 0 1 Thus, under the identiﬁcation of C∞ with P (C), the homomorphism

θ : GL2(C) → S(C∞) a b corresponding to this action sends the matrix to the M¨obiusmap repre- c d az+b sented by z 7→ cz+d , and so Im θ = M. Thus M is a subgroup of S(C∞). a b Moreover ker θ consists of invertible matrices ﬁxing every line through c d the origin in C2. Now 1 a b Stab = ∈ GL ( ) | c = 0 , GL2(C) 0 c d 2 C

0 a b Stab = ∈ GL ( | b = 0 and GL2(C) 1 c d 2 C

1 a b Stab = ∈ GL ( ) | a + b = c + d . GL2(C) 1 c d 2 C Since a M¨obiustransformation that ﬁxes three distinct points is the identity, ker θ is the intersection of these three sets i.e. a 0 ker(GL ( ) → M) = | a 6= 0 2 C 0 a is the group of non-zero scalar matrices.61 Thus P GL2(C) := GL2(C)/{λI | λ ∈ C 6= 0}'M. It is not hard to see that a similar argument shows that PSL2(C) := SL2(C)/{±I}'M. We can summarize this discussion with the following theorem.

1 Theorem. The action of GL2(C) on P (C) induces an isomorphism from P GL2(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 Möbius 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 to n Pn the matrix A and e1, . . . , en is the standard basis for F then A(ei) = j=1 Ajiej. n If f1, . . . , fn is another basis for F then there is an invertible linear map P such that P (ei) = fi for i = 1, . . . , n. i.e. P corresponds to the matrix P whose columns Pn f1, . . . , fn and fi = j=1 Pjiej for i = 1, . . . , n. It follows that for j = 1, . . . , n, n n n n X −1 X −1 X X −1 Pkj fk = Pkj Plkel = (PP )ljel = ej. k=1 k=1 l=1 l=1 Then

A(fi) = AP (ei) n X = (AP )jiej j=1 n n X X −1 = (AP )ji( Pkj fk) j=1 k=1 n X −1 = (P AP )kifk k=1

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

Proposition. GLn(F) acts on Mn(F) by conjugation. Proof. Consider −1 ·: GLn(F) × Mn(F) → Mn(F); (P,X) 7→ PXP .

Then for X ∈ Matn(F), −1 In · X = InXIn = 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) represent the same linear map with respect to diﬀerent bases if and only if they are in the same GLn(F)-orbit under this conjugation action. Example (See Vectors and Matrices). If A: C2 → C2 is a linear map then precisely one 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 0 0 µ

62 with λ, µ ∈ C distinct — in this case {λ, µ} is determined by A but they may 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 0 0 λ

2 with λ ∈ C — in this case λ is determined by A indeed A = λ idC and A is represented 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 1 0 λ again λ is determined by A.63 We may interpret this group-theoretically: every GL2(C)-orbit in M2(C) with respect to the conjugation action is one of the following: λ 0 O := Orb with λ, µ ∈ distinct, λ,µ GL2(C) 0 µ C λ 0 O(1) := Orb with λ ∈ and λ GL2(C) 0 λ C λ 1 O(2) := Orb ( with λ ∈ . λ GL2(C) 0 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}, (1) Oλ = {λI2} and (2) 2 Oλ = {A ∈ M2(C) | det(tI − A) = (t − λ) for all t ∈ C,A 6= λI2}. We can also compute λ 0 a b aλ bλ = 0 µ c d cµ dµ and a b λ 0 aλ bµ = c d 0 µ cλ dµ so that for λ 6= µ, λ 0 a 0 Stab = | ad 6= 0 GL2(C) 0 µ 0 d and Stab (λI ) = GL ( ). GL2(C) 2 2 C Similarly λ 1 a b aλ + c bλ + d = 0 λ c d cλ dλ and a b λ 1 aλ a + bλ = c d 0 λ cλ c + dλ so λ 1 a b Stab = | a 6= 0 . GL2(C) 0 λ 0 a All other stabilisers are conjugate to these ones. We can easily read oﬀ the conjugacy classes and centralisers in GL2(C) by restricting to the case λ, µ 6= 0. 63 it is the unique eigenvalue of A and A 6= λ idC. GROUPS 39

Exercise. Deduce that in M' P GL2(C), ccl(z 7→ az) = ccl(z 7→ z 7→ bz) if and only if b ∈ {a, 1/a} thus provide a description of all the conjugacy classes in M and compute centralisers of suitable representatives of each class.

Lecture 19 5.4. The orthogonal and special orthogonal groups. Recall that any (square) T T T matrix A has a transpose A with Aij = Aji and det A = det A. Moreover if A, B are square matrices of the same size then (AB)T = BT AT .

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

Lemma. O(n) is a subgroup of GLn(R). T −1 Proof. In ∈ O(n) and if A, B ∈ O(n) then B = B so −1 T −1 −1 T T −1 −1 T T −1 T T (AB ) AB = (B ) A AB = (B ) IB = (BB ) = In = In −1 −1 T and similarly (AB )(AB ) = In. n Pn Recall that R comes with an inner product v · w = i=1 viwi which deﬁnes a length function on Rn via |v| = (v·v)1/2. We also recall the deﬁnition of Kronecker’s delta 1 if i = j; δ := ij 0 if i 6= j. n 64 A basis f1, . . . , fn of R is said to be orthonormal if fi · fj = δij. Lemma. n (a) If {f1, . . . , fn} ⊂ R such that fi ·fj = δij for all 1 6 i, j 6 n, then {f1, . . . , fn} is an orthonormal basis for Rn. n 1 2 2 (b) If v, w ∈ R then v · w = 4 (|v + w| − |v − w| ). Pn Proof. (a) Suppose i=1 λifi = 0. Then for j = 1, . . . , n, n ! n X X 0 = λifi · fj = λi(fi · fj) = λj. i=1 i=1

Thus {f1, . . . , fn} is linearly independent. Since it has n elements it must span 65 Rn. (b) We compute (v + w) · (v + w) = v · v + v · w + w · v + w · w and (v − w) · (v − w) = v · v − v · w − w · v − w · w subtracting and using symmetry of the inner product we see that |v + w|2 − |v − w|2 = 4v · w as required.

64There is a little apparent notational ambiguity here since we use subscripts to index the n basis vectors as well as to index the coordinates of a vector. Each fi is in R so can be written Pn Pn as k=1(fi)kek and fi · fj = k=1(fi)k(fj )k. 65 n Pn In fact if v ∈ R , v = i=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. n Proof. Suppose that A ∈ On and v, w ∈ R . Then n X Av · Aw = (Av)i(Aw)i i=1 n  n  n ! X X X =  Aijvj Aikwk i=1 j=1 k=1 n X T = vj(A A)jkwk j,k=1 X = vjwj = v · w. j Thus (i) =⇒ (ii). n Suppose now that Av·Aw = v·w for all v, w ∈ R . Then in particular Aei ·Aej = 66 n ei · ej = δij for each 1 6 i, j 6 n and Ae1, . . . , Aen is an orthonormal basis for R 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 X T = AkiAkj = (A A)ij k=1 T and A A = In. Thus (iii) =⇒ (i). (ii) =⇒ (iv) is immediate from the deﬁnitions. To see that (iv) =⇒ (ii) we use (b) of the lemma: 1 Av · Aw = (|A(v + w)|2 − |A(v − w)|2 4 and 1 v · w = (|v + w|2 − |v − w|2). 4 If (iv) holds then RHSs of these equations to be equal for all v, w and thus the LHSs 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 that preserve the scalar product or equivalently to the subgroup of S(Fn) consisting of linear maps that preserve length. The conjugation action GLn(R) on Mn(R) restricts to an action of O(n) on Mn(R). The equivalence of (i) and (iii) in the propostion shows that two distinct matrices in Mn(R) are in the same O(n)-orbit if and only if they represent the same linear 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

T T 2 Proof. If A ∈ O(n) then det A = det A so 1 = det In = det AA = (det A) and det A = ±1. Since −1 0   0 1 0   ∈ O(n)  ..   .  0 1 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) and O(n)/SO(n) ' C2. There are complex versions of the orthogonal group and the special orthogonal group called the unitary group and the special unitary group. We won’t have time to 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 plane m m normal to n is the function Rn : R → R 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); m (b) R 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; 2 m (c) (Rn) = idR and; (d) det Rn = −1. m Proof. (a) First we show that Rn is linear: if x, y ∈ R 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). m Now we show that Rn preserves the inner product: if x, y ∈ R 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.

67 n I have not deﬁned an orientation of R . One way would be as an SO(n)-orbit of orthonormal n bases for R which would make this completely tautological. There are more sophisticated ways that make it less so. With this deﬁnition the next sentence gives that there are exactly two n orientations of R . 42 SIMON WADSLEY

m So we may ﬁnd a basis f1, . . . , fm for R such that f1 = n and f2, . . . , fm ∈ U. Then Rn(f1) = −f1 and Rn(fi) = fi for 2 6 i 6 m as required. −1 −1 2 (c), (d) If P is the change of basis matrix so that P RnP = D then P (Rn) P = −1 2 2 −1 (P RnP ) = Im. Thus (Rn) = PImP = Im. Moreover −1 −1 = det D = (det P ) det Rn det P = det Rn as required.

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

(x−y) 68 Proof. Let n = ((x−y)·(x−y))1/2 so that n · n = 1. 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 the 3 69 form Rn with n ∈ R of length 1.

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

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

Next consider Rn1 A(e2). If Rn1 A(e2) = e2 let n2 = 0. Otherwise

Rn1 A(e2) · e1 = e2 · e1 = 0 since Rn1 A is orthogonal and fixes e1. Thus we may use the proposition to find n2 of unit length parallel to Rn1 A(e2) − e2 such that Rn2 Rn1 A(e2) = e2. Morever n2 · e1 = 0 since n2 is parallel to Rn1 A(e2) − e2 and Rn1 A(e2) and e2 are both orthogonal to e1. Thus Rn2 Rn1 A fixes e1 and e2 in every case.

Now Rn2 Rn1 A(e3) has length 1 and is orthogonal to both e1 and e2 since e3 is orthogonal to both e1 and e2, and Rn2 Rn1 A is orthogonal and ﬁxes e1 and e2.

Thus Rn2 Rn1 A(e3) = ±e3. If Rn2 Rn1 A(e3) = e3 let n3 = 0, otherwise let n3 = e3.

In either case we can compute that Rn3 Rn2 Rn1 A ﬁxes e1, e2 and e3. Thus as it is a linear map it must be the identity. We conclude A = Rn1 Rn2 Rn3 since each Rni has order 1 or 2 according as ni = 0 or ni 6= 0. In particular A is a product of at most three reﬂections. Proposition. If A ∈ O(2) then either (i) A = SO(2) and there is some 0 6 θ < 2π such that cos θ sin θ A = 70 or − sin θ cos θ

2 (ii) A 6∈ SO(2) and A = Rn for some n ∈ R 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 b A = c d for some a, b, c, d ∈ R such that a b a c = I . c d b d 2

Thus a2 + b2 = c2 + d2 = 1, ac + bd = 0 and ad − bc = ±1. Let a = cos θ, b = sin θ, c = sin φ and d = cos φ with 0 6 θ, φ < 2π. Then 0 = cos θ sin φ + cos φ sin θ = sin(θ + φ) and ad − bc = cos θ cos φ − sin θ sin φ = 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 θ sin θ/2 sin −θ/2 sin θ/2 = = − sin θ − cos θ − cos θ/2 cos θ/2 − cos θ/2 and cos θ sin θ cos θ/2 cos θ/2 = . sin θ − cos θ sin θ/2 sin θ/2 Since sin θ/2 cos θ/2 · = 0 − cos θ/2 sin θ/2 sin θ/2 we see that A = R for n = .71 n − cos θ/2

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

Proof. By the last theorem A is a product of at most three reﬂections Rn. Since det A = 1 and each Rn has determinant −1, A must be a product of either 0 or 2 3 reﬂections. If 0 then A = idR and any v will work. If A = Rn1 Rn2 with n1, n2 of unit length. Then consider the linear map 3 2 x · n1 R → R ; x 7→ . x · n2

By the rank-nullity theorem it has non-trivial kernel. i.e. there is v ∈ R3 such that v · 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). 72 3 That is every rotation in R has an axis. 44 SIMON WADSLEY

Lecture 21 Corollary. Every A in SO(3) is conjugate in SO(3) to a matrix of the form 1 0 0  0 cos θ sin θ . 0 − sin θ cos θ

3 Proof. Suppose A ∈ SO(3) and v1 ∈ R non-zero such that Av1 = v1. By replacing 3 v1 by v1/|v1| we may assume |v1| = 1. Let U := ker(R → R; x 7→ x · v1). By the rank-nullity theorem dim U = 2. We can choose an orthonormal basis v2, v3 3 for U. Then v1, v2, v3 is an orthonormal basis for R and so the matrix P with columns v1, v2 and v3 is in O(3). If det P = −1 then we may swap v2 and v3 so that P ∈ SO(3). We claim that B := P −1AP is of the required form. −1 −1 Certainly B ∈ SO(3) and B(e1) = P Av1 = P (v1) = e1. Moreover

e1 · Be2 = Be1 · Be2 = e1 · e2 = 0 and similarly e1 · Be3 = 0. So 1 0 0  B = 0 C11 C12 0 C21 C22 for some C ∈ M2(R). T T 73 Then B B = I3 gives C C = I2. Since det B = 1, det C = 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 is an invertible function X → X. In this chapter we will study permutations of finite sets. More particularly we will study permutations of [n] := {1, 2, . . . , n}. Since there 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 action of the cyclic subgroup of Sn generated by σ has precisely one orbit of size greater than one. Example. If σ : [5] → [5] such that σ(1) = 3, σ(2) = 2, σ(3) = 5, σ(4) = 4 and σ(5) = 1 then σ ∈ S5. We can draw σ as follows:

y 1 2 q 4 3 - 4 4 5 .

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

73 Alternatively we could’ve shown that B acts on the span of e2 and e3 by a length preserving transformation C and so C ∈ O(2). GROUPS 45

Suppose that σ is a cycle of order k. Then σ generates the subgroup hσi := {id, σ, . . . , σk−1}. For any b ∈ [n] in an orbit of size 1 i σ (b) = b for all i ∈ Z j and if a ∈ [n] is in the orbit of size greater than one then for c = σ (a) ∈ Orbhσi(a), σi(c) = σi+j(a) = σj(σi(a)). i i i i Thus σ (c) = c whenever σ (a) = a. i.e. σ ∈ Stabhσi(a) only if σ = id. Thus Stabhσi(a) = {e} and | Orbhσi(a)| = k. Notation. If σ is a cycle of order k such that the orbit of size greater than one contains 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 distinct and exhaust the orbit of a under hσi. Thus (aσ(a) ··· σk−1(a)) uniquely determines σ 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 start the 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). Thus to 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 Orbhτi(τ(k)) = Orbhτi(k) is the non-trivial hτi-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 (possibly empty) product of disjoint cycles. Moreover the representation of π as a product of disjoint cycles is unique up to reordering.

Example. Consider (135) and (145) in S5. How is (135)(145) expressed as a product of 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 the cycles 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 follows that (135)(145) = (14)(35). Pictorally 1 2 3 4 5 1 2 3 4 5

1 s 2 3 ) 4 5 =

1s 2' 3 4' 5 1z 2 3Ð $ 4 5 whereas 1 2 3 4 5 1 2 3 4 5

1 s 2 ' 3 4 ' 5 =

× 1s 2 3) 4 5 1Ð 2 3 4 5 i.e. (145)(135) = (13)(45).

Lecture 22

Proof of disjoint cycle decomposition theorem. Suppose that π ∈ Sn. The hπi- orbits partition [n], m [ [n] = Oi i=1 say. By re-ordering the orbits we may assume that ki := |Oi| > 1 for i = 1, . . . l 74 and |Oi| = 1 for i = l + 1, . . . m. . For 1 6 i 6 l pick ai ∈ Oi and deﬁne ki−1 σi := (aiπ(ai) ··· π (ai)). Ql 75 Since the Oi are disjoint, the σi are disjoint cycles. We claim that π = i=1 σi.

To prove the claim suppose that a ∈ [n]. Since the Oi partition [n], a ∈ Oia for precisely one ia ∈ [m]. If ia > l then π(a) = a and σi(a) = a for all i ∈ [l]. Thus l ! Y π(a) = a = σi (a). i=1

If ia 6 l then σi(a) = a for i ∈ [l]\{ia} and σia (a) = π(a) by deﬁnition. Thus l ! l Y Y σi (a) = σia σi(a) = σia (a) = π(a) i=1 i=1 i6=ia as required.

74we allow l = 0 or l = m in which cases one of these lists is empty 75 Since 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 appear in the product if and only if its non-trivial orbit O is a non-trivial orbit of hπi and σ(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). Qk n Proof. Let π = i=1 σi with σi pairwise disjoint and o(σi) = ni. Then π = Qk n n n i=1 σi since the σi commute. Moreover π = id if and only if σi = id for each 76 n i. Thus π = id if and only if ni divides n for each i = 1, . . . k i.e. if and only if n 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 that every 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. For example (12)(23)(34) = (1234) = (14)(13)(12).

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

Theorem. Given π ∈ Sn let l(π) be the number of orbits of hπi in [n]. For any π ∈ 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 π Qm as a product of disjoint cycles π = i=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)+1 in this case.77 Suppose instead that a and b lie in diﬀerent orbits in π. Then some σi = 78 (ax2 ··· xk) and some σj = (by2 ··· yl). 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)−1 in this case. 79 Thus in every case l(π(ab)) = l(π) ± 1.

76 n This can be seen by considering what π 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

: Sn → ({±1}, ·) such that (π) = 1 if π is a product of an even number of transpositions and (π) = −1 if π is a product of an odd number of transpositions. Moreover, for n > 2, Im = {±1}. Qm Proof. Suppose that π = i=1 τi with each τi a transposition. By induction on m Qm0 we see that l(π) ≡ l(id)+m mod 2. Thus if also π = i=1 σi with each σi a transposition then m ≡ m0 mod 2 and the function as deﬁned in the statement makes sense.80 It is now easy to verify that is a homomorphism and, since ((12)) = −1, that Im = {±1} for n > 2.

Deﬁnition. Given π ∈ Sn we say that π is even if (π) = 1 and that π is odd if (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(: Sn → {±1}) is the normal subgroup of Sn consisting of all even permutations.

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

Lecture 23

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

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

−1 σ(a1 ··· am)σ = (σ(a1) ··· σ(am)). Proof. If a ∈ [n] then consider −1 σ(ai+m1) if a = ai σ(a1 ··· am)σ (σ(a)) = σ(a) if a 6∈ {a1, . . . , am}. Since every b ∈ [n] is uniquely σ(a) for some a ∈ [n] we deduce that

−1 σ(a1 ··· am)σ (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 only if 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

Qk Proof. The lemma tells us that if π = i=1(ai1ai2 ··· aim(i)) is a decomposition of π into a product of disjoint cycles then m −1 Y −1 σπσ = (σ(ai1ai2 ··· aim(i))σ ) i=1 m Y = (σ(ai1)σ(ai2) ··· σ(aim(i)). i=1 Since the right-hand-side is a product of disjoint cycles we see that anything conjugate to π must be a product of the same number of disjoint cycles of the same length. Qk Suppose now that τ = i=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 83 {a01, . . . , a0m(0)} = [n]\{aij | 1 6 i 6 k, 1 6 j 6 m(i)} and 84 {b01, . . . b0m(0)} = [n]\{bij | 1 6 i 6 k, 1 6 j 6 m(i)}. Then we deﬁne σ(aij) = bij for 0 6 i 6 k and 1 6 j 6 m(i). Thus σ ∈ Sn and −1 τ = σπσ by the lemma.

Example. Conjugacy classes in S4

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

Corollary (Conjugacy classes in An). If π ∈ An then its conjugacy class in An is equal 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 two conjugacy classes in An of equal size.

|An| |Sn| |CSn (π)| | OrbAn (π)| = = = | OrbSn (π)| . |CAn (π)| 2|(CAn (π))| 2|CSn (π) ∩ An| 1 Thus if CSn (π) ⊂ An then we see that | OrbAn (π)| = 2 | OrbSn (π)|. Otherwise consider | : C (π) → {±1}. Since (C (π)) 6= {1} we see CSn (π) Sn Sn that Im | = {±1} and CSn (π) C (π)/C (π) = C (π)/ ker | ' {±1}. Sn An Sn CSn(π)

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

83 Pn So i=0 m(i) = n and every element of [n] is equal to precisely one aij . 84 So every element of [n] is also equal to precisely one bij . 50 SIMON WADSLEY

Example (Conjugacy classes in A4). 4 2 The even cycle types in S4 are 1 , 2 and 3.1. Now (12) ∈ CS4 (e) and (12) ∈ 4 CS4 ((12)(34)) so the centralisers of elements of conjugacy classes of cycle type 1 2 and 2 contain elements of odd order. Thus these classes are the same in A4 and S4.

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 of equal size. representative element e (12)(34) (123) (132) cycle type 14 22 3.1 3.1 number of elements in class 1 3 4 4 size 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) and so be a union of conjugacy classes (Example Sheet 3 Q3). But |A4|/2 = 12/2 = 6 and H must contain id. No set of conjugacy classes including {e} has a total of 6 elements.

Lecture 24 6.4. Simple groups. Definition. We say a non-trivial group G is simple if G has no normal subgroups except itself and its trivial subgroup. Since if N is a normal subgroup of G one can view G as ‘built out of’ N and G/N, one way to try to understand all groups is to first understand all simple groups and then how they can fit together.

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

Theorem. A5 is simple. 5 2 2 Proof. The cycle types of even elements of S5 are 1 , 2 .1, 3.1 and 5. These 5 have 1, 5 × 3 = 15, 2 × 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 so these 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)) = h(12345)i ⊂ A5. Thus the class with 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 5 number of elements in class 1 15 20 12 12 size of centraliser 60 4 3 5 5

Now if N is a non-trivial proper normal subgroup of A5 then it is a union of conjugacy 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 is straightforward to check that no union of conjugacy classes including {e} can have these 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). Using our 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 of An is a product of an even number of transpositions. So we must show that every product of two transpositions is also a product of 3-cycles. Suppose a, b, c and d are distinct. Then (ab)(ab) = id, (ab)(bc) = (abc) and (ab)(cd) = (abc)(bcd). Thus An is indeed generated by 3-cycles. Now it suffices to show that every non-trivial normal subgroup N of An contains a 3-cycle — if N contains one it must contain all and so it generates An and so it is An. To this end we pick π ∈ N\{e} with the most fixed points. We show that π is a 3-cycle. If π has two transpositions in its decomposition as a product of disjoint cycles say π swaps a and b and also swaps c and d. Let σ = (cde) for some e 6∈ {a, b, c, d}. Then −1 −1 τ := π (σπσ ) ∈ 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 τ has more fixed points than π which contradicts our definition of π. Now we suppose that hπi 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) −1 −1 and τ = (σπσ )π ∈ N since N is a normal subgroup of An this time we compute τ(π(a)) = π(a), τ(π2(a)) = b 6= π2(a), and if i is fixed by π then i is also fixed by τ. Thus τ 6= id has more fixed points than π. A triumph of late 20th century mathematics was the classification of all finite simple 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 there is no non-abelian simple group of odd order that first appeared as a circa 250 page paper in 1963. The first proof of the whole classification theorem was annouced in the early 1980s. It ran to over ten of thousand pages spread across a large number of journal articles by over 100 authors. It turned out not to be quite complete. In 2004 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 would be 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 using the language of groups. But even when considering only ﬁnite groups there is much more to learn.