Abstract Algebra Theory and Applications - Computer Science ...

5.2 LAGRANGE’S THEOREM 93The converse of Lagrange’s Theorem is false. The group A 4 has order12; however, it can be shown that it does not possess a subgroup of order6. According to Lagrange’s Theorem, subgroups of a group of order 12 canhave orders of either 1, 2, 3, 4, or 6. However, we are not guaranteed thatsubgroups of every possible order exist. To prove that A 4 has no subgroupof order 6, we will assume that it does have a subgroup H such that |H| = 6and show that a contradiction must occur. The group A 4 contains eight3-cycles; hence, H must contain a 3-cycle. We will show that if H containsone 3-cycle, then it must contain every 3-cycle, contradicting the assumptionthat H has only 6 elements.Theorem 5.9 Two cycles τ and µ in S n have the same length if and onlyif there exists a σ ∈ S n such that µ = στσ −1 .Proof. Suppose thatDefine σ to be the permutationτ = (a 1 , a 2 , . . . , a k )µ = (b 1 , b 2 , . . . , b k ).σ(a 1 ) = b 1σ(a 2 ) = b 2σ(a k ) = b k .Then µ = στσ −1 .Conversely, suppose that τ = (a 1 , a 2 , . . . , a k ) is a k-cycle and σ ∈ S n . Ifσ(a i ) = b and σ(a (i mod k)+1 ) = b ′ , then µ(b) = b ′ . Hence,µ = (σ(a 1 ), σ(a 2 ), . . . , σ(a k ))..Since σ is one-to-one and onto, µ is a cycle of the same length as τ.□Corollary 5.10 The group A 4 has no subgroup of order 6.Proof. Since [A 4 : H] = 2, there are only two cosets of H in A 4 . Inasmuchas one of the cosets is H itself, right and left cosets must coincide; therefore,gH = Hg or gHg −1 = H for every g ∈ A 4 . By Theorem 5.9, if H containsone 3-cycle, then it must contain every 3-cycle, contradicting the order of H.□