1. a. D₄ = {1, a, a², a³, b, ba, ba², ba³} where o(a) = 4, o(b) = 2 and aba = b. The classes are {1}, {a, a³}, {a²}, {b, ba²}, {ba, ba³}. Since normal subgroups are of orders 1, 2, 4, 8, they are {1}, {1, a²}, {1, a, a², a³}, {1, b, a², ba²}, {1, a², ba, ba³} and D₄.

3. If a^m = g⁻¹ag with g G, then

5. Let H = class a₁ ∪ . . . ∪ classa_n. Given g G and h H, let h classa_i. Then g⁻¹hg classa_i ⊆ H, so g⁻¹hg H. Then g⁻¹Hg ⊆ H, as required.

7. Let K = g⁻¹Hg, so H = gKg⁻¹. We claim N(K) = g⁻¹N(H)g. Let a N(K) so a⁻¹Ka = K. To show a g⁻¹N(H)g it suffices to show gag⁻¹ N(H). But

9. If |class a| = 2 then |G : N(a)| = 2 by Theorem 2. Hence N(a) is normal in G. Since N(a) ≠ G we are done if N(a) ≠ {1}. But N(a) = {1} implies |G| = |G : N(a)| = 2 so G is abelian and every conjugacy class is a singleton, a contradiction.

11. We have H ⊆ N(H) ⊆ G and H has finite index m in G. So

13. Write X = {g⁻¹Hg g G} and Y = {N(H)g g G}. Define ϕ : X → Y by ϕ(g⁻¹Hg) = N(H)g. Then:

15. Let α = γ₁γ₂. . . γ_r in S_n where the γ_i are disjoint cycles. Given σ S_n,

17. a. S₄ consists of ε, three permutations of type (a b)(c d), eight 3-cycles, six 2-cycles and six 4-cycles. These are the conjugacy classes by Exercise 15. So, by Theorem 1, each normal subgroup H S₄ is a union of these classes. Since |H| divides 24 = |S₄|, and since ε H, the only possibilities are H = ε, H = K, H = A₄ and H = S₄.

19. It suffices to show that G is abelian (then there are |G| conjugacy classes). So assume G is not abelian. Then there must be a nonsingleton conjugacy class, say class a. Thus there is only one singleton class, {1}. This means |G| = 1 + |class a| = 1 + |G : N(a)|. Write m = |G : N(a)| so that |G| = md, d ≥ 2. But then md = 1 + m gives 1 = m(d − 1) so m = 1, that is |class a| = 1, contrary to assumption.

21. If G/K and K are p-groups, let g G. Then in G/K for some n, so . Since K is itself a p-group, for some m; that is . Thus o(g) is a power of p, so G is a p-group. Conversely, subgroups of p-groups are clearly p-groups, as are images because if then

23. We have |G| = pⁿ for some n ≥ 1 so for all g G. It follows that in for all [g_i) in , so is a p-group. If 1 ≠ g and 1 ≠ h then [g, 1, 1, . . .), [1, g, 1, 1, . . .), . . . are all distinct in . So .

25. By Theorem 1, H is a union of G-conjugacy classes; suppose there are m of these which are singletons. The remaining classes in H have order a multiple of p by Theorem 3 (since G is a p-group) and p divides |H| (since H ≠ {1}). Hence p|m. Since m > 0 (because {1} ⊆ H) this means m > 1 and there exists a ≠ 1, such that class a = {a} ⊆ H. But class a = {a} means a Z(G), so a H ∩ Z(G).

26. a. Let |G| = p³, G nonabelian. Write Z = Z(G). Then Z ≠ {1} by Theorem 6 and Z ≠ G because Gisnonabelian so |Z| = p or p². If |Z| = p² then |G/Z| = p so G/Z is cyclic and G is abelian by Theorem 2 §2.9, contrary to assumption. So |Z| = p. Then |G/Z| = p² so G/Z is abelian by Theorem 8. Hence G′ ⊆ Z so G′ = {1} or G′ = Z. But G′ = {1} implies G is abelian. So G′ = Z. Finally, if K G, |K| = p, then G/K is abelian so G′ ⊆ K. Thus |K| = p = |Z| = |G′| implies K = G′.

27. We proceed by induction on n where |G| = pⁿ. If n = 1 it is clear. In general |H| = p^m. If H = {1} it is clear. If H ≠ {1} let K = H ∩ Z(G). Then K G and K ≠ {1} by Exercise 25. Let |K| = p^b. We have H/K G/K so, by induction, let where and for all i. Thus G_i G and |G_i/G_i+1| = p for all i. Similarly

29. Since C G, let Z[G/C] = K/C. Since |G/C| > 1, Theorem 6 shows C ⊂ K. But K G so K H by Exercise 26 §2.8. If k K then kC is in the center of G/C, so h⁻¹k⁻¹hk C H. Hence k⁻¹Hk ⊆ H, and similarly kHk⁻¹ ⊆ H. Thus k N(H) and we have shown K ⊆ N(H).

31. The associativity is verified as follows:

33.