Section 1.7. Group Actions 73 If σ = (1 2)(3 4), define τ = (1 2)(3 5). Now H contains (τστ −1)σ−1, because it is a normal subgroup, and τστ −1σ−1 = (3 5 4), as the reader should check. If σ = (1 2 3 4 5), define ρ = (1 3 2) now H contains ρσρ−1σ−1 = (1 3 4), as the reader should also check. We have shown that if H = {(1)}, then H contains a 3-cycle and, hence, H = A5. Therefore, A5 is simple. • Without much more effort, we can prove that the alternating groups An are simple for all n ≥ 5. Observe that A4 is not simple, for the four-group V is a normal subgroup of A4. Lemma 1.122. A6 is a simple group. Proof. Let H = {(1)} be a normal subgroup of A6 we must show that H = A6. Assume that there is some α ∈ H with α = (1) that fixes some i, where 1 ≤ i ≤ 6. Define F = {σ ∈ A6 : σ(i) = i}. Note that α ∈ H ∩ F , so that H ∩ F = {(1)}. The Second Isomorphism Theorem gives H ∩ F F . But F is simple, for F ∼ = A5, and so the only normal subgroups in F are {(1)} and F . Since H ∩ F = {(1)}, we have H ∩ F = F that is, F ⊆ H. It follows that H contains a 3-cycle (for F does), and so H = A6, by Exercise 1.99 on page 75. We may now assume that there is no α ∈ H with α = (1) that fixes some i with 1 ≤ i ≤ 6. If we consider the cycle structures of permutations in A6, however, any such α must have cycle structure (1 2)(3 4 5 6) or (1 2 3)(4 5 6). In the first case, α2 ∈ H is a nontrivial permutation that fixes 1 (and also 2), a contradiction. In the second case, H contains α(βα−1β−1), where β = (2 3 4), and it is easily checked that this is a nontrivial element in H which fixes 1, another contradiction. Therefore, no such normal subgroup H can exist, and so A6 is a simple group. • Theorem 1.123. An is a simple group for all n ≥ 5. Proof. If H is a nontrivial normal subgroup of An, then H = {(1)} and we must show that H = An. By Exercise 1.99 on page 75, it suﬃces to prove that H contains a 3-cycle. If β ∈ H is nontrivial, then there exists some i that β moves say, β(i) = j = i. Choose a 3-cycle α that fixes i and moves j. The permutations α and β do not commute: βα(i) = β(i) = j, while αβ(i) = α(j) = j. It follows that γ = (αβα−1)β−1 is a nontrivial element of H. But βα−1β−1 is a 3-cycle, by Theorem 1.9, and so γ = α(βα−1β−1) is a product of two 3-cycles. Hence, γ moves at most 6 symbols, say, i1,...,i6 (if γ moves fewer than 6 symbols, just adjoin others so we have a list of 6). Define F = {σ ∈ An : σ fixes all i = i1,...,i6}. Now F ∼ = A6 and γ ∈ H ∩ F . Hence, H ∩ F is a nontrivial normal subgroup of F . But F is simple, being isomorphic to A6, and so H ∩ F = F that is, F ⊆ H. Therefore, H contains a 3-cycle, and so H = An the proof is complete. •

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