Solutions to Assignment 4 1. Let G be a group and x, y ∈ G such that [x, y] = xyx−1y−1 ∈ Z(G). Prove that for every positive integer n, we have (a) [xn, y] = [x, y]n, n n n n (b) x y = (xy) [x, y](2). Solution: (a) Using induction on n, [x, y]n = [x, y]n−1[x, y] = [xn−1, y][x, y] = xn−1[x, y]yx1−ny−1 = xn−1xyx−1y−1yx1−ny−1 = [xn, y]. n n−1 (b) The statement is easily verified for n = 1, 2. We note that 2 − 2 = n − 1, and use induction on n. n n n−1 n−1 n−1 n−1 n−1 n−1 (xy) [x, y](2) = (xy)x y [x, y] = x[x, y] yx y = x[xn−1, y]yxn−1yn−1 = xxn−1yx1−ny−1yxn−1yn−1 = xnyn. 2. For H, K < G, we use [H, K] to denote the subgroup of G generated by all elements of the form hkh−1k−1 where h ∈ H and k ∈ K. Prove that (a) [H, K] C hH, Ki, where hH, Ki denotes the subgroup of G generated by H and K, (b) [G, H]H C G, (c) [G, H]H is the smallest normal subgroup of G which contains H. Solution: (a) For h0 ∈ H and [h, k] as above, note that −1 −1 −1 −1 −1 −1 −1 −1 −1 h0(hkh k )h0 = (h0hkh h0 k )(kh0k h0 ) = [h0h, k][h0, k] ∈ [H, K]. −1 Similarly for k0 ∈ K we have k0[h, k]k0 ∈ [H, K]. Lastly, note that in general −1 −1 −1 y(x1 . xn)y = (yx1y ) ... (yxny ). (b) [G, H] C G by (a), so [G, H]H is indeed a subgroup of G. For g, x ∈ G and h, h0 ∈ H, note that −1 −1 −1 −1 −1 −1 −1 −1 −1 g[x, h]h0g = gxhx h h0g = gxhx g h hgh h0g −1 −1 −1 = [gx, h]hgh h0g = [gx, h][g, h] [g, h0]h0. (c) Any normal subgroup containing H contains all elements of the form xhx−1 for x ∈ G and h ∈ H, and hence contains [G, H]. 3. Let G be a group. Set Z0(G) = {e} and inductively define Zi+1(G) to be the inverse image of Z(G/Zi(G)) under the canonical surjection G −→ G/Zi(G). This gives a sequence of normal subgroups of G, Z0(G) C Z1(G) C Z2(G) C ··· . Set C0(G) = G and inductively define Ci+1(G) = [Ci(G),G]. This gives us C0(G) B C1(G) B C2(G) B ··· . 1 Solutions to Assignment 4 (a) If Zk(G) = G for some integer k, show that Ci(G) ⊆ Zk−i(G) for all 1 ≤ i ≤ k. (b) If Ck(G) = {e} for some integer k, show that Ck−i(G) ⊆ Zi(G) for all 1 ≤ i ≤ k. (c) Conclude that for an integer k, we have Zk(G) = G if and only if Ck(G) = {e}. In this case the group G is said to be nilpotent, and the least such k is called the nilpotency class of G. Solution: We write Ci = Ci(G) and Zi = Zi(G) for notational convenience. (a) We use induction on i. Note that Zk = G implies C0 ⊆ Zk, which starts the induction. Assuming the result for i − 1, we have Ci−1 ⊆ Zk−i+1, and so Ci = [Ci−1,G] ⊆ [Zk−i+1,G]. But Zk−i+1/Zk−i is the center of G/Zk−i, so [Zk−i+1,G] ⊆ Zk−i. (b) Again we use induction on i, and Ck = {e} ⊆ Z0 starts the induction. Assuming the result for i − 1, we have Ck−i+1 ⊆ Zi−1, and so [Ck−i,G] ⊆ Zi−1. This implies that the image of Ck−i in G/Zi−1 lies in Z(G/Zi−1) = Zi/Zi−1, i.e., Ck−i ⊆ Zi. (c) is immediate from (a) and (b). 4. Let G be the subgroup of GL2(R) consisting of all matrices of the form a c . 0 b Show that G is a solvable group. Is G nilpotent? Prove or disprove. Solution: Routine matrix multiplication shows that 1 x 1 0 G(1) = : x ∈ and G(2) = , 0 1 R 0 1 and so G is solvable. On the other hand, 1 x a c 1 x − ax/b , = , 0 1 0 b 0 1 (1) and so C2(G) = G = C1(G). It follows that Cn(G) = C1(G) for all n ≥ 1, and so G is not nilpotent. 2 Solutions to Assignment 4 5. Let G be a nilpotent group and H be the set of elements of G of finite order. (a) Prove that H is a subgroup of G. Hint: Use induction on the nilpotency class of G. (b) Prove that every finite subset of H generates a finite subgroup. (c) Prove that any two elements of H of relatively prime order commute. Solution: (a) If a, b have finite order, we need to show ab has finite order. Since a subgroup of a nilpotent group is nilpotent, there is no loss of generality in taking G = ha, bi. We use induction on k, the nilpotency class of G. If k = 1 then G is abelian and so ab has finite order. The group G/Ck−1(G) has nilpotency class at most k − 1, so the induction hypothesis implies n that abCk−1(G) ∈ G/Ck−1(G) has finite order, i.e., that (ab) ∈ Ck−1(G) for some n. It now suffices to show that every element of Ck−1(G) has finite order. Note that Ck−1(G) ⊆ Z(G) by Problem 3, and so for x ∈ Ck−2(G) and y, z ∈ G we have [x, y][x, z] = xyx−1[x, z]y−1 = xyx−1xzx−1z−1y−1 = [x, yz]. Since G is generated by a, b, it now follows that Ck−1(G) is generated by elements of the form m m [x, a] and [x, b] for x ∈ Ck−2(G). If |a| = m, then Problem 1 implies that [x, a] = [x, a ] = e, and so [x, a], and similarly [x, b], have finite order. Since Ck−1(G) is abelian and is generated by elements of finite order, it follows that every element of Ck−1(G) has finite order. (b) It suffices to show that if G is a nilpotent group generated by finitely many elements of finite order, than G is finite. We use induction on the nilpotency class k of G. If k = 1 then G is abelian, and the result is easy. As in the above argument, G/Ck−1(G) has nilpotency class at most k − 1, so the induction hypothesis implies that G/Ck−1(G) is finite. Let x1, . , xn ∈ G be coset S representatives, i.e., G = xiCk−1(G). Since Ck−1(G) ⊆ Z(G), every commutator of elements of G equals [xi, xj] for some 1 ≤ i, j ≤ n. Consequently Ck−1(G) is an abelian group generated by finitely many elements of finite order, and hence is finite. But then |G| = |G/Ck−1(G)||Ck−1(G)| is finite as well. (c) Let a, b ∈ H be elements of relatively prime order. By part (b), the subgroup ha, bi is a finite nilpotent group, and so its Sylow subgroups are normal. The elements a, b belong to different Sylow subgroups of ha, bi, say P and Q. But then aba−1b−1 ∈ P ∩ Q = {e}. 6. G be a finite group, and ϕ : G −→ G a homomorphism. (a) Prove that there exists a positive integer n such that for all integers m ≥ n, we have Im ϕm = Im ϕn and Ker ϕm = Ker ϕn. (b) For n as above, prove that G is the semidirect product of the subgroups Ker ϕn and Im ϕn. Solution: (a) We have a sequence of subgroups, Ker ϕ ⊆ Ker ϕ2 ⊆ Ker ϕ3 ··· , which must stabilize since G is finite. Similarly, the sequence Im ϕ ⊇ Im ϕ2 ⊇ Im ϕ3 ··· must stabilize as well. (b) Let x ∈ Ker ϕn ∩ Im ϕn. Then x = ϕn(y) for some y ∈ G, and ϕn(x) = ϕ2n(y) = e. But then y ∈ Ker ϕ2n = Ker ϕn, and so x = e. This shows that Ker ϕn ∩ Im ϕn = {e}. Since G/ Ker ϕn ≈ Im ϕn, we have |G| = | Ker ϕn|| Im ϕn|. Since (Ker ϕn)(Im ϕn) is a subgroup of G of order |G|, it must equal G. Consequently G is a semidirect product of Ker ϕn and Im ϕn. 3 Solutions to Assignment 4 7. Let G be a finite group, and p a prime integer dividing |G| such that the map ϕ : G −→ G, where ϕ(x) = xp, is a homomorphism. (a) Prove that G has a unique p-Sylow subgroup P . (b) Prove that there is a normal subgroup N C G such that N ∩ P = {e} and G = PN. (c) Show that G has a nontrivial center. Solution: (a) By the previous problem, there exists n such that Im ϕm = Im ϕn and Ker ϕm = Ker ϕn for all m ≥ n. But then Ker ϕn is the subgroup of G consisting of all elements of G whose order is a power of p. Consequently P = Ker ϕn is the unique p-Sylow subgroup of G. (b) Let N = Im ϕn. By the previous problem, N ∩ P = {e} and G = PN.