Mathematics 352 – Abstract Algebra II Solutions for Problem Set 3 February 4, 2008

Section 8.4

6. Let H ⊆ K be subgroups of G such that H is normal as a subgroup of K. This means that for all x ∈ K, x^{−1}Hx = H. By the definition, this means x ∈ N(H), so K ⊆ N(H).

7. (a) Let A ⊆ G be a subgroup. Let x ∈ A. Then x^{−1}Ax ⊆ A since A is closed under inverses and products. On the other hand if y ∈ A is an arbitrary element, then xyx^{−1} ∈ A since A is a subgroup and hence x^{−1}(xyx^{−1})x = y ∈ x^{−1}Ax. Therefore A ⊆ x^{−1}Ax, so combining the two inclusions, A = x^{−1}Ax. This shows A ⊂ N(A).

(b) We have

g ∈ N(A) ⟺ g^{−1}Ag = A by definition
⟺ Ag = gA (multiply by g on left).

12. Let K be a Sylow p-subgroup of G, and N be a normal subgroup of G. Assume that K is a normal subgroup of N. We claim first that this means that K is also a Sylow p-subgroup of N. Assume |G| = p^{t}m with (p, m) = 1, so |K| = p^{t}. But by Lagrange’s theorem, |N| divides |G|, and so the largest power of p that can divide N is p^{t}. Since K ⊆ N, we must have |N| = p^{t}m′ where (p, m′) = 1, and hence K is also a Sylow p-subgroup of G. Now let x be any element of G. Since N is normal in G, x^{−1}Nx = N. Since K ⊆ N, this implies x^{−1}Kx ⊆ N. But since |x^{−1}Kx| = |K|, it follows that x^{−1}Kx = K is also a Sylow p-subgroup of N. Since K is normal in N, Sylow 2 implies that there is just one such Sylow p-subgroup of N, namely K. This means that x^{−1}Kx = K. Therefore, K is also normal in G.

Section 8.5

1. We have |G| = p^{2}q with p, q prime, p < q and q ≢ 1 mod p. The only hypothesis we are missing to apply Theorem 8.30 is the one that says p^{2} ≢ 1 mod q. Suppose that q|(p^{2} − 1). Then since q is prime, q|(p + 1) or q|(p − 1). The second is impossible since p < q so p − 1 < q as well. The first is also impossible. Suppose that q|(p + 1). There are two cases to consider. First, if p = 2, then q = 3 and this contradicts q ≢ 1 mod p. Second, if p is an odd prime, then p + 1 is even, p + 1 = 2ℓ for some ℓ < p. It must be the case that q|ℓ since q > p, so q is also an odd prime. But this is also impossible since p < q. Hence the hypothesis for Theorem 8.30 is satisfied, and G must be abelian.

6. The n for which these theorems are sufficient are

Theorem 7.28 (prime order) applies to n = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
applies to 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
Corollary 8.29 (order p^{2}) applies to n = 4, 9, 25, 49
Theorem 8.33 (order 2p) applies to n = 6, 10, 14, 22, 26, 34, 38, 46, 58, 62, 74, 82, 86, 94
Corollary 8.18 (order pq) applies to n = 15, 33, 35, 51, 65, 69, 77, 85, 87, 91, 95
Theorem 8.30 (order p^{2}q) applies to n = 45, 99
Theorem 8.34 (order 8) applies to n = 8
Theorem 8.35 (order 12) applies to n = 12
none of these apply to n = 16, 18, 20, 21, 24, 27, 28, 30, 32, 36, 39,
40, 42, 44, 48, 50, 52, 54, 55, 56, 57, 60, 63
64, 66, 68, 70, 72, 75, 76, 78, 80, 81, 84, 88
90, 92, 93, 96, 98, 100.

8. (a) We have

(\begin{matrix}±1 & a \\ 0 & 1 \end{matrix}) (\begin{matrix}±1 & b \\ 0 & 1 \end{matrix}) = (\begin{matrix}±1 & a ± b \\ 0 & 1 \end{matrix}) .

Moreover,

(\begin{matrix}±1 & a \\ 0 & 1 \end{matrix})^{−1} = (\begin{matrix}±1 & ∓a \\ 0 & 1 \end{matrix})

Hence this set of matrices is closed under products and inverses and forms a subgroup of GL(2, ℤ_{n}).

(b) Let R = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) . Then R^{n} = I (the identity matrix). Similarly let D = (\begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix}) so

D^{2} = I. We have

DR = (\begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) = (\begin{matrix}−1 & −1 \\ 0 & 1 \end{matrix}) = (\begin{matrix}1 & −1 \\ 0 & 1 \end{matrix}) (\begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix}) = R^{−1}D.

Therefore the mapping given by R^{i}D^{j} ↦ r^{i}d^{j} is an isomorphism from this group of matrices to D_{n}.

10. We have D_{n} = ⟨r, d : d^{n} = e, r^{2} = e, dr = r^{−1}d⟩, and every element can be written uniquely as r^{i}d^{j} with 0 ≤ i ≤ n − 1 and 0 ≤ j ≤ 1. (a) If n = 2k, then r^{i}r^{k} = r^{i+k} = r^{k}r^{i}, so r^{k} commutes with all the elements of the subgroup generated by r. Moreover, applying the relation dr = r^{−1}d repeatedly,

(r^{i}d)r^{k} = r^{i}(dr^{k}) = r^{i}(r^{−1})^{k}d = r^{i}(r^{k})^{−1}d = (r^{i}r^{k})d = (r^{k}r^{i})d = r^{k}(r^{i}d).

Therefore r^{k} ∈ Z(D_{n}).

(b) We show that every other element of D_{n} besides e, r^{k} does not commute with something, which will imply that Z(D_{n}) = {e, r^{k}}. Consider the elements r^{i} for i ≠ 0, k first. We have dr^{i} = (r^{−1})^{i}d = (r^{i})^{−1}d = r^{n−i}d. Since i ≠ k, i ≢ n − i mod n, and this shows that r^{i} ∉ Z(D_{n}). Similarly, the element r^{i}d does not commute with r:

(r^{i}d)r = r^{i}(dr) = r^{i}r^{−1}d = r^{i−1}d but r(r^{i}d) = r^{i+1}d.

Therefore r^{i}d ∉ Z(D_{n}) for all i.

(c) The same reasoning as in part (b) shows that when n is odd, Z(D_{n}) = {e} since there is no element such that (r^{i})^{−1} = r^{i} except r^{0} = e.

11. By consulting the group table, Z(Q) = {1, −1}.