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


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1 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 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 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 kcycle. A 2cycle 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 3cycle 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 kcycle 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 = ( ) 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.
3 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 3cycles. (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 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 nonabelian. 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 nonabelian. [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 }.
5 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 nonzero, 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 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 Gorbits. Thus X gets partitioned into distinct Gorbits. 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] [OrbitStabilizer 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 welldefined 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.
7 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 Example1: 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 welldefined 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 OrbitStabilizer Theorem, G = [x] C G (x) and so the result follows. [3.5.3] Let G be a finite (nonabelian) 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 nonabelian 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 Example2: 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 welldefined 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 nonnormal 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 8 T.K.SUBRAHMONIAN MOOTHATHU [3.5.5] [Major Example3: the action of SL(2, R) on the upper halfplane] This is a group action that is important in advanced Analysis. Let X = {(x + iy) ( C : y > ) 0} be the upper halfplane 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 3cycle σ A 4 (note that a 3cycle 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 3cycles in A 4. Three distinct elements can be chosen from 1,2,3,4 in 4 ways, and if σ is a 3cycle, then σ 2 is also a 3cycle. Hence the number of 3cycles 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.
9 (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 (nontrivial) 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 pgroup if G = p n for some n N. [3.6.2] Let p be a prime and G be a pgroup. 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 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 nonabelian 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. Case1 : 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. Case2 : 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 nonabelian. 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 pgroup, the converse of Lagrange s theorem is true. In fact, we have: [3.6.6] Let p be a prime and G be a pgroup 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 psubgroup of G. Moreover, any subgroup of order p k of G (where k n), will be referred to as a psubgroup 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}.
11 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. Case1 : 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. Case2 : 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 nonabelian, then using the class equation G = Z(G) + k i=1 [G : C G(x i )] and the assumption of case2, 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 psubgroups. The next two theorems of Sylow will give information about the nature and number of Sylow psubgroups. 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 psubgroup of G is contained in a Sylow psubgroup of G. (ii) Any two Sylow psubgroups of G are conjugate. Proof. Let K G be a psubgroup and H G be a Sylow psubgroup 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 Orbitstabilizer 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 ( Orbitstabilizer theorem), so O K (xh) = p j for some j 0 since K is a psubgroup. 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 psubgroup is unique, then it is normal.
12 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 psubgroups of G, then k divides m and k 1 (mod p). Proof. Let H G be a Sylow psubgroup 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 psubgroups 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 psubgroups 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 psubgroups of N i, and this implies H = H i since H i N i and since Sylow psubgroups 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 qsubgroups 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 psubgroups 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 psubgroup of G and K be the unique Sylow qsubgroup 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 selfstudy: Classification of finite groups of low orders. *****
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