Groups. Chapter The symmetric group

Transcription

1 Chapter 1 Groups In this chapter we ll cover pretty much all of group theory. This material is roughly the same as Rotman s chapters 2 and 5, but beware there are some extra things not in Rotman... You should *know* the material in Rotman chapter 2 well already. 1.1 The symmetric group Groups arise naturally as symmetries of things. More formally, for any object A in a category C, the set Aut C (A) of all automorphisms of A (= isomorphisms from A to A) is a group under composition of morphisms. The first example arises in exactly this way: take any set X and let A(X) be the group of all automorphisms of X in the category of sets. This is the symmetric group on X, consisting of all invertible maps from X to itself. In the special case that X = {1,..., n}, we denote X by S n. The order S n of the symmetric group S n is n!. Now let G be any group. Then, A(G) is also a group. Define λ : G A(G), g λ g where λ g A(G) is the function x gx (i.e. λ g is left multiplication by g ). Note λ is a group homomorphism. Moreover, it s injective, for if λ g = id G, then λ g (1 G ) = 1 G = g so ker λ = {1 G }. Hence, λ defines an isomorphism between G and im λ A(G). We ve proved: Cayley s theorem. Every group is isomorphic to a subgroup of the symmetric group A(X) for some set X. Every finite group, is isomorphic to a subgroup of S n for some n. By the way, you can also define ρ : G A(G), g ρ g, where ρ g is the function right multiplication by g. But ρ satisfies ρ(gh) = ρ(h)ρ(g), the wrong way round to be a homorphism. In fact, ρ is an antihomomorphism from G to A(G). I ve already started using the language of subgroups, homomorphisms, kernels etc... Let s review this stuff quickly. Let G, H be groups. Recall that a homomorphism f : G H means a function such that f(g 1 g 2 ) = f(g 1 )f(g 2 ) for all g 1, g 2 G. A homomorphism is called a monomorphism if it is injective, i.e. there exists a set function f : H G such that f f = id G. It s an epimorphism if its surjective, i.e. there exists a set function f : H G such that f f = id H. It s an isomorphism if it s bijective, i.e. both injective and surjective. Given a homomorphism f : G H, we set im f = f(g) = {f(g) g G}, ker f = f 1 ({1 H }) = {g G f(g) = 1 H }, its image and kernel. A subgroup of a group G is a non-empty subset K of G such that k 1 k 1 2 K for every k 1, k 2 K. This condition implies that K is itself a group in its own right, the operation being the restriction of the operation on G to K. We usually write K G to indicate that K is a subgroup of G. For example, im f H (and f is an epimorphism if and only if im f = H); ker f G (and f is a monomorphism if and only if ker f = {1 G }). Actually, ker f is more than just a subgroup of G: it is a normal subgroup. By definition, a subgroup N G is normal if any of the following equivalent conditions hold: 11

2 12 CHAPTER 1. GROUPS (N1) gng 1 N for all g G, n N (the element gng 1 here is the conjugate of n by g, often denoted g n); (N2) gng 1 = {gng 1 n N} = N for all g G (the subgroup gng 1 is the conjugate of N by g, often denoted g N); (N3) Ng = gn for all g G (the subsets Ng = {ng n N} are the right cosets of N in G; the subsets gn = {gn n N} are the left cosets). We usually write N G to indicate that N is a normal subgroup of G. The sets Ng and gn right cosets and left cosets are both equivalence classes for certain equivalence relations on G, hence they partition G into disjoint subsets. In case G is a finite group and K G, we let [G : K] = G/K denote the number of left (equivalently, right) cosets of K in G, the index of K in G. Now, all the left cosets have the same size, namely, K. Since equivalence classes partition G into disjoint subsets, we deduce: Lagrange s theorem. G = K [G : K]. Lagrange s theorem shows in particular that the order of any subgroup of a finite group G divides the order of the group. Now let s return to discussing the finite group S n in more detail. So it s the group of all permutations of {1,..., n}. Let x 1,..., x a be distinct elements of {1,..., n}. Denote the permutation which maps x a x 1, x i x i+1 for each i = 1,..., a 1 and fixes all other points, by (x 1 x 2... x a ). This is called an a-cycle. A 2-cycle is called a transposition. Disjoint cycle notation. Every permutation f S n can be written as a product of disjoint cycles f = (x 1 x 2... x a )(y 1 y 2... y b ).... (z 1 z 2... z c ). Moreover, this representation of the permutation f is unique up to deleting 1-cycles and reordering the product (disjoint cycles commute!). You should be quite familiar with using the disjoint cycle notation. Now define a function sgn : S n {±1} as follows. Take g S n. Write g = c 1... c m as a product of disjoint cycles, where c i is an o i -cycle. Set m sgn(g) = ( 1) oi 1. i=1 So for example, sgn(1) = 1, sgn(t) = 1 for t a transposition,... Note this is well defined, as follows from the uniqueness of the disjoint cycle decomposition. However if we had tried to define sgn in this way for a not necessarily disjoint product of cycles we would have run into trouble Lemma. Every element g S n can be expressed as a product of transpositions. Proof. Using the disjoint cycle decomposition, it suffices to show that any a-cycle can be written as a product of transpositions. Then, for example, the cycle (12... n) equals (12)(23)(34)... (n 1 n) Theorem. sgn is a group homomorphism.

3 1.1. THE SYMMETRIC GROUP 13 Proof. We first prove: Claim. If s is a transposition and w S n is arbitrary, then sgn(sw) = sgn(w). To see this, say s = (a b) and w = c 1... c m as a product of disjoint cycles (including 1-cycles). Assume that a appears in c 1. If b appears in c 1 too, then c 1 = (a x 1... x k b y 1... y l ) and (a b)c 1 = (b y 1... y l )(a x 1... x k ) which is a product of disjoint cycles. In then follows by definition of sgn that sgn((a b)w) = sgn(w). The other possibility is if b appears in another of the cycles, say c 2 (allowing 1-cycles). Now another similar explicit calculation with cycle notation expresses (a b)c 1 c 2 as a product of disjoint cycles, and the conclusion again follows using the definition of sgn. Now we can prove the theorem. We need to show that sgn(xw) = sgn(x) sgn(w) for any x, w S n. Write x = s 1... s m as a product of transpositions, applying Lemma 1.1.1, and proceed by induction on m, the case m = 1 being the claim. For m > 1, set y = s 1 x = s 2... s m. Then, by the claim, sgn(x) = sgn(y); by the induction hypothesis, sgn(yw) = sgn(y) sgn(w) = sgn(x) sgn(w). Hence, using the claim once more, sgn(xw) = sgn(s 1 (yw)) = sgn(yw) = sgn(x) sgn(w) Corollary. If w S n is written as a product of m transpositions, then sgn(w) = ( 1) m. We define the alternating group A n to be the kernel of the homomorphism sgn : S n {±1}. Providing n > 1, A n is a normal subgroup of S n of index 2. The elements of A n are called even permutations. A group G is called simple if it has no proper normal subgroups. For instance, the cyclic group C p where p is a prime is a simple group for Lagrange s theorem implies that C p has no proper subgroups at all. The goal in the remainder of the section is to prove that A n is simple for n 5. (On the other hand, A 4 is not simple, for it contains the Klein 4 group V 4 = {1, (12)(34), (13)(24), (14)(23)} as a normal subgroup of index 3.) As a first step to the goal, we need to understand the conjugacy classes of the group S n, that is, the sets of the form G x := { g x g G} for x G. The conjugacy classes of G partition it into disjoint subsets (because conjugacy classes are the equivalence classes of the equivalence relation is conjugate to ). Observe that a subgroup N G is a normal subgroup if and only if it is a union of conjugacy classes of G: so once we understand conjugacy classes we can quite easily test subgroups for normality Lemma. Take g, x S n and write as a product of disjoint cycles. Then, x = (a 1... a s )(b 1... b t )... gxg 1 = (ga 1... ga s )(gb 1... gb s ).... Proof. Calculate what each side does to an integer i. Define the cycle-type of x S n to be the ordered tuple (o 1, o 2,..., o l ) consisting of the orders of the disjoint cycles of x in cycle notation, arranged so that o 1 o 2 o l > 0. You should include all the trivial 1-cycles so that in addition l i=1 o i = n. For instance (1 2) S 5 has cycle-type (2, 1, 1, 1). The lemma immediately implies the description of the conjugacy classes of S n : Theorem. The conjugacy classes of S n are precisely the {x S n x has cycle-type λ} as λ runs over all conceivable cycle-types also known as the partitions of n.

4 14 CHAPTER 1. GROUPS Now we can prove that A n is simple for n 5. We proceed with a series of lemmas. The first lemma proves the result for A 5 itself; there are quicker ways to do this but this approach illustrates the usefulness of thinking about conjugacy classes Lemma. The group A 5 is simple. Proof. Let us list the conjugacy classes in S 5. Cycle-type Size sgn Splits in A 5? (5) 24 + yes (4, 1) 30 (3, 2) 20 (3, 1, 1) 20 + no (2, 2, 1) 15 + no (2, 1, 1, 1) 10 (1, 1, 1, 1, 1) 1 + no Now an S n -conjugacy class that is contained in A n can either be equal to a single A n -conjugacy class, or else it can be a union of two A n -conjugacy classes of the same size. (Proof?) We have listed in the table which of the S 5 -classes split as A 5 -classes in this way. Therefore, the conjugacy classes of A 5 have sizes 12, 12, 20, 15, 1. Now, a normal subgroup of A 5 must be a union of these conjugacy classes, must contain 1 and must be of order dividing 60. There s no way to do this by elementary arithmetic based on the orders of the classes in A 5! Lemma. Let n 5. Then all 3-cycles are conjugate in A n. Proof. Let i, j, k, l, m,... be arbitrary (distinct) numbers from {1, 2,..., n}. We can conjugate (1, 2, 3) to (i, j, k) using the permutation g which maps 1 to i, 2 to j, and 3 to k. If g happens to be odd, then use (l, m)g instead Lemma. Let n 3. Then A n is generated by the 3-cycles. Proof. Any element of A n is a product of an even number of transpositions. Consider a product of two transpositions (i, j)(k, l). If all numbers i, j, k, l are distinct, then (i, j)(k, l) = (i, j)(j, k)(j, k)(k, l) = (i, j, k)(j, k, l). Otherwise the product looks like (i, j)(j, k), which equals (i, j, k). The lemma follows Lemma. If H A n for n 5 and H contains a 3-cycle, then H = A n. Proof. We ve just seen that A n is generated by 3-cycles and all 3-cycles are conjugate in A n Lemma. The group A 6 is simple. Proof. Let {1} H A 6. Suppose 1 g H has a fixed point. Without loss of generality, suppose g6 = 6. Then, g lies in the naturally embedded subgroup A 5 < A 6. So, {1} H A 5 A 5, so H A 5 = A 5 by simplicity of A 5. Hence, H contains a 3-cycle, so H = A 6 by the previous lemma. This reduces to the case that all elements 1 g H move all 6 points. Now consider the two possible cycle types one by one and get a contradiction. For instance, if (12)(3456) H, then its

5 1.2. ISOMORPHISM THEOREMS 15 square is a non-identity element with a fixed point. Otherwise, (123)(456) H (or an element of the same cycle-type). In this case, conjugating gives that But this is a non-identity element that fixes Theorem. For n 5, A n is simple. (123)(456)(234)(123)(456)(243) H. Proof. We may assume that n 7. Take {1} = H A n. Using induction on n, the argument in the proof of the previous lemma reduces to the case when all 1 g H move all of 1,..., n. Take such a g such that, without loss of generality, g1 = 2. Consider (234)g(243)g 1 H. Its a non-trivial element since it sends 2 to 3. But its a product of two 3 cycles, so it can move at most 6 points, so as n 7 it must have a fixed point Remark. A permutation of the set {1, 2,... } is called finitary if it fixes all but finitely many points. Denote by A the finitary alternating group, i.e. the group of all even finitary permutations. Using the fact that A = n 1 A n, it is easy to see that A is simple, giving us an example of an infinite simple group. 1.2 Isomorphism theorems Let N be a normal subgroup of G. Then, (N3) says that G/N = N\G, i.e. left cosets are the same as right cosets so we can just call them simply cosets. Define a multiplication on the set of cosets G/N by (g 1 N)(g 2 N) = (g 1 g 2 )N. The fact that this is well-defined depends on N being a normal subgroup. This multiplication gives G/N the structure of a group in its own right, called the quotient group of G by the normal subgroup N. Note that there is a canonical homomorphism π : G G/N, g gn, which is an epimorphism of G onto G/N with kernel exactly N. The group G/N together with the map π : G G/N has the following universal property: Universal property of quotients. Let N G and π : G G/N be the canonical epimorphism. Given any homomorphism f : G H with N ker f, there exists a unique homomorphism f : G/N H such that f = f π. Now let me state the isomorphism theorems for groups. These are all really consequences of the universal property of quotients: you should be able to prove them for yourselves. First isomorphism theorem. Let f : G H and N = ker f. Then, N G and f factors through the quotient G/N to induce an isomorphism f : G/N im f. Second isomorphism theorem. Let K G, N G. Then, KN = {kn k K, n N} is a subgroup of G, N KN, K N K and K/K N = KN/N. Third isomorphism theorem. Let K N G with K G, N G. Then, N/K G/K and G/N = (G/K)/(N/K). There is one other important result traditionally included with the isomorphism theorems: the lattice isomorphism theorem. First, recall that a relation on a set X is called a partial ordering (and X is called a partially ordered set) if for all x, y, z X

6 16 CHAPTER 1. GROUPS (R1) x x; (R2) x y and y x implies y = x; (R3) x y and y z implies x z. If in addition we have that for all x, y X one of x < y, x = y, x > y holds, then < is called a total order or a linear order (and X is called a totally ordered set). Now let X be a partially ordered set and A X. An element a X is called the least upper bound of A if a is the unique minimal element of {x X a x a A}. The least upper bound of A may or may not exist: for instance there could be many such minimal elements or no minimal element at all. Similarly, an element a X is called the greatest lower bound of A if a is the unique maximal element of {x X a x a A}. A partially ordered set is called a lattice if every pair of elements of X has both a least upper bound and a greatest lower bound in X. A partially ordered set is called a complete lattice if every non-empty subset of X has both a least uppoer bound and a greatest lower bound in X. Let G be any group and X be the set of all subgroups of G, partially ordered by inclusion. Then, X is a complete lattice. The greatest lower bound of a set of subgroups of G is simply their intersection, also a subgroup of G. The least upper bound of a set of subgroups of G is the subgroup generated by the subgroups, that is, the intersection of all subgroups of G that contain all the given subgroups. More generally, given any subset A of G, we write A for the subgroup of G generated by A, namely, the intersection of all subgroups of G that contain A. In the special case A = {a} for some a G, we write simply a for the subgroup of G generated by the element a. As a variation, let G be any group and X be the set of all normal subgroups of G, partially ordered by inclusion. One checks that the intersection of a family normal subgroups is normal, and that the group generated by a family of normal subgroups is normal. Hence, X in this case is again a complete lattice. Lattice isomorphism theorem. Let f : G G be an epimorphism with kernel K. Then, the map H f(h) gives an isomorphism between the lattice of subgroups (resp. normal subgroups) of G containing K and the lattice of subgroups (resp. normal subgroups) of G. (The inverse map sends a subgroup H of G to its preimage f 1 (H ) in G.) Continuing the theme of subgroups generated by things, let G be any group and g G an element. Then, g denotes the subgroup of G generated by the element g. So, g = {1 G, g ±1, g ±2, g ±3,... }, and is called the cyclic subgroup generated by g. It may or may not be a finite group; if it is finite, its order is denoted g, the order of the element g. Note Lagrange s theorem implies that in a finite group, the order of every element divides the order of the group. A group G is called cyclic if G = g for some element g G. We then say that g is a generator of the cyclic group G. For example, the group (Z, +) of integers under addition is an infinite cyclic group generated by the integer 1 (the identity element of Z is the integer 0 of course we use the additive notation in this Abelian group!). Actually, any infinite cyclic group is isomorphic to the group (Z, +). We can classify all subgroups of the infinite cyclic group: Lemma. If H (Z, +), then H = n for some integer n 0. Proof. Take n > 0 minimal subject to the condition that n H (if no such n exists, then H = {0} and there is nothing to prove). Then, n H. Now take any m H, and write m = qn + r for integers q, r with 0 r < n. Then, r = m qn H, so by minimality of n we have that r = 0. Hence, m = qn, so that m n. This shows that H = n. Note that n m if and only if m n. It follows that the lattice of subgroups of Z is isomorphic to the opposite of the lattice of non-negative integers partially ordered by divides. The non-negative integer n corresponds under the isomorphism to the subgroup n. Given n N, we define the group (Z n, +), the group of integers modulo n under addition, to be the quotient group Z/ n. Thus, the elements of the group Z n are the cosets of n in Z, namely,

7 1.2. ISOMORPHISM THEOREMS 17 {[0], [1],..., [n 1]} where [i] = {i + jn j Z}. We have that [i] + [j] = [i + j]. The group Z n is the cyclic group of order n: it is generated by the element [1] which has order n in Z n. Sometimes we denote the cyclic group of order n instead by C n, to indicate that we re writing the operation multiplicatively instead. So C n = {1, x, x 2,..., x n 1 } where x is any element of C n of order n. By the lattice isomorphism theorem, the lattice of subgroups of Z n is isomorphic to the opposite of the lattice of divisors of n. The divisor d of n corresponds under the isomorphism to the cyclic subgroup of Z n generated by [d], which is a subgroup of order n/d. In other words: Lemma. The group Z n has a unique subgroup of order d for each divisor d of n, namely, the cyclic subgroup generated by [n/d]. Given groups G 1, G 2, their (external) direct product is the set G 1 G 2 (Cartesian product) with coordinatewise multiplication. Actually, G 1 G 2, with the obvious projections π i : G 1 G 2 G i is the product of G 1 and G 2 in the categorical sense. More generally given a family G i (i I) of groups, their product i I G i is simply their Cartesian product as sets with coordinatewise multiplication. In other words, the category of groups possesses arbitrary products (see (0.3.2)) Lemma. If n = st with (s, t) = 1, then Z n = Zs Z t. Proof. The element ([1] s, [1] t ) Z s Z t has order st as (s, t) = 1. Hence it generates a cyclic subgroup of Z s Z t of order st. Hence, it generates all of Z s Z t, which is therefore isomorphic to the cyclic group of order n. The Euler φ function is defined by φ(n) = #{x Z n x = n} = #{1 k < n (k, n) = 1}. By the lemma, if n = st with (s, t) = 1, then φ(n) = φ(s)φ(t). It follows immediately that to compute φ(n) it suffices to know φ(p n ) for each prime power p n. In this special case, it is an exercise to show that ( φ(p n ) = p n 1 1 ). p Here s an important number theoretic fact about the Euler φ function: Lemma. For n N, n = d n φ(d). Proof. We have from Lagrange s theorem that n = Z n = d n (the number of elements of Z n of order d). Now if x Z n has order d, it generates a subgroup of order d, isomorphic to Z d. By Lemma 1.2.2, Z n has a unique such subgroup of order d. Hence, the number of elements of Z n of order d equals the number of elements of Z d of order d, namely φ(d). The other basic example of a finite group you should be familiar with at this point is the dihedral group D n of order 2n. I leave this to the exercises... Note many people denote it by D 2n instead so you need to watch for this. For us, D 4 is the group of symmetries of the square of order 8.

8 18 CHAPTER 1. GROUPS 1.3 Group actions We say a group G acts (on the left) on a set X if there is a given map G X X, (g, x) gx such that (A1) 1x = x for all x X; (A2) (gh)x = g(hx) for all x X, g, h G. We call X a (left) G-set if G acts on X. Equivalently, X is a G-set if the map α : G A(X), g α g, where α g (x) = gx for x X, is a group homomorphism. Then, the G-set X is called faithful if this map α is injective. So the action is faithful if and only if gx = x for all x X implies g = 1. The action of G on X is called transitive if for every x, y X, there exists at least one g G such that gx = y. More generally, let X be a G-set and define an equivalence relation on X by x y if there exists a g G such that gx = y. The equivalence classes are called the orbits of G on X; the orbit of x X is denoted Gx = {gx g G}. So the action is transitive if and only if there exists just one orbit. Elements x X with gx = x for all g G (i.e. the orbit of x is just {x}) are called fixed points, and we write F ix(x) for the set of all fixed points. Given x X, its stabilizer G x, or the isotropy group of x, is defined to be G x = {g G gx = x}. Note G x is a subgroup of G. So, x is a fixed point if and only if G x = G. Given x X, the resulting orbit map is the map f : G X, g gx. Clearly, f(g) = f(g ) if and only if g 1 g G x, i.e. gg x = g G x, i.e. g and g lie in the same left G x -coset. Hence, the orbit map induces a well-defined map G/G x Gx, gg x gx. This map is clearly surjective, and we checked above that it is injective. In other words, the orbit map induces a bijection between the cosets G/G x of G x in G and the orbit Gx of x. In particular, if Gx is finite, we get that Gx = [G : G x ]. So if G itself is finite, the size of any orbit of G divides the order of G Lemma. If x and y lie in the same orbit, then G x and G y are conjugate in G, i.e. there exists g G such that gg x g 1 = G y. Proof. Pick g G such that gx = y, i.e. x = g 1 y. Then, for h G x, we have that ghg 1 y = ghx = gx = y, so gg x g 1 G y. Similarly, g 1 G y g G x. Hence, G y gg x g 1. Now we have the language, we are ready for examples Given any G-set X, a subset Y X is called G-stable if gy Y for each g G, y Y. So, Y is G-stable if and only if it is a union of orbits. In that case, Y is itself a G-set in its own right via the restriction of the action of G on X to Y If H < G is any subgroup, then H acts on G by left multiplication, i,e, (h, g) hg. The orbits are the right cosets H\G If H < G is any subgroup, then H acts on G by (h, g) gh 1. The orbits are the left cosets G/H If H, K < G, then H K acts on G by ((h, k), g) hgk 1. The orbits are the double cosets K\G/H, subsets of the form HgK = {hgk h H, k K}.

9 1.3. GROUP ACTIONS G acts on itself by conjugation, (g, x) gxg 1. The orbits are just the conjugacy classes of G. In this case, given x G, the stabilizer G x = {g G gxg 1 = x} is called the centralizer of x in G. We often write C G (x) for the centralizer of x in G. The order of the conjugacy class G x is the index of C G (x) in G: G x = [G : C G (x)]. The fixed points for this action are precisely the elements in the center Z(G). An element x G lies in Z(G) if and only if C G (x) = G More generally, G acts on the set of subsets of G by conjugation. So for a subset S G, the action is by (g, S) gsg 1 = {gsg 1 s S}. The stablizer G S of S under this action is usually called the normalizer of S in G, denoted N G (S). So, N G (S) = {g G gsg 1 = S}. In particular, if H is a subgroup of G, we have its normalizer N G (H), and H N G (H). In particular, H G if and only if G = N G (H). For an arbitrary subgroup H G, the number of distinct conjugates of H in G is given by the formula is equal to [G : N G (H)] The symmetric group S n acts faithfully on the numbers {1,..., n} by the very definition of S n! This action is transitive so that there is a single orbit, and each point stabilizer is isomorphic to the symmetric group S n The dihedral group D 4 of symmetries of the square acts naturally on the set of vertices (labelled {1, 2, 3, 4} say) of the square. This yields a homomorphism α : D 4 S 4 which is in fact injective, hence D 4 embeds as a subgroup of S 4. In other words, the action of D 4 on the vertices of the square is faithful. Given a finite G-set X, we have the class equation X = F ix(x) + x [G : G x ] where the sum is over a set of representatives x of the non-trivial orbits. Applying this in particular to the conjugation action of G on itself, we obtain G = Z(G) + x [G : G x ] summing over a set of representatives for the conjugacy classes of non-central elements. Let p be a prime. We call a finite group G a p-group if the order of G has order a power of p. (More generally, a possibly infinite group G is called a p-group if every element of G has order a power of p). Now if X is a G-set and x X is not a fixed point, then G x is a proper subgroup of G, so [G : G x ] is a divisor of G strictly greater than 1. So if G is a finite p-group, [G : G x ] is a proper power of p, so [G : G x ] 0 (mod p). Taking both sides of the class equation above modulo p, we deduce: Lemma. If G is a finite p-group and X is a finite G-set, then F ix(x) X (mod p). Applying this to the conjugation action of G on itself, this shows in particular: Corollary. If G is a finite p-group, then Z(G) {1}. Proof. By the lemma, Z(G) G 0 (mod p). But 1 Z(G), so Z(G) 1. But 1 0 (mod p) so we cannot have Z(G) = 1. You should be able to prove now that if G = p 2 for a prime p, then G is Abelian.

10 20 CHAPTER 1. GROUPS Lemma. Let G be a finite Abelian group and p be a prime dividing the order of G. Then, there exists x G with x = p. Proof. Say G = pm. If m = 1, then G = C p and the result is obvious. If m > 1, take 1 x G. If x = pk for some k, then x k has order p and we re done. So we may assume that p x. Then, p G/ x, so by induction there exists an element y G such that y / x but y p x. Then, y ph = 1 for some h coprime to p (indeed, h divides the order of x). So the order of y h divides p, so either equals p, as required, or 1. But in the latter case, we have that y h = 1. Write 1 = ah+bp for integers a, b. Then, y = y ah+bp = y bp which lies in x as y p x. Hence, y x, a contradiction. Cauchy s theorem. If G is a finite group and p is a prime dividing the order of G, then there exists an element x G with x = p. Proof. Let G = pm. If m = 1, the conclusion is obvious. Otherwise, we have the class equation: G = Z(G) + x [G : C G (x)]. For each x in the sum, C G (x) < G, so if p C G (x) for any such x we get the conclusion by induction. Hence, p C G (x) for any such x, in other words, p divides each [G : C G (x)]. Since p also divides G, we deduce that p divides Z(G). Then Z(G) contains an element of order p by the lemma. Finally we have the Orbit counting lemma. If X is a finite G-set then #{Orbits} = 1 G X g where X g = {x X gx = x}, the set of fixed points of g. Thus, the number of orbits is the average number of fixed points of elements of G. Proof. Let S = {(g, x) g G, x X, gx = x}. We count S in two different ways. One way, S = g G Xg. On the other hand, We re done. S = O an orbit = O an orbit x O g G {(g, x) g G, x O, gx = x} G x = O an orbit O G x = O an orbit G. 1.4 Finite subgroups of SO(3) I want next to mention some important examples of infinite groups. Suppose V is an n-dimensional vector space over a field K. Then, GL(V ) denotes the group of all invertible linear maps from V to V. Picking a basis for V and representing elements of GL(V ) as n n matrices with respect to this basis we get that GL(V ) is isomorphic to the group GL n (K) of n n invertible matrices with entries in K. The group GL n (K) has many interesting subgroups. For instance, we have SL n (K), the normal subgoup consisting of all n n matrices of determinant 1. Note det : GL n (K) K is a group homomorphism with kernel SL n (K), so GL n (K)/SL n (K) = K. As long as the field K is infinite, these are all infinite groups...

11 1.4. FINITE SUBGROUPS OF SO(3) 21 Another very common situation is if the vector space V is equipped with a non-degenerate symmetric or skew-symmetric bilinear form (.,.). Then we have the subgroup Cl(V ) = {g GL(V ) (gv, gw) = (v, w) for all v, w V } of GL(V ) consisting of all linear transformations which preserve the given bilinear form. My notation Cl here is short for classical groups of this form are often referred to as the classical groups. Don t know why. You get the orthogonal groups if the form is symmetric, the symplectic groups if the form is skew-symmetric... Let me just mention the most classical of all the classical groups! This arises if the field is R and you take the standard inner product on R n, then you get the group O(n). In terms of matrices with respect to the standard basis, O(n) is the group of all n n matrices with A T A = I n. Note this implies that det A = ±1. The special orthogonal group SO(n) is the group of all n n complex matrices with A T A = I n and det A = +1. So SO(n) O(n) of index 2. (Another very important classical group is the unitary group U(n): the group of all n n matrices over C with A T Ā = I n where Ā is the matrix obtained by taking complex conjugate of all the entries of A. This is the classical group corresponding to the standard Hermitian form on C n.) I ll say a bit more about these things in class and in the exercises... Let s just focus our attention here on SO(2) and SO(3). The first is a very easy group: it is easy to see (just consider matrices) that it is the group of all rotations of R 2. So it acts on the unit circle S 1 in R 2. Pick any base point x S 1 and consider the orbit map SO(2) S 1, g gx. This is a bijection. So you can just identify SO(2) with S 1 then it inherits a topology from the usual topology on S 1. This makes SO(2) into a compact Lie group... Consider instead SO(3), which is rather more interesting from an algebraic point of view. Again it acts on the unit sphere S 2 in R 3. Pick any base point x S 2 and consider the orbit map SO(3) S 2, g gx. The stabilizer of x also stabilizes x so you see that the stabilizer of x is SO(2) so just consists of rotations in the plane orthogonal to x. Thus the orbit map induces a bijection SO(3)/SO(2) S 2... Indeed the sphere S 2 is what is often called a homogeneous space: the quotient of a Lie group by a closed subgroup... Let s classify the elements of SO(3). Take any g SO(3). It is a 3 3 matrix. Consider its eigenvalues. They are roots of a cubic with real coefficients. Let λ C be an eigenvalue. Then, writing for complex conjugation, there is 0 x C 3 such that x T x = x T ḡ T gx = λλ x T x. Since x T x is not zero, this shows that λλ = 1. So if λ is real it is ±1. Now there are either three real eignevalues, all ±1, multiplying to 1 or one real eigenvalue and a complex conjugate pair of complex eigenvalues. Either way there must be an eigenvalue equal to 1. So there is a fixed axis v 1. Then extend to an orthonormal basis v 1, v 2, v 3 you get that the matrix of g looks like cos θ sin θ 0 sin θ cos θ because the action of g on v1 lies in SO(2) which is just rotations. So: elements of SO(3) are just rotations about some axis. (Elements of O(3) itself are either rotations or the composite of a rotation and a reflection in the plane orthogonal to the axis of rotation.) Let s focus now on one simple algebraic question: the problem of classifying the finite subgroups of SO(3). This is an excuse to think about some nice finite group actions arising in nature. First some examples. (1) Symmetries of a Tetrahedron. By symmetry here I mean bijective isometry (distance preserving map). Let G be the group of symmetries of a regular tetrahedron T drawn so its vertices lie on the unit sphere. Then G acts on the set {1, 2, 3, 4} of vertices. Hence we have a homomorphism G S 4. Note this map is injective, i.e. the action of G on the vertices is faithful. Because if g fixes all four vertices, then it fixes a basis v 1, v 2, v 3 for R 3 so since it is an isometry for any vector v R 3 we have that (gv, gv i ) = (v, v i ) = (gv, v i ) for each i = 1,..., 3. So gv = v and g is the identity. Now let s compute G : can send pick any of four vertices for first corner, then any of three for its neighbour and then finally two choices

12 22 CHAPTER 1. GROUPS fix the third neighbour which determines everything. So there are 24 symmetries. Hence G S 4. The subgroup G 0 = G SO(3) is then just the rotations, 12 of these, so G 0 = A4. This gives our first finite subgroup of SO(3), A 4. (2) Symmetries of a Cube. Let G be the group of symmetries of a regular cube drawn so its vertices lie on the unit sphere. Let G 0 = G SO(3). Note G = = 48 and G 0 = 24. I ll show G 0 = S4 and G = S 4 C 2. Let X be the set of 4 pairs of diagonally opposite vertices {v, v}. Note G acts on X, giving a homomorphism G X. Suppose g lies in the kernel of this homomorphism. For each pair of vertices {v, v}, we have that g{v, v} = {v, v} so either gv = v or gv = v. If gv = v for some v then g is a rotation about that axis and you see that it must be the identity already. So for some vertex v we have that gv = v, so g = 1, a reflection. Thus the kernel of our map is {1, 1}. In particular it is an isomorphism from G 0 to S 4 by orders, so G 0 = S4. Finally consider G: let K = {1, 1}. Then G = G 0 K since elements of G 0 and K commute. This gives our second finite subgroup of SO(3), S 4. (3) Symmetries of a Dodecahedron. This is a bit harder, but you show that G 0 = A5, G = A 5 C 2. So we ve got a third finite subgroup of SO(3), A 5. You probably remember the platonic solids: Tetrahedron; Four triangles. Cube; Six squares. Octahedron; Six triangles. Dodecahedron; Twelve pentagons. Icosahedron; Twenty triangles. Note cube and octahedron are dual, dodecahedron and icosahedron are dual, so these give rise to just three finite subgroups of SO(3). Now let s attempt the general classification. Let G be a finite subgroup of SO(3) of order n > 1. Let X = {unit vectors x R 3 gx = x for some 1 g G}. Thus X is the set of poles left fixed by some non-identity element of G. Each 1 g G is a rotation about some axis so it has exactly two poles. So X 2n 2. Note G acts on X (if x is a pole of g then hx is a pole of hgh 1 ). Let s count the number of orbits of G on X by the orbit counting lemma: N := #{Orbits} = 1 ( X + 2(n 1)). n (There are X points fixed by 1, all other elements of G fix 2 points.) Hence 1 n (2 + (n 1)2) N 1 (2n 2 + (n 1)2) n that is 2 N < 4. Suppose first that N = 2. Then 2n = X + 2n 2 so X = 2. Hence G consists just of rotations about a single axis. So G = C n is cyclic with generator a rotation os smallest angle. Now suppose that N = 3. Let the three orbits have orders n/n i, i = 1, 2, 3, so that n 1 n 2 n 3. Note also that for an element in an orbit with n/n i elements, its stabilizer has order n i so this stabilizer must be the cyclic group consisting of rotations about this axis of order n i. We have that 3n = n + n + n + 2(n 1). n 1 n 2 n 3 Hence = n 1 n 2 n 3 n. From this you get the following possibilities:

13 1.5. THE SYLOW THEOREMS 23 (1) (n 1, n 2, n 3 ) = (2, 2, r). In this case n = 2r and the orbit sizes are (r, r, 2). (2) (n 1, n 2, n 3 ) = (2, 3, 3). In this case n = 12 and the orbit sizes are (6, 4, 4). (3) (n 1, n 2, n 3 ) = (2, 3, 4). In this case n = 24 and the orbit sizes are (12, 8, 6). (4) (n 1, n 2, n 3 ) = (2, 3, 5). In this case n = 60 and the orbit sizes are (30, 20, 12). Consider (1). The orbit of size 2 must have rotation of 360/r about this axis in its stablizer. The orbits of size r have rotation of 180 in their stablizers, must be about axis orthogonal to the previous axis. You get that the orbits are {x, x} and then the coners and midpoints of edges of a regular r-gon drawn around the circumference of the sphere around this axis. (Picture). This must therefore be the dihedral group D r of symmetries of a regular r-gon... The rotations are clear, th reflections of the polygon are realized as rotations using the third dimension. Consider (2). For an orbit of size 4 there is a rotation of 120 degrees about each pole in the orbit. The other three poles in the orbit must therefore lie in an equilateral triangle... You get that the group is the symmetry group of the tetrahedron. The orbits are: the 4 midpoints of the faces, the 4 vertices, the 6 midpoints of the edges. Similarly you can see (3) must be the symmetry group of the cube. The orbits are the 12 midpoints of the edges, the 8 vertices and the 6 midpoints of the faces. Finally (4) is the dodecahedron. Classification of finite subgroups of SO(3).. There are 2 infinite families of finite subgroups of SO(3): cyclic groups and dihedral groups. Then three more exceptional conjugacy classes of finite subgroups: A 4, S 4 and A The Sylow theorems Now suppose that G is a finite group of order p n k for a prime p with (p, k) = 1. A subgroup H G with H = p n is called a Sylow p-subgroup. Equivalently, H is a Sylow p-subgroup of G if H is a p-group and p [G : H] Lemma. If H is a p-subgroup of G, then [N G (H) : H] [G : H] (mod p). In particular, if p [G : H] (i.e. H is not a Sylow p-subgroup of G) then N G (H) H. Proof. Let H act on G/H by left multiplication. Then, ah F ix(g/h) if and only if HaH = ah which is if and only if a 1 Ha = H which is if and only if a N G (H). Hence, [N G (H) : H] = F ix(g/h). Now apply Lemma First Sylow theorem. Let H < G be a p-subgroup of G that is not a Sylow p-subgroup. Then, there exists a p-subgroup K G with H K and K = p H. In particular, H can be embedded into a Sylow p-subgroup of G. Proof. Since p [G : H], we have by the preceeding lemma that H N G (H), H N G (H) and N G (H)/H is a group of order divisible by p. Pick any x N G (H)/H of order p. Then, the pre-image in N G (H) of x is a subgroup of order p H which contains H as a normal subgroup. Second Sylow theorem. All Sylow p-subgroups of G are conjugate. Proof. Let H, K be two Sylow p-subgroups of G. Let H act on G/K by left multiplication. Then, ak F ix(g/k) if and only if HaK = ak which is if and only if a 1 Ha K, i.e. a 1 Ha = K since they have the same order. So to prove that H and K are conjugate in G, it suffices to show that F ix(g/k) 0. But F ix(g/k) G/K 0 (mod p) by Lemma

14 24 CHAPTER 1. GROUPS Third Sylow theorem. Let m be the number of distinct Sylow p-subgroups of G. Then, m 1 (mod p) and m G. Proof. Let H be a Sylow p-subgroup (exists by the first Sylow theorem). Applying the second Sylow theorem, m is the number of conjugates of H, which is [G : N G (H)]. This gives that m G. Now let X be the set of all Sylow p-subgroups of G and let H act on X by conjugation. Then, F ix(x) m (mod p). It therefore suffices to show that F ix(x) = 1. Now, K F ix(x) if and only if hkh 1 = K for all h H, i.e. H N G (K). But H N G (K) if and only if H is a Sylow p-subgroup of N G (K). By the second Sylow theorem, K is the unique Sylow p-subgroup of N G (K), so this is if and only if H = K. Hence, F ix(x) = {H}. We give some examples to illustrate applying Sylow theorems Let G = pq with p > q prime and G not Abelian. Then, q p 1 and G = a, b a q = b p = 1, aba 1 = b r for some 1 < r < p with r q 1 (mod p). Proof. Let K = b be a Sylow p-subgroup of G. We have that n p, the number of Sylow p- subgroups, divides pq and is congruent to 1 mod p. Since p > q, this means that n p = 1. Hence, K must be normal in G. Now let H = a be a Sylow q-subgroup of G. Since G is not Abelian, we cannot have that H G. (If it was then for any h H, k K we would have that hkh 1 k 1 H K = {1}. Hence hk = kh for all h H, k K which implies that G = H K is Abelian.) One deduces that the number of Sylow q-subgroups of G must be p, and that q p 1 by the third Sylow theorem. Now, since K G, aba 1 K so equals b r for some 1 < r < p (cannot have r = 1 since G is not Abelian). Next, a q ba q = b rq = b implies that r q 1 (mod p). We ve now shown that G satisfies the given relations. Hence, there exists an epimorphism from the group with the given relations to G, which is injective since you easily check that any group satisfying the given relations has order at most pq As a special case of the previous example, we get that a group of order 2p for p an odd prime is either isomorphic to C 2p or to D p There is no simple group of order 12. Proof. Let G be a simple group of order 12. Consider n 2, the number of Sylow 2-subgroups. It is either 1 or 3. In the former case, G has a normal Sylow 2-subgroup so is not simple. Hence, n 2 = 3. Now, G acts on the Sylow 2-subgroups by conjugation, giving a non-constant homomorphism G S 3. The kernel is a proper normal subgroup of G since G = 12, S 3 = 6. There are plenty of questions of this nature on past qual exams! 1.6 Semidirect products In (1.5.2), we showed that if G is a non-abelian group of order G = pq with p > q prime, then q p 1 and G = a, b a q = b p = 1, aba 1 = b r for some 1 < r < p with r q 1 (mod p). We stopped short of showing that there actually exists such a group G for all possible choices of p, q, r. However, we showed in the proof that G contained a normal subgroup K = b = C p. The quotient group G/K is isomorphic to C q. So you can think of G as being built by glueing the group C q on top of the group C p. In other words, G is an extension of C q by C p. For the general definition, let H, K be two groups. We call a group G an extension of H by K if G contains a copy of K as a normal subgroup and G/K = H. We can represent this by the following exact sequence : 1 K G H 1

15 1.6. SEMIDIRECT PRODUCTS 25 (the precise meaning of this will be given later when studying modules.) It is important to change the point of view and ask: given H and K what possible extensions G of H by K can we form? For instance, G = H K is an example of an extension of H by K: the kernel of the canonical projection G H is isomorphic to K. (Equally well, you could call G = H K an extension of K by H.) But this extension H K is not really very interesting! How can we build more complicated extensions? We discuss here one important class of extensions which are not too complicated to understand, known as semidirect products. This is really a generalization of the construction of the direct product there are in general other extensions which are not semidirect products. So now let H and K be given groups and suppose we also have a group homomorphism θ : H Aut K, h θ h (where Aut K is the group of all automorphisms of the group K). Define G = K θ H, the external semidirect product, to be the group equal to the Cartesian product K H as a set, with multiplication defined by (k, h)(k, h ) = (kθ h (k ), hh ). You need to check that this really is a multiplication making K H into a group! For instance, (k, h) 1 = (θ h 1(k 1 ), h 1 ). Now, (k, 1)(k, 1) = (kk, 1), so the set K = {(k, 1) k K} is a subgroup of G isomorphic to K. The map G H determined by projection onto the second coordinate is a surjective group homomorphism, with kernel K. This verifies that G has a copy of K as a normal subgroup and the quotient group is isomorphic to H. Hence: the semidirect product G is an extension of H by K. Observe moreover that: (1, h)(k, 1)(1, h 1 ) = (θ h (k), h)(1, h 1 ) = (θ h (k), 1). This shows that the homomorphism θ : H Aut K can be recovered from the semidirect product: θ h is precisely the homomorphism determined by conjugating by the element (1, h) of G. Actually, K θ H is a very special sort of extension: the map π : G H determined by projection onto the second coordinate is split. This means that there is a group homomorphism σ : H G called a splitting such that π σ = id H, namely, the map h (1, h). In fact an extension of H by K is a semidirect product if and only if such a splitting map exists; more general extensions need not have a splitting map (i.e. need not be semidirect products). This last observation is the key to recognizing whether or not a given group G is isomorphic to a semidirect product of groups H and K. Indeed, given a group G and subgroups H and K, we say that G is the internal semidirect product of H and K if (1) K G; (2) H K = {1}; (3) G = KH = {kh k K, h H}. If G is the internal semidirect product of H and K, we define a map θ : H Aut K by setting θ h (k) = hkh 1 for all k K. Then, G is isomorphic to the external semidirect product K θ H. Conversely, any external semidirect product K θ H is the internal semidirect product of {1} H and K {1}. Take the very special case of external semidirect product where θ h = id K for each h H. Then, K θ H is exactly the same as the definition of the direct product K H: so semidirect products are a generalization of direct products. The previous paragraph gives in this special case that a group G is isomorphic to the direct product of subgroups H, K G if H G, K G, H K = {1} and G = HK. We return to the example (1.5.2). So we have primes p > q with q p 1 and 1 < r < p with r q 1 (mod p). We wish to prove that there exists a non-abelian group G of order pq. Define θ 1 : Z p Z p, k rk ( multiplication by r ). It is obvious that this is an automorphism of Z p. Since Z is the free group on generator 1 we get a well-defined homomorphism θ : Z Aut Z p such that 1 θ 1 ; then n (θ 1 ) n which is the map k r n k. Hence θ q is the identity. So this

16 26 CHAPTER 1. GROUPS homomorphism factors through the quotient Z p of Z to induce a well-defined group homomorphism θ : Z q Aut Z p. Finally, define G = Z p θ Z q. It s an extension of Z p by Z q : it is exactly a non-abelian group satisfying the conditions of (1.5.2). Let me give some more examples of extensions that are and are not semidirect products... (1) S n = A n C 2. (2) Let Q 3 be the quaternion group of order 2 3 = 8, Q 3 = {±1, ±i, ±j, ±k} < H. (Here H is the division algebra of quaternions, that is, the R-vector space on basis 1, i, j, k such that i 2 = j 2 = k 2 and ij = k, jk = i, ki = j.) Let x = i and y = j. Then the following relations hold: x 4 = 1, x 2 = y 2, yxy 1 = x 1. Observe that C 4 := x is a normal subgroup of Q 3, and Q 3 /C 4 = C2. That is, Q 3 is an extension of C 4 by C 2. However, Q 3 = C4 C 2. (3) Let V be a finite dimensional vector space over a field F. The affine group of V is defined to be the semidirect product AGL(V ) := V GL(V ), where the action of GL(V ) on V is taken to be natural. Elements of AGL(V ) are called affine transformations of V. (4) Let us describe the group Isom(R n ) if all isometries of the metric space R n, i.e. all normpreserving bijections where the norm x is defined to be (x, x) for the usual inner product (.,.) on R n. Let g be any such isometry. Let v = g(0). Consider g defined by g (x) = g(x) v; so g (0) = 0. Let A be the matrix with ith column equal to g (e i ) where e i is the ith standard basis vector. Since g is an isometry, the columns of A form an orthonormal basis for R n, so A T A = I, so A O(n). Finally let g be defined by g (x) = g (x) Ax. Then g is an isometry fixing 0 and all e i. It follows that g is the identity (because (g (x), e i ) = (x, e i ) for all i). Hence: g(x) = Ax + v. We ve almost proved that Isom(R n ) = R n O(n), the semidirect product of the abelian group R n (translations) by the orthogonal group O(n), acting naturally on R n. You can fill in the details. (5) At this point in class I discussed question 6 from the first midterm: Let G = p(p + 1) where p is prime. Prove that G has either a normal subgroup of order p or a normal subgroup of order (p + 1). The solution to this showed that if G does not have a normal subgroup of order p then either G has a normal subgroup of order p or else p + 1 is a prime power q r and every element of G is of order 1, p or q. Using this result and thinking some more about semidirect products, we can classify the non-abelian groups of order 12. Such a group G has either a normal Sylow 3-subgroup, or else there is no normal Sylow 3-subgroup but then there is definitely a normal V 4. So G is either a semidirect product C 3 C 4, C 3 V 4 or V 4 C 3. Now you analyse the possible θ s needed to define these semidirect products to get the classification of non-abelian groups of order There are exactly 3. Either C 3 V 4 = D3 C 2 = D6, or V 4 C 3 = A4, or C 3 C 4. The last one is a new group of order 12, usually denoted T.

17 1.7. PROJECTIVE UNIMODULAR GROUPS Projective unimodular groups The only simple groups we have seen so far are cyclic groups of prime order and the alternating groups A n for n 5. We will now show that certain groups of matrices are also simple. Let F be an arbitrary field. Consider the group SL n (F ) of all n by n matrices over F with determinant 1. The most important case will be when F = F q is a finite field with q elements (which only happens if q is a prime power). Let e i,j denote the ij-matrix unit, i.e. the matrix with 1 in the ijth position and 0 s elsewhere. Let I n denote the n n identity matrix. A transvection is a matrix of the form t i,j (a) := I n + ae i,j (a F, 1 i j n). Note t i,j (a)t i,j (b) = t i,j (a + b), so {t i,j (a) a F } is a subgroup of SL n (F ) isomorphic to the additive group of the field Lemma. The group SL n (F ) is generated by transvections. Proof. This is essentially equivalent to Gaussian elimination. One just needs to observe that multiplying an n n matrix A on the left by t ij (a) leads to the elementary row transformation which adds to the ith row of A the jth row of A multiplied by a, and multiplying A by t ij (a) on the right leads to a similar elementary column transformation. But be careful, as the operation of multiplying rows or columns with non-zero scalars is not available! Lemma. The center of SL n (F ) consists of the scalar matrices of determinant 1. Proof. The scalar matrices are clearly central. Conversely, let A Z(SL n (F )). Then At ij (1) = t ij (1)A or, equivalently, Ae ij = e ij A. However, e ij A is the matrix whose ith row coincides with the jth row of A and other rows are zero, while Ae ij is the matrix whose jth column coincides with the ith column of A and other columns are zero. This implies that a ii = a jj and that all other elements in the ith row and jth column of A are zero Lemma. The commutator subgroup G = [G, G] of G = SL n (F ) is equal to G again, unless n = 2 and F is a finite field with F 3. Proof. Denote by diag(b 1,..., b n ) the diagonal matrix with entries b 1,..., b n down the diagonal. It is easy to check that [t ik (a), t kj (b)] = t ij (ab) (i, j, k all distinct), [t ij (a), diag(b 1,..., b n )] = t ij (a ab i b 1 j ) (i j) for all a, b F, b 1,..., b n F. Using the first of these, it follows that G contains all transvections for n > 2 hence G = G in these cases by Lemma Lemma Using the second one, note that if q > 3 we can find b 1, b 2 F q such that b 1 b 2 and b 1 b 2 = 1. Hence even if n = 2 but q > 3 we can still get all transvections in G. Now define the projective unimodular group P SL n (F ) to be the quotient P SL n (F ) := SL n (F )/Z(SL n (F )) of SL n (F ) by the scalar matrices of determinant 1. We ll be interested in these groups when they are finite: if F = F q is the finite field with q elements, we have (by homework exercises) that n P SL n (F q ) = q n(n 1)/2 (q i 1)/(n, q 1). i=2