Abstract Algebra Theory and Applications - Computer Science ...

9.2 GROUP HOMOMORPHISMS 159Proof. We will first show that A n is generated by 3-cycles of the specificform (ijk), where i and j are fixed in {1, 2, . . . , n} and we let k vary. Every3-cycle is the product of 3-cycles of this form, since(iaj) = (ija) 2(iab) = (ijb)(ija) 2(jab) = (ijb) 2 (ija)(abc) = (ija) 2 (ijc)(ijb) 2 (ija).Now suppose that N is a nontrivial normal subgroup of A n for n ≥ 3 suchthat N contains a 3-cycle of the form (ija). Using the normality of N, wesee that[(ij)(ak)](ija) 2 [(ij)(ak)] −1 = (ijk)is in N. Hence, N must contain all of the 3-cycles (ijk) for 1 ≤ k ≤ n. ByLemma 9.5, these 3-cycles generate A n ; hence, N = A n .□Lemma 9.7 For n ≥ 5, every normal subgroup N of A n contains a 3-cycle.Proof. Let σ be an arbitrary element in a normal subgroup N. There areseveral possible cycle structures for σ.• σ is a 3-cycle.• σ is the product of disjoint cycles, σ = τ(a 1 a 2 · · · a r ) ∈ N, where r > 3.• σ is the product of disjoint cycles, σ = τ(a 1 a 2 a 3 )(a 4 a 5 a 6 ).• σ = τ(a 1 a 2 a 3 ), where τ is the product of disjoint 2-cycles.• σ = τ(a 1 a 2 )(a 3 a 4 ), where τ is the product of an even number of disjoint2-cycles.If σ is a 3-cycle, then we are done. If N contains a product of disjointcycles, σ, and at least one of these cycles has length greater than 3, sayσ = τ(a 1 a 2 · · · a r ), thenis in N since N is normal; hence,(a 1 a 2 a 3 )σ(a 1 a 2 a 3 ) −1σ −1 (a 1 a 2 a 3 )σ(a 1 a 2 a 3 ) −1