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. Let xpn ∈ N and g ∈ G. Then gxpn g−1 = (gxg−1)pn ∈ N, and so N is normal. (c) Since N C G and P C G, we have [n, h] ∈ N ∩ P for all elements n ∈ N and h ∈ P . But N ∩ P = {e}, so [n, h] = e. By the previous problem G is a semi-direct product of P and N but, since elements of N commute with those of P , we actually have G ≈ P × N. The p-group P has a nontrivial center which is contained in the center of G.

8. Let Q = {±1, ±i, ±j, ±k} be the quaternion group, i.e., (−1)2 = 1 is the identity element, i2 = j2 = k2 = −1, and

ij = k = −ji, jk = i = −kj, ki = j = −ik.

(a) Determine all subgroups of Q and prove that they are normal. (b) What is the order of Aut(Q)?

Solution: (a) Aside from e, the group Q consists of six elements of order 4 and one element of order 2, namely −1. Consequently the only subgroups of Q are

Q, hii, hji, hki, h−1i, {e}.

(b) Any two elements of order four, which are not powers of each other, constitute a generating set for Q. Any automorphism ϕ ∈ Aut(Q) is determined by its behavior on a generating set, and must take a generating set to a generating set. Consider the generating set {i, j}. Then

ϕ(i) ∈ {±i, ±j, ±k} and ϕ(j) ∈ {±i, ±j, ±k} \ {±ϕ(i)}.

Consequently | Aut(Q)| = 24.

9. Let p < q be prime numbers such that p divides q − 1. If G is a non-abelian group of order pq, show that Z(G) = {e}. Solution: If Z(G) is nontrivial, then |Z(G)| equals p or q. But then G/Z(G) is cyclic, in which case G must be abelian.

4 Solutions to Assignment 4

10. Determine, up to isomorphism, all groups of order 63. Solution: Let |G| = 63. Since 63 = 32 × 7, by an earlier homework problem either the 3-Sylow or the 7-Sylow subgroup is normal in G. Suppose the 3-Sylow subgroup P is normal, then

G ≈ P oα Z/7 for α : Z/7 −→ Aut(P ).

If P ≈ Z/9 then | Aut(P )| = 8 and α is trivial. In this case, G ≈ Z/9 × Z/7 ≈ Z/63. If P ≈ Z/3 × Z/3 then | Aut(P )| = 48 and once again α must be trivial. This implies that G ≈ Z/3 × Z/3 × Z/7 ≈ Z/3 × Z/21. If P is not normal then the 7-Sylow subgroup Q ≈ Z/7 is normal, and so

G ≈ Z/7 oβ P where β : P −→ Aut(Z/7)

is a nontrivial homomorphism. Note that Aut(Z/7) ≈ Z/6, and so it has a unique subgroup of order 3, which must be the image of β. Consequently, up to isomorphism, there are exactly two other groups of order 63, namely

G ≈ Z/7 oβ Z/9 and G ≈ Z/7 oβ (Z/3 × Z/3).