1. Since |S₄| = 2³ · 3, the Sylow 3-subgroups are all cyclic of order 3, and thus have the form P = γ , γ = (i j k). Now σ(123)σ⁻¹ = (σ(1)σ(2) σ(3)) for all σ S₄ (Lemma 3 §2.8) so let where {1, 2, 3, 4} = {i, j, k, x}. Then σ(123)σ⁻¹ = (ijk) so γ and (1 2 3) are conjugate. Hence P and (123) are conjugate.

3. P is a Sylow p-subgroup of N(P), being a p-subgroup of maximal order. It is unique because it is normal in N(P).

5. Let |G| = 1001 = 7 · 11 · 13. We have n₇ = 1, 11, 13, 143 and n₇ ≡ 1 (mod 7), so n₇ = 1. Similarly n₁₁ = 1, 7, 13, 91 and n₁₁ ≡ 1 (mod 11), so n₁₁ = 1; and n₁₃ = 1, 7, 11, 77 and n₁₃ ≡ 1 (mod 13) so n₁₃ = 1. Thus let H G, K G, L G have order |H| = 7, |K| = 11 and |L| = 13. Then H ∩ K = {1} so HK ≅ H × K ≅ C₇₇. Now HK ∩ L = {1} so (HK)L ≅ HK × L ≅ C₁₀₀₁. Thus G = HKL ≅ C₁₀₀₁ is unique up to isomorphism.

7.

8. If |G| = 520 = 2³ · 5 · 13, then n₁₃ = 1, 2, 4, 8, 18, 20, 40, 80 and n₁₃ ≡ 1 (mod 13), so n₁₃ = 1, 40. Similarly n₅ = 1, 2, 4, 8, 13, 26, 52, 104, and n₅ ≡ 1 (mod 5) so n₅ = 1, 26. If either n₁₃ = 1 or n₅ = 1 we are done. Otherwise there are 40 · 12 = 480 elements of order 13 and 26 · 4 = 104 elements of order 5, giving 480 + 104 = 584 in all, a contradiction.

9.

10. a. If |G| = 385 = 5 · 7 · 11 then n₁₁ = 1 and n₇ = 1, so let P G, Q G satisfy |P| = 11 and |Q| = 7. Then

11. a. If |G| = 105 = 3 · 5 · 7, then n₇ = 1, 15 and n₅ = 1, 21. Let |P| = 7 and |Q| = 5. If neither P nor Q is normal in G, then G has 15 · 6 = 90 elements of order 7 and 21 · 4 = 84 elements of order 5, a contradiction. So P G or Q G, whence PQ is a subgroup and |PQ| = |P||Q| = 35 because P ∩ Q = {1}. Since |G : PQ| = 3, let θ : G → S₃ have ker θ ⊆ PQ. Clearly | ker θ| ≠ 1, 5, 7 so ker θ = PQ and PQ G. Finally |PQ| = 35 = 7 · 5 means P PQ and Q PQ, so PQ ≅ P × Q ≅ C₃₅.

13. Let P be a Sylow p-subgroup of G. Then |P| = pⁿ, and |G : P| = m because p > m. Thus, by Theorem 1 §8.3 there is a homomorphism θ : G → S_m with ker θ ⊆ P. If | ker θ| = p^k then p^n−k = |G/ker θ| divides m ! . Since p > m this means that k = n, whence P = ker θ G.

15. α(P) is clearly a p-subgroup of G so α(P) ⊆ a⁻¹Pa for some a G by Theorem 2. But a⁻¹Pa = P here because P G.

17. For convenience, write N = N(P) . Let gP N/P have order p^k, g N. We must show that g P. We have so o(g) is a power of p. Thus g is a p-subgroup of N. Since P is a Sylow p-subgroup of N, Theorem 2 gives g ⊆ a⁻¹Pa for some a N. But P N so a⁻¹Pa = P. Thus g P, and gP = P.

18. a. We have P ⊆ N(P) ⊆ H. If a N(H) then a⁻¹Pa ⊆ a⁻¹Ha = H, so P and a⁻¹Pa are both Sylow p -subgroups of H. By Sylow's second theorem, h⁻¹(a⁻¹Pa)h = P for some h H. Thus ah N(P) ⊆ H, whence a H.

19. If Q is also a Sylow p-subgroup of G, then Q = a⁻¹Pa by Sylow's second theorem. If g N(Q) then Q = g⁻¹Qg; that is a⁻¹Pa = g⁻¹a⁻¹Pag. This implies that aga⁻¹ N(P) = P, whence g a⁻¹Pa = Q.

21. K is clearly a p-subgroup. If a G and P is a Sylow p -subgroup, then aPa⁻¹ is also a Sylow p-subgroup, so K ⊆ aPa⁻¹. Hence a⁻¹Ka ⊆ P; since P was an arbitrary Sylow p -subgroup, a⁻¹Ka ⊆ K. Thus K G. Now let H be any normal p-subgroup of G. If H is a normal p-subgroup of G, then H ⊆ aPa⁻¹ for some a G. Thus H = a⁻¹Ha ⊆ P; whence H ⊆ K.

23. If k|n, let n = kd and a₁ = a^d. Then o(a₁) = k and a₁ba₁ = a^dba^d = b. Finally, let k = 2m₁, 2(m₁d) = kd = n = 2m, so m₁d = m. Thus , and so a₁, b ≅ Q_k.

25. If p and q are distinct primes, any group of one of the following orders is not simple: pⁿ (Theorem 8 §8.2), pq (Example 5), p²q (Exercise 14) and p²q² (Example 9). The only remaining orders (apart from primes) in the range 2–59 are as follows, with the reason that such a group G is not simple (We also include |G| = 36 —see Theorem 4).