Section 1.7. Group Actions 75 Hint. If α = (1 2 3 4 5), then |CS5 (α)| = 5 because 24 = 120/|CS5 (α)| hence CS5 (α) = α . What is CA5 (α)? (iv) Prove that the conjugacy classes in A5 have sizes 1, 12, 12, 15, and 20. 1.98. (i) Prove that every normal subgroup H of a group G is a union of conjugacy classes of G, one of which is {1}. (ii) Use part (i) and Exercise 1.97 to give a second proof of the simplicity of A5. ∗ 1.99. (i) For all n ≥ 5, prove that all 3-cycles are conjugate in An. Hint. Show that (1 2 3) and (i j k) are conjugate by considering two cases: they are not disjoint (so they move at most 5 letters) they are disjoint. (ii) Prove that if a normal subgroup H of An contains a 3-cycle, where n ≥ 5, then H = An. (Remark. We have proved this in Lemma 1.120 when n = 5.) 1.100. Prove that the only normal subgroups of S4 are {(1)}, V, A4, and S4. Hint. Use Theorem 1.9, checking the various cycle structures one at a time. 1.101. Prove that A5 is a group of order 60 that has no subgroup of order 30. ∗ 1.102. (i) Prove, for all n ≥ 5, that the only normal subgroups of Sn are {(1)}, An, and Sn. Hint. If H Sn is a proper subgroup and H = An, then H ∩ An = {(1)}. (ii) Prove that if n ≥ 3, then An is the only subgroup of Sn of order 1 2 n!. Hint. If H is a second such subgroup, then H Sn and (H ∩ An) An. (iii) Prove that S5 is a group of order 120 having no subgroup of order 30. Hint. Use the representation on the cosets of a supposed subgroup of order 30, as well as the simplicity of A5. (iv) Prove that S5 contains no subgroup of order 40. ∗ 1.103. (i) Let σ, τ ∈ S5, where σ is a 5-cycle and τ is a transposition. Prove that S5 = σ, τ . (ii) Give an example showing that Sn, for some n, contains an n-cycle σ and a trans- position τ such that σ, τ = Sn. ∗ 1.104. Let G be a subgroup of Sn. (i) If G ∩ An = {1}, prove that |G| ≤ 2. (ii) If G is a simple group with more than two elements, prove that G ⊆ An. ∗ 1.105. (i) If n ≥ 5, prove that Sn has no subgroup of index r, where 2 r n. (ii) Prove that if n ≥ 5, then An has no subgroup of index r, where 2 ≤ r n. 1.106. (i) Prove that if a simple group G has a subgroup of index n 1, then G is isomorphic to a subgroup of Sn. Hint. Kernels are normal subgroups. (ii) Prove that an infinite simple group (such do exist) has no subgroups of finite index n 1. Hint. Use part (i).

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