72 Chapter 1. Groups I We are now going to show that A5 is a nonabelian simple group (indeed, it is the smallest such there is no nonabelian simple group of order less than 60). This was first proved by Galois, and it is the key to showing that quintic polynomials are not solvable by radicals. Suppose that an element x ∈ G has k conjugates. If there is a subgroup H ⊆ G with x ∈ H ⊆ G, how many conjugates does x have in H? Since xH = {hxh−1 : h ∈ H} ⊆ {gxg−1 : g ∈ G} = xG, we have |xH | ≤ |xG|. It is possible that there is strict inequality |xH | |xG|. For example, if G = S3, x = (1 2), and H = x , then |xH | = 1 (because H is abelian) and |xG| = 3 (because all transpositions are conjugate in S3). Let us now consider this question for G = S5, x = (1 2 3), and H = A5. Lemma 1.119. All 3-cycles are conjugate in A5. Proof. Let G = S5, α = (1 2 3), and H = A5. We know that |αS5 | = 20, for there are twenty 3-cycles in S5, as we saw in Table 2 on page 10. Therefore, 20 = |S5|/|CS5 (α)| = 120/|CS5 (α)|, by Corollary 1.109, and so |CS5 (α)| = 6 that is, there are exactly six permutations in S5 that commute with α. Here they are: (1), (1 2 3), (1 3 2), (4 5), (4 5)(1 2 3), (4 5)(1 3 2). The last three of these are odd permutations, so that |CA5 (α)| = 3. We conclude that |αA5 | = |A5|/|CA5 (α)| = 60/3 = 20 that is, all 3-cycles are conjugate to α = (1 2 3) in A5. • This lemma can be generalized from A5 to all An for n ≥ 5 see Exercise 1.99 on page 75. Lemma 1.120. If n ≥ 3, every element in An is a product of 3-cycles. Proof. If α ∈ An, then α is a product of an even number of transpositions: α = τ1τ2 · · · τ2q−1τ2q. Of course, we may assume that adjacent τ’s are distinct. As the transpositions may be grouped in pairs τ2i−1τ2i, it suﬃces to consider products ττ , where τ and τ are transpositions. If τ and τ are not disjoint, then τ = (i j), τ = (i k), and ττ = (i k j) if τ and τ are disjoint, then ττ = (i j)(k ) = (i j)(j k)(j k)(k ) = (i j k)(j k ). • Theorem 1.121. A5 is a simple group. Proof. We shall show that if H is a normal subgroup of A5 and H = {(1)}, then H = A5. Now if H contains a 3-cycle, then normality forces H to contain all its conjugates. By Lemma 1.119, H contains every 3-cycle, and by Lemma 1.120, H = A5. Therefore, it suﬃces to prove that H contains a 3-cycle. As H = {(1)}, it contains some σ = (1). We may assume, after a harmless relabeling, that either σ = (1 2 3), σ = (1 2)(3 4), or σ = (1 2 3 4 5). As we have just remarked, we are done if σ is a 3-cycle.

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