Abstract Algebra Theory and Applications - Computer Science ...

208 CHAPTER 12 GROUP ACTIONSExample 10. For S n it takes a bit of work to find the conjugacy classes. Webegin with cycles. Suppose that σ = (a 1 , . . . , a k ) is a cycle and let τ ∈ S n .By Theorem 5.9,τστ −1 = (τ(a 1 ), . . . , τ(a k )).Consequently, any two cycles of the same length are conjugate. Now letσ = σ 1 σ 2 · · · σ r be a cycle decomposition, where the length of each cycle σ iis r i . Then σ is conjugate to every other τ ∈ S n whose cycle decompositionhas the same lengths.The number of conjugate classes in S n is the number of ways in whichn can be partitioned into sums of positive integers. For example, we canpartition the integer 3 into the following three sums:3 = 1 + 1 + 13 = 1 + 23 = 3;therefore, there are three conjugacy classes. The problem of finding thenumber of such partitions for any positive integer n is what computer scientistscall NP-complete. This effectively means that the problem cannot besolved for a large n because the computations would be too time-consumingfor even the largest computer.Theorem 12.4 Let G be a group of order p n where p is prime. Then Ghas a nontrivial center.Proof. We apply the class equation|G| = |Z(G)| + n 1 + · · · + n k .Since each n i > 1 and n i | G, p must divide each n i . Also, p | |G|; hence, pmust divide |Z(G)|. Since the identity is always in the center of G, |Z(G)| ≥1. Therefore, |Z(G)| ≥ p and there exists some g ∈ Z(G) such that g ≠ 1.□Corollary 12.5 Let G be a group of order p 2 where p is prime. Then G isabelian.Proof. By Theorem 12.4, |Z(G)| = p or p 2 . If |Z(G)| = p 2 , then weare done. Suppose that |Z(G)| = p. Then Z(G) and G/Z(G) both haveorder p and must both be cyclic groups. Choosing a generator aZ(G) for