DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3. Contents

DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Cayley s Theorem 1 2. The permutation group S n 2 3. Center of a group, and centralizers 4 4. Group actions - basics 5 5. Group actions - three major examples 7 6. When a prime p divides the order of a group 8 7. Sylow theorems 10 1. Cayley s Theorem Now we give a converse to the observation that we made just before defining a group. We show below that any group G can be realized as a subgroup of the group of bijections on the set G. [3.1.1] [Cayley s Theorem] If G is a group, then G is isomorphic to the subgroup {L a : a G} of S G, where L a : G G is the left multiplication L a (x) = ax. Proof. We know already that L a S G. Clearly L e = Id G and L ab = L a L b. Also note that if a b, then L a L b since L a (e) = a b = L b (e). Hence the map a L a from G to S X is an injective homomorphism, and therefore G is isomorphic to the subgroup {L a : a G} of S G. Remark: For a G, let R a : G G be the right multiplication, R a (x) = xa. The map a R a from G to S G is not a homomorphism in general since R a R b = R ba. If G is a finite group with G = n, then S G may be identified with S n, the permutation group on n symbols. Therefore, we obtain a corollary of [3.1.1] in the following form. [3.1.2] [Cayley s Theorem for finite groups] If G is a finite group with G = n, then G is isomorphic to a subgroup of S n. Remark: Thus, to understand all finite groups, it is enough to study S n s and their subgroups. Remark: We also mention here a functional interpretation of a group homomorphism in terms of the left multiplication. Let G, G be groups and let f : G G be a map. Then f is a homomorphism iff f(ab) = f(a)f(b) for every a, b G iff f L a = L f(a) f for every a G. 1

2 T.K.SUBRAHMONIAN MOOTHATHU 2. The permutation group S n Notation and definitions: If 2 k n and if x 1,..., x k {1,..., n} are distinct, the notation (x 1,..., x k ) stands for the permutation α S n given by α(x j ) = x j+1 for 1 j < k, α(x k ) = x 1 and α(y) = y for every y {1,..., n} \ {x 1,..., x k }; and we say α is a k-cycle. A 2-cycle α S n is called a transposition. Note that the inverse of a transposition is itself. Cycles α = (x 1 x k ) and β = (y 1 y m ) in S n are disjoint if {x 1,..., x k } {y 1,..., y m } =. Example: If α = (1 2 4) S 6, then α is a 3-cycle with values α(1) = 2, α(2) = 4, α(3) = 3, α(4) = 1, α(5) = 5, α(6) = 6, and α is disjoint with the transposition β = (3 5) S n. Remark: While dealing with permutations, product means composition, and αβ stands for α β. [3.2.1] (i) Every α S n can be expressed as a product of finitely many disjoint cycles in S n. (ii) Every cycle β S n of length k can be written as a product of k 1 transpositions. (iii) Every α S n can be expressed as a product of finitely many transpositions. (iv) If α = δ 1 δ m and α = γ 1 γ k are two representations of α S n as a product of transpositions, then m k is even (i.e, m is odd iff k is odd). Proof. (i) Since S n = n! <, we have order(α) <. Let x {1,..., n} and let k N be the smallest such that α k (x) = x. This gives a k-cycle β 1 = (x 0 x 1 x k 1 ) S n, where x j = α j (x). By induction, the restriction of α to the remaining n k elements can be written as a product β 2 β m of finitely many disjoint cycles. Hence α = β 1 β m. (ii) (x 1 x k ) = (x 1 x 2 )(x 2 x 3 ) (x k 1 x k ) = (x 1 x k )(x 1 x k 1 ) (x 1 x 3 )(x 1 x 2 ). (iii) This follows from parts (i) and (ii). (iv) Define f α (x 1,..., x n ) = f(x α(1),..., x α(n) ) if f is a polynomial in n variables, and α S n. Note that f αβ = (f α ) β. Now suppose f = 1 i<j n (x j x i ). It may be checked that f δ = f if δ S n is a transposition. Hence if α is as given in the hypothesis, then ( 1) k f = f α = ( 1) m f, and therefore m k is even. Definition: Suppose α S n is written as a finite product of transpositions as α = δ 1 δ m. We say α is an even permutation or an odd permutation depending upon whether m is even or odd respectively. This definition makes sense because of [3.2.1](iv). Examples: (i) The identity permutation is an even permutation. If n 2, then Id = (1 2)(1 2) S n. (ii) α = (1 2 3 4) S 6 is an odd permutation since α = (1 2)(2 3)(3 4). Remark: Let α, β S n. (i) If both of α, β are even or both are odd, then αβ is an even. (ii) If one of α, β is odd and the other is even, then αβ is odd. (iii) β 1 is even iff β is even. Definition: The nth alternating group A n is defined as A n = {α S n : α is an even permutation}. That it is indeed a group is proved below. [3.2.2] A n S n for every n N, and [S n : A n ] = 2 for n 2.

DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3 3 Proof. Let α, β A n, then α, β are even permutations, hence αβ 1 is even. That is, αβ 1 A n, and this shows that A n is a subgroup. If n = 1, then A n = S n. Now suppose n 2 so that A n S n. If δ, γ S n are odd, then δ 1 γ is even, hence δ 1 γ A n, or δa n = γa n. So A n has exactly two disjoint left cosets, A n and δa n, implying [S n : A n ] = 2. Consequently, A n S n (this can be seen directly also). Remark: It can be shown with some effort that if n 5, then A n has no proper normal subgroup, and that the only proper normal subgroup of S n is A n. Essentially because of this fact, polynomials of degree 5 are in general hard to solve. More of this can be learned from Galois Theory. [3.2.3] (i) For n 2, S n is generated by {(1 k) : 2 k n}. (ii) For n 2, S n is generated by {(k, k + 1) : 1 k < n}. (iii) For n 3, S n is not abelian (hence not cyclic), but S n can be generated by the two element set {(1 2), (1 2 n)}. (iv) If p 3 is a prime and 1 a < b p, then S p is generated by {(a b), (1 2 p)} [This fact is important in Galois Theory]. (v) For n 3, A n is generated by 3-cycles. (vi) For n 3, A n is generated by {(1 2 k) : 3 k n}. Proof. In the proofs below we will use the observation that (a b) = (a c)(b c)(a c) when ab, c {1,..., n} are distinct. Also the integers appearing below are to be considered modulo n (or p) and are to be identified with the corresponding number in {1,..., n}. (i) Note that (a b) = (1 a)(1 b)(1 a) and use [3.2.1](iii). (ii) Suppose H = {(k, k + 1) : 1 k < n}. To prove (a, b) H for a < b, we use induction on b a. Let m 1, and assume that H contains all transpositions of the form (k, k + j) for j m. Then by induction hypothesis we get (k, k + m + 1) = (k, k + m)(k + m, k + m + 1)(k, k + m) H. (iii) (1 2)(2 3) = (1 2 3) (1 3 2) = (2 3)(1 3), and so S n is not abelian. Let α = (1 2) and β = (1 2 n). To see S n = α, β, observe that (k + 1, k + 2) = β k αβ k for k 0 and use part (ii). (iv) Let H = α, β, where α = (a b), β = (1 2 n). Put r = b a and note that (a + k, b + k) = β k αβ k H. Hence (i, i + r) H for every i. Next, (i, i + 2r) = (i, i + r)(i + r, i + 2r)(i, i + r) H, and inductively (i, i + qr) H for every i, q. Since p is prime and r Z U p, there is q such that qr = 1 in Z U p. Hence (i, i + 1) H for every i and in particular (1 2) H. Now use (iii). (v) (a b)(b c) = (a b c); and (a b)(c d) = (a b)(b c)(b c)(c d) = (a b c)(b c d) if a, b, c, d are distinct. (vi) Let H = {(1 2 k) : 3 k n}. Clearly (1 k 2) = (1 2 k) 2 H. Now consider distinct k, m > 2. Then (1 m)(1 k) = (1 k m). Therefore, in view of part (i), it suffices to show that (1 k m) H. And (1 k m) = (1 k 2)(1 m 2)(1 2 k) H. [3.2.4] Any two transpositions in S n are conjugate to each other, i.e., if α, β S n are transpositions, then there is δ S n such that β = δαδ 1.

4 T.K.SUBRAHMONIAN MOOTHATHU Proof. We assume α β, and let α = (a b), β = (c d). First suppose a, b, c, d are distinct. After a relabeling, we may assume α = (1 3), β = (2 4). Then take δ = (1 2 n) and note that β = δαδ 1. If b = c, then after a relabeling assume α = (1 2), β = (2 3). Then also β = δαδ 1. Remark: If G is a group generated by k elements, and if H is a subgroup of G, then it may not be possible to generate H by k elements (later we will see that it is possible if G is abelian). In fact H may not even be finitely generated (an example is: let G be the free group generated by two elements a, b and let H be the subgroup generated by {a n ba n : n N} - we skip the details). 3. Center of a group, and centralizers Definition: The center Z(G) of a group G is Z(G) = {a G : ax = xa for every x G}. Remark: We have Z(G) = G iff G is abelian. If Z(G) is very small, then it means G is highly non-abelian. Also note that Z(G) = {a G : axa 1 = x for every x G} = {a G : xax 1 = a for every x G}. The following is easy. [3.3.1] If G is a group, then Z(G) is an abelian, normal subgroup of G. Remark: For n 5, it can be shown (with some effort) that A n has no proper normal subgroups and hence Z(A n ) = {e} by the above result. Definition: If G is a group and x G, the centralizer of x in G is C G (x) = {a G : ax = xa} = {a G : axa 1 = x}. [3.3.2] Let G be a group and x G. Then C G (x) is a subgroup of G (may not be abelian or normal - see the example below), and Z(G) = x G C G(x). [3.3.3] Let n 3, G = S n, and δ = (1 2) G. Then, (i) Z(G) = {e}. (ii) C G (δ) = {β G : β({1, 2}) = {1, 2}}. (iii) C G (δ) is not normal in S n. (iv) If n 5, then C G (δ) is not abelian. Proof. (i) If α G \ {e}, there are a b such that α(a) = b. Let c {1,..., n} \ {a, b} and β = (b c) G. Then αβ(a) = b c = βα(a). (ii) The inclusion is clear. If β S n is not a member of RHS, then there is a > 2 such that b = β(a) {1, 2}. Now δβ(a) = δ(b) b = βδ(a). Thus β / C G (δ). This proves. (iii) (2 3)(1 2)(2 3) = (1 3) / C G (δ). (iv) Use part (ii) and the fact that S {3,4,5} is non-abelian. [3.3.4] Let n 3 and G = D n be the nth dihedral group generated by α, β with the relations α n = e = β 2 and βα = α 1 β = α n 1 β. (i) If n is odd, then Z(G) = {e}. (ii) If n is even, then Z(G) = {e, α n/2 }.

DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3 5 Proof. If α i β Z(G), then (α i β)α = α(α i β), or α i 1 β = α i+1 β, or e = α 2, a contradiction since order(α) = n 3. Hence α i β / Z(G) for any i. Next note that for 1 i < n, we have α i Z(G) iff α i (α j β k ) = (α j β k )α i for every j, k iff α i+j β k = α j i β k for every j, k iff α 2i = e. ( ) 1 a [3.3.5] Let G = GL(2, R), H = SL(2, R), and A =, where a R \ {0}. Then, 0 1 (i) Z(G) = {bi ( : b R \ ) {0}} and Z(H) = { I, I}, where I is the ( 2 2 ) identity matrix. p q p q (ii) C G (A) = { : p R \ {0}, q R}, and C H (A) = { : p = ±1, q R}. 0 p 0 p ( ) ( ) b 0 c1 c 2 Proof. Consider matrices B = and C =, and assume B, C GL(2, R) (so 0 b c 3 c 4 that b, b 0 for instance). Statement (i) can be deduced from the following two observations: if b = b, then BC = CB; if b b and at least one of c 2, c 3 is non-zero, then BC CB. Similar arguments will establish (ii). Remark: If G is a group and H is a subgroup, then C H (x) = C G (x) H for any x H. The following fact will be used later. [3.3.6] Let G be a group. If G/Z(G) is cyclic, then G is abelian and hence Z(G) = G. Proof. Let a G be such that G/Z(G) is generated by az(g). Consider x, y G. Then there are b, c Z(G) and integers i, j such that x = a i b, y = a j c. Since each of b, c commutes with every element of G, we have xy = a i ba j c = a i+j bc = a i+j cb = a j ca i b = yx. 4. Group actions - basics In Cayley s Theorem [3.1.1], we saw that if G is a group, then G is isomorphic to a subgroup of S G, i.e., there is an injective homomorphism π : G S G. Now, given a group G, there may exist many sets X such that there is a homomorphism π : G S X (here π may not be injective). This abstract concept is quite useful. So we formalize it. Definition: Let G be a group and X be a nonempty set. An action of G on X is a homomorphism π : G S X. Sometimes we will write π a for the map π(a) S X for a G. Note that by definition π a : X X is a bijection for each a G even if π is not injective. Definition: Let π be the action of a group G on a nonempty set X. (i) The action is said to be faithful if π is injective, equivalently if ker(π) = {e} (for example, by Cayley s Theorem every group has a faithful action on itself). (ii) F ix(g) = {x X : π a (x) = x for every a G} X. (iii) If x X, the orbit and stabilizer of x are defined respectively as O G (x) = {π a (x) : a G} X, Stab(x) = {a G : π a (x) = x} G. It may be checked that Stab(x) is a subgroup of G for every x X. Also, ker(π) = x X Stab(x).

6 T.K.SUBRAHMONIAN MOOTHATHU [3.4.1] Let π be the action of a group G on a nonempty set X. Define a relation R on X by the condition that (x, y) R iff there is a G such that π a (x) = y. Then R is an equivalence relation on X, and the equivalence classes are precisely the distinct G-orbits. Thus X gets partitioned into distinct G-orbits. Definition: Let π be the action of a group G on a nonempty set X. The action π is said to be transitive if for every x, y X, there is a G such that π a (x) = y. In view of [3.4.1], we may note that π is transitive O G (x) = X for some x X O G (x) = X for every x X. It may also be noted that if the cardinality of G is strictly less than the cardinality of X, then π cannot be transitive since the cardinality of O G (x) cannot exceed that of G. Some examples of group actions: (i) [Action of (Z, +) on R] Let π : Z S R be π m (x) = x + m for m Z and x R. Then ker(π) = {0} (so the action is faithful) and F ix(z) = {x R : x + m = x for every m Z} =. If x R, then O G (x) = x + Z R (so the action is not transitive) and Stab(x) = {0}. More generally, if we fix t R \ {0}, then π : Z S R given by π m (x) = x + tm is an action of Z on R. We have ker(π) = {0}, F ix(z) =, O G (x) = x + tz and Stab(x) = {0}. On the other hand, the attempts π m (x) = mx and π m (x) = 2x + m do not define actions of (Z, +) on R (why?). (ii) [Action of (R, +) on [0, 1)] Let π : R S [0,1) be π a (x) = x + a (mod 1) for a R and x [0, 1). Then ker(π) = Z (so the action is not faithful) and F ix(r) =. If x [0, 1), then O G (x) = [0, 1) (so the action is transitive) and Stab(x) = Z. (iii) [Action of S 3 on N 3 ] Let G = S 3, X = N 3 and π : G S X be π σ (n 1, n 2, n 3 ) = (n σ(1), n σ(2), n σ(3) ). Then ker(π) = {e} (so the action is faithful) and F ix(s 3 ) = {(n 1, n 2, n k ) N 3 : n 1 = n 2 = n 3 }. Since S 3 is a finite group and N 3 is an infinite set, the action cannot be transitive. If x = (5, 7, 5) N 3, then O G (x) = {(5, 5, 7), (5, 7, 5), (7, 5, 5)} and Stab(x) = {e, (1 3)} S 3, and in particular, we note that O G (x) Stab(x) = 3 2 = 6 = G. [3.4.2] [Orbit-Stabilizer Theorem] Let π be the action of a finite group G on a nonempty set X. Then G = O G (x) Stab(x) for every x X. Proof. Fix x X and let H = Stab(x). Since G / H = [G : H] = G/H, it suffices to produce a bijection between G/H and O G (x). Define ψ : G/H O G (x) as ψ(ah) = π a (x). For a, b G, we have ah = bh iff a 1 b H iff x = π a 1 b(x) = πa 1 (π b (x)) iff π a (x) = π b (x). Therefore ψ is well-defined and injective. Clearly ψ is surjective also. [3.4.3] [Corollary] Let p be a prime and let G be a finite group with G = p n acting on a nonempty set X. Then for each x X, there is k {0, 1,..., n} such that O G (x) = p k. [3.4.4] Let π be the action of a group G on a set X. If x, y X and a G are such that π a (x) = y, then Stab(y) = a Stab(x) a 1. In particular, if the action is transitive, then Stab(x) and Stab(y) are conjugate subgroups for any two x, y X.

DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3 7 Proof. b Stab(x) π b (x) = x π b (π a 1(y)) = π a 1(y) π aba 1(y) = y aba 1 Stab(y). 5. Group actions - three major examples Definition: If G is a group, the conjugacy relation R given by (x, y) R iff there is a G such that axa 1 = y is an equivalence relation on G and the equivalence classes are called conjugacy classes. We denote the conjugacy class of x G as [x]. Note that [x] = {x} iff x Z(G). [3.5.1] [Major Example-1: conjugacy action of a group on itself] Let G be a group and X = G. Let π : G S G be π a (x) = axa 1 for a, x G. Easy to verify that π is indeed a well-defined homomorphism. We have ker(π) = {a G : π a = Id G } = {a G : axa 1 = x for every x G} = Z(G) (so in general π is not faithful), F ix(g) = {x G : [x] = {x}} = Z(G), O G (x) = [x], and Stab(x) = {a G : axa 1 = x} = {a G : ax = xa} = C G (x). Since O G (e) = {e}, the conjugacy action is transitive iff G = {e}. [3.5.2] Let G be a finite group and x G. Then, [x] = [G : C G (x)]. In particular [x] divides G. Proof. Consider the conjugacy action of G on G. Then O G (x) = [x] and Stab(x) = C G (x). Hence by the Orbit-Stabilizer Theorem, G = [x] C G (x) and so the result follows. [3.5.3] Let G be a finite (non-abelian) group and suppose [x 1 ],..., [x k ] is a listing of distinct conjugacy classes in G having more than one element. Then, (i) [Class equation] G = Z(G) + k i=1 [G : C G(x i )]. (ii) 1 < [G : C G (x i )] < G (equivalently, G > C G (x i ) > 1) for 1 i k. Proof. We know that G = Z(G) [x 1 ] [x k ] is a disjoint union, and [x i ] = [G : C G (x i )]. This gives (i). We have [x i ] > 1 by hypothesis; and [x i ] < G since e / [x i ]. This gives (ii). The class equation is useful in getting some insight into the structure of a finite non-abelian group (if the group is abelian, then Z(G) = G, so the class equation becomes trivial and does not give any new information). We will discuss this in the next section. Project topic: Study about conjugacy classes in the groups S n, A n and D n. [3.5.4] [Major Example-2: conjugacy action of a group on its subgroups] Let G be a group, let X be the collection of all subgroups of G, and π : G S X be π a (H) = aha 1 for a G and any subgroup H of G. It may be checked that π is a well-defined homomorphism. Note that F ix(g) = {all normal subgroups of G}, O G (H) = {all subgroups of G conjugate to H}, and Stab(H) = {a G : aha 1 = H} is the largest subgroup of G containing H in which H is normal. Therefore, for this particular group action, Stab(H) is called the normalizer of the subgroup H in G, and is denoted as N G (H). If N G (H) is small in G and not much bigger than H, then it means H is highly non-normal in G. Remark: Suppose G = S 3, H = {e, (1 2)} and H = {e, (1 2 3), (1 3 2)}. Since N G (H) {2, 6} and since H is not normal, N G (H) = G. Since [G : H ] = 2, we have H G and so N G (H ) = H.

8 T.K.SUBRAHMONIAN MOOTHATHU [3.5.5] [Major Example-3: the action of SL(2, R) on the upper half-plane] This is a group action that is important in advanced Analysis. Let X = {(x + iy) ( C : y > ) 0} be the upper half-plane a b and let G = SL(2, R). Define π : G S X as follows: if A = G, then π A : X X is c d given by π A (z) = (az + b)/(cz + d). (i) How do we know π A (z) X? First note by direct calculation that the imaginary part of π A (i) is 1/(c 2 + d 2 ) > 0 and hence π A (i) X. This observation together with some facts from Complex Analysis (which you will study later) will imply that π A (z) X for every z X. (ii) It can be checked that π AB = π A π B, and in particular π A π A 1 = π I = Id X = π A 1π A. So indeed π A S X, and π is a homomorphism. (iii) A ker(π) (az + b)/(cz + d) = z for every z X cz 2 + (d a)z + b = 0 for every z X c = d a = b = 0. Since we also have det[a] = 1, we conclude that ker(π) = {±I}. ( ) 2 0 (iv) Consider A = G. If π A (z) = z, then 4z = z or z = 0 / X. Hence F ix(g) =. 0 1/2 ( y ) x/ y (v) If z = x + iy X, then A = 0 1/ G and π A (i) = z. So O G (i) = X and hence the y action is transitive (and O G (z) = X for every z X). ( ) a b (iv) If (ai + b)/(ci + d) = i, then a = d and c = b. Therefore, Stab(i) = { : a, b b a R, a 2 + b 2 = 1}. Since the action is transitive, Stab(z) = C Stab(i) C 1 if C G is such that π C (i) = z. 6. When a prime p divides the order of a group The converse of Lagrange s theorem is not true: if G is a finite group and if d is a natural number dividing G, then G may not have a subgroup of order d. This is illustrated by the following example. [3.6.1] We show that A 4 has no subgroups of order 6 even though 6 divides A 4 = 12. Let if possible, H A 4 be a subgroup with H = 6. Consider a 3-cycle σ A 4 (note that a 3-cycle is an even permutation since (a b c) = (a b)(b c)). Then at least two of the cosets H, σh, σ 2 H must coincide since [A 4 : H] = 2; and in each case, it is not difficult to see that σ H (or use [2.4.3] to see that σ = σ 4 = (σ 2 ) 2 H). Thus H contains all 3-cycles in A 4. Three distinct elements can be chosen from 1,2,3,4 in 4 ways, and if σ is a 3-cycle, then σ 2 is also a 3-cycle. Hence the number of 3-cycles in A 4 is 4 2 = 8. This implies H 8, a contradiction. However, the converse of Lagrange s theorem is true in certain special cases. Suppose G is a finite group and d divides G. Then we can say that G has a subgroup of order d if at least one of the following holds: (i) G is abelian.

(ii) d is a prime power. DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3 9 The converse of Lagrange s theorem with G abelian we may not prove in our course. We remark that it follows essentially from the (non-trivial) fact that any finite abelian group is isomorphic to a product C 1 C k of cyclic groups where each C i = p r i i not be distinct). for some prime p i (here p i s need The converse of Lagrange s theorem with the second assumption (i.e., when d is a prime power) is Sylow s first theorem that we will prove in the last section. In this section, we consider some results which can be considered as stepping stones to Sylow s theorems. Definition: Let p be a prime. A finite group G is said to be a p-group if G = p n for some n N. [3.6.2] Let p be a prime and G be a p-group. Then, Z(G) > 1. Proof. In the class equation G = Z(G) + k i=1 [G : C G(x i )], we note that p divides [G : C G (x i )] for 1 i k by [3.4.3]. So p divides Z(G). [3.6.3] Let p be a prime and G be a group of order p 2. Then G is abelian. Moreover, G is isomorphic to either Z p 2 or Z p Z p. Proof. We have Z(G) = p or p 2 by [3.6.2]. Hence G/Z(G) = p or 1, and thus G/Z(G) is cyclic. So G is abelian by [3.3.6] (thus Z(G) = G and the case Z(G) = p really does not arise). If G is cyclic, then G = Z p 2. Next suppose G is not cyclic. Then order(x) = p for every x G \ {e}. Let a G \ {e}, H = a, b G \ H and let K = b. Clearly H K = {e}. The map f : H K G given by f(x, y) = xy is a homomorphism since G is abelian. Also kerf = {(e, e)} since H, K are subgroups with H K = {e}. Thus f is injective. Since H K = p 2 = G, we conclude that f is also surjective, and thus f is an isomorphism. So G = H K = Z p Z p. (Another argument is the following: note that HK is greater than p and divides G = p 2, so HK = p 2 and hence HK = G; now use [2.5.4].) [3.6.4] If a prime p divides G for a finite abelian group G, then G has an element of order p. Proof. Let G = np and use induction on n, where the case n = 1 is trivial (since G becomes cyclic in this case). In the general case, let H G be a maximal proper subgroup. If p divides H, we are done by induction. If p H, let a G \ H and K = a. Then HK is a subgroup of G since G is abelian, and hence HK = G by the maximality of H. The map f : H K G given by f(x, y) = xy is a surjective homomorphism with kerf = {(x, x 1 ) : x H K}. Therefore, as a consequence of the first isomorphism theorem, G = H K / kerf = H K / H K, and this implies that p divides K. Then, the element a K /p K G has order p. We may drop the abelian hypothesis from the above. The following can be considered as a baby converse of Lagrange s theorem.

10 T.K.SUBRAHMONIAN MOOTHATHU [3.6.5] [Cauchy s theorem] If G is a finite group and if a prime p divides G, then G has an element of order p (and hence a subgroup of order p). Proof. Suppose G = np and as before we use induction on n. The case n = 1 is easy. Now assume the result for values up to n 1 and consider G with G = np. We may assume G is non-abelian since the abelian case was already done in [3.6.4]. Suppose the class equation is G = Z(G) + k i=1 [G : C G(x i )]. Note that C G (x i ) < G for 1 i k. Case-1 : p divides C G (x i ) for some i. Then C G (x i ) = mp for some m < n and hence there is an element y C G (x i ) G of order p by induction hypothesis. Case-2 : p does not divide C G (x i ) for 1 i k. Then p divides G / C G (x i ) = [G : C G (x i )] for 1 i k and hence p divides Z(G) by the class equation. We have Z(G) G since G is assumed to be non-abelian. Consequently, Z(G) = mp for some m < p and hence there is an element y Z(G) G of order p by induction hypothesis. When G is a p-group, the converse of Lagrange s theorem is true. In fact, we have: [3.6.6] Let p be a prime and G be a p-group with G = p n. Then there is a finite sequence {e} = H 0 H 1 H n 1 H n = G of normal subgroups of G such that H i = p i for 0 i n (hence H i /H i 1 is cyclic of order p for 1 i n). Proof. We use induction on n. The case n = 1 is trivial. Now assume the result for values up to n 1 and consider G with G = p n. We have Z(G) > 1 by [3.6.2], hence Z(G) = p k for some k {1,..., n}, and therefore there is y Z(G) of order p by Cauchy s theorem. Let H = y and note that H G since H Z(G). Since G/H = p n 1, by the induction hypothesis applied to G/H, we can find a finite sequence {H} = K 0 K 1 K n 1 = G/H of normal subgroups of G/H such that K i = p i for 0 i n 1. Take H 0 = {e} and H i = q 1 (K i 1 ) for 1 i n, where q : G G/H is the quotient map. 7. Sylow theorems Sylow theorems are very powerful since they give us information (sometimes complete information) about the structure of a finite group G just based on a single number, namely G. Definition: Let p be a prime and G be a finite group with G = p n m, where n, m N and p does not divide m. Then any subgroup of order p n of G is called a Sylow p-subgroup of G. Moreover, any subgroup of order p k of G (where k n), will be referred to as a p-subgroup of G. Here is the promised converse to Lagrange s theorem in a special case: [3.7.1] [Sylow s first theorem] Let p be a prime and G be a finite group with G = p n m, where n, m N and p does not divide m. Then G has a subgroup of order p k for each k {1,..., n}.

DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3 11 Proof. It suffices to show that G has a subgroup H with H = p n, for then we may apply [3.6.6] to H. We use induction on G. In the initial case G = p, we have nothing to prove. Now assume that we have covered all groups of order < G, and consider G. Case-1 : G has a proper subgroup K such that p does not divide [G : K]. Then K = p n m for some m < m. Since K < G, there is a subgroup H of order p n for K, and hence for G, by induction hypothesis. Case-2 : p divides [G : K] for any proper subgroup K of G. If G is abelian, then Z(G) = G and so p divides Z(G). If G is non-abelian, then using the class equation G = Z(G) + k i=1 [G : C G(x i )] and the assumption of case-2, we conclude that p divides Z(G). Let y Z(G) be an element of order p ( Cauchy s theorem), let H = y, and as before note that H G since H Z(G). By induction hypothesis, G/H has a subgroup H of order p n 1. If K = q 1 (H ), where q : G G/H is the quotient map, then K is a subgroup of G with K = p n. The first theorem of Sylow establishes the existence of Sylow p-subgroups. The next two theorems of Sylow will give information about the nature and number of Sylow p-subgroups. We essentially follow Artin s approach for the proofs. [3.7.2] [Sylow s second theorem] Let p be a prime and G be a finite group with G = p n m, where n, m N and p does not divide m. Then, (i) Any p-subgroup of G is contained in a Sylow p-subgroup of G. (ii) Any two Sylow p-subgroups of G are conjugate. Proof. Let K G be a p-subgroup and H G be a Sylow p-subgroup of G. We claim that there is x G such that K xhx 1. If this claim is proved, then (i) and (ii) follow easily. In order to prove the claim, as a first step, we will consider a group action and will realize K xhx 1 as the stabilizer of some element. Let X = G/H, the space of left cosets of H, and consider the action of K on X by left multiplication. That is, π : K S X is given by π a (xh) = axh. If xh X, then Stab(xH) = {a K : axh = xh} = {a K : x 1 ax H} = {a K : a xhx 1 } = K xhx 1. Hence, to prove the claim, it is enough to show that there is x G such that Stab(xH) = K. In view of the Orbit-stabilizer theorem, it now suffices to show there is x G such that O K (xh) = 1. For any x G, O K (xh) must divide K ( Orbit-stabilizer theorem), so O K (xh) = p j for some j 0 since K is a p-subgroup. On the other hand, X = G/H = m is not divisible by p. Since X gets partitioned into distinct orbits under the action of K, there is at least one orbit whose cardinality is not divisible by p. That is, there is x G such that O K (xh) = p 0 = 1, and we are done. Remark: A corollary of [3.7.2](ii) is that if the Sylow p-subgroup is unique, then it is normal.

12 T.K.SUBRAHMONIAN MOOTHATHU [3.7.3] [Sylow s third theorem] Let p be a prime and G be a finite group with G = p n m, where n, m N and p does not divide m. If k is the number of Sylow p-subgroups of G, then k divides m and k 1 (mod p). Proof. Let H G be a Sylow p-subgroup and consider its normalizer N = N G (H) = {a G : aha 1 = H}. Observe that an = bn a 1 b N a 1 bhb 1 a = H bhb 1 = aha 1. Therefore, the number of left cosets of N in G is equal to the number of conjugate subgroups of H. But the second number is k since any two Sylow p-subgroups are conjugate. Thus [G : N] = k. Since H N G and [G : H] = m, we conclude that k divides m by the product rule [2.4.4]. Let H 1,..., H k G be the Sylow p-subgroups and let N i be the normalizer of H i in G. Also write H = H 1. Let X = {H 1,..., H k } and let H act on X by conjugation. That is, π : H S X is given by π a (H i ) = ah i a 1. Consider i {1,..., k}. Since O H (H i ) must divide H = p n, we have either O H (H i ) = 1, or p divides O H (H i ). Moreover, O H (H i ) = 1 iff ah i a 1 = H i for every a H iff H N i. If H N i, then both H and H i are Sylow p-subgroups of N i, and this implies H = H i since H i N i and since Sylow p-subgroups must be conjugate by [3.7.2]. That is, O H (H i ) = 1 iff H i = H = H 1. Since X gets partitioned into distinct orbits, we conclude that k = X is congruent to 1 (mod p). [3.7.4] [Corollary] Let n = pq, where p < q are primes and p does not divide q 1. Then any group of order n is cyclic. All such n with n 100 are n = 15, 33, 35, 51, 65, 69, 77, 85, 87, 91, 95. Proof. Consider a group G with G = pq. Let k be the number of Sylow q-subgroups of G. Using [3.7.3], we have that k p, and k 1 (mod q). Since p < q and p is prime, we immediately get k = 1. Next, let j be the number of Sylow p-subgroups of G. By [3.7.3] we have j q, and j 1 (mod p). Since q is prime and p does not divide q 1, we conclude that j = 1. Let H be the unique Sylow p-subgroup of G and K be the unique Sylow q-subgroup of G. By [3.7.2], H, K are normal subgroups of G. We have H K = {e} since p, q are distinct primes. Using the proof of [2.5.4], note that xy = yx for every (x, y) H K. Hence f : H K G defined as f(x, y) = xy is a homomorphism, and f is injective since H K = {e}. Since H K = pq = G, f is surjective and thus an isomorphism. So G = H K = Z p Z q = Zn since p and q are distinct primes. (Another argument is: note that H K = p + q 1 < pq, so there exists x G \ H K, and the order of x is necessarily pq, implying G is cyclic.) Topic for self-study: Classification of finite groups of low orders. *****