Abstract Algebra Theory and Applications - Computer Science ...

13.1 THE SYLOW THEOREMS 221Proof. We will use induction on the order of G. If |G| = p, then clearly Gmust have an element of order p. Now assume that every group of order k,where p ≤ k < n and p divides k, has an element of order p. Assume that|G| = n and p | n and consider the class equation of G:|G| = |Z(G)| + [G : C(x 1 )] + · · · + [G : C(x k )].We have two cases.Case 1. The order of one of the centralizer subgroups, C(x i ), is divisibleby p for some i, i = 1, . . . , k. In this case, by our induction hypothesis, weare done. Since C(x i ) is a proper subgroup of G and p divides |C(x i )|, C(x i )must contain an element of order p. Hence, G must contain an element oforder p.Case 2. The order of no centralizer subgroup is divisible by p. Then pdivides [G : C(x i )], the order of each conjugacy class in the class equation;hence, p must divide the center of G, Z(G). Since Z(G) is abelian, it musthave a subgroup of order p by the Fundamental Theorem of Finite AbelianGroups. Therefore, the center of G contains an element of order p. □Corollary 13.2 Let G be a finite group. Then G is a p-group if and onlyif |G| = p n .Example 1. Let us consider the group A 5 . We know that |A 5 | = 60 =2 2 · 3 · 5. By Cauchy’s Theorem, we are guaranteed that A 5 has subgroupsof orders 2, 3 and 5. The Sylow Theorems give us even more informationabout the possible subgroups of A 5 .We are now ready to state and prove the first of the Sylow Theorems.The proof is very similar to the proof of Cauchy’s Theorem.Theorem 13.3 (First Sylow Theorem) Let G be a finite group and p aprime such that p r divides |G|. Then G contains a subgroup of order p r .Proof. We induct on the order of G once again. If |G| = p, then we aredone. Now suppose that the order of G is n with n > p and that the theoremis true for all groups of order less than n. We shall apply the class equationonce again:|G| = |Z(G)| + [G : C(x 1 )] + · · · + [G : C(x k )].First suppose that p does not divide [G : C(x i )] for some i. Thenp r | |C(x i )|, since p r divides |G| = |C(x i )| · [G : C(x i )]. Now we can applythe induction hypothesis to C(x i ).