Abstract Algebra Theory and Applications - Computer Science ...

160 CHAPTER 9 HOMOMORPHISMS AND FACTOR GROUPSis also in N. Sinceσ −1 (a 1 a 2 a 3 )σ(a 1 a 2 a 3 ) −1= σ −1 (a 1 a 2 a 3 )σ(a 1 a 3 a 2 )= (a 1 a 2 · · · a r ) −1 τ −1 (a 1 a 2 a 3 )τ(a 1 a 2 · · · a r )(a 1 a 3 a 2 )= (a 1 a r a r−1 · · · a 2 )(a 1 a 2 a 3 )(a 1 a 2 · · · a r )(a 1 a 3 a 2 )= (a 1 a 3 a r ),N must contain a 3-cycle; hence, N = A n .Now suppose that N contains a disjoint product of the formσ = τ(a 1 a 2 a 3 )(a 4 a 5 a 6 ).Thensinceσ −1 (a 1 a 2 a 4 )σ(a 1 a 2 a 4 ) −1 ∈ N(a 1 a 2 a 4 )σ(a 1 a 2 a 4 ) −1 ∈ N.Soσ −1 (a 1 a 2 a 4 )σ(a 1 a 2 a 4 ) −1= [τ(a 1 a 2 a 3 )(a 4 a 5 a 6 )] −1 (a 1 a 2 a 4 )τ(a 1 a 2 a 3 )(a 4 a 5 a 6 )(a 1 a 2 a 4 ) −1= (a 4 a 6 a 5 )(a 1 a 3 a 2 )τ −1 (a 1 a 2 a 4 )τ(a 1 a 2 a 3 )(a 4 a 5 a 6 )(a 1 a 4 a 2 )= (a 4 a 6 a 5 )(a 1 a 3 a 2 )(a 1 a 2 a 4 )(a 1 a 2 a 3 )(a 4 a 5 a 6 )(a 1 a 4 a 2 )= (a 1 a 4 a 2 a 6 a 3 ).So N contains a disjoint cycle of length greater than 3, and we can applythe previous case.Suppose N contains a disjoint product of the form σ = τ(a 1 a 2 a 3 ), whereτ is the product of disjoint 2-cycles. Since σ ∈ N, σ 2 ∈ N, andσ 2 = τ(a 1 a 2 a 3 )τ(a 1 a 2 a 3 )= (a 1 a 3 a 2 ).So N contains a 3-cycle.The only remaining possible case is a disjoint product of the formσ = τ(a 1 a 2 )(a 3 a 4 ),where τ is the product of an even number of disjoint 2-cycles. Butσ −1 (a 1 a 2 a 3 )σ(a 1 a 2 a 3 ) −1