Abstract Algebra Theory and Applications - Computer Science ...

13.1 THE SYLOW THEOREMS 223Proof. We define a bijection between the conjugacy classes of K andthe right cosets of N(K) ∩ H by h −1 Kh ↦→ (N(K) ∩ H)h. To show thatthis map is a bijection, let h 1 , h 2 ∈ H and suppose that (N(K) ∩ H)h 1 =(N(K) ∩ H)h 2 . Then h 2 h −11 ∈ N(K). Therefore, K = h 2 h −11 Kh 1h −12 orh −11 Kh 1 = h −12 Kh 2, and the map is an injection. It is easy to see that thismap is surjective; hence, we have a one-to-one and onto map between theH-conjugates of K and the right cosets of N(K) ∩ H in H.□Theorem 13.6 (Second Sylow Theorem) Let G be a finite group and pa prime dividing |G|. Then all Sylow p-subgroups of G are conjugate. Thatis, if P 1 and P 2 are two Sylow p-subgroups, there exists a g ∈ G such thatgP 1 g −1 = P 2 .Proof. Let P be a Sylow p-subgroup of G and suppose that |G| = p r mand |P | = p r . LetP = {P = P 1 , P 2 , . . . , P k }consist of the distinct conjugates of P in G. By Lemma 13.5, k = [G : N(P )].Notice that|G| = p r m = |N(P )| · [G : N(P )] = |N(P )| · k.Since p r divides |N(P )|, p cannot divide k. Given any other Sylow p-subgroup Q, we must show that Q ∈ P. Consider the Q-conjugacy classes ofeach P i . Clearly, these conjugacy classes partition P. The size of the partitioncontaining P i is [Q : N(P i ) ∩ Q]. Lagrange’s Theorem tells us that thisnumber is a divisor of |Q| = p r . Hence, the number of conjugates in everyequivalence class of the partition is a power of p. However, since p does notdivide k, one of these equivalence classes must contain only a single Sylowp-subgroup, say P j . Therefore, for some P j , x −1 P j x = P j for all x ∈ Q. ByLemma 13.4, P j = Q.□Theorem 13.7 (Third Sylow Theorem) Let G be a finite group and letp be a prime dividing the order of G. Then the number of Sylow p-subgroupsis congruent to 1 (mod p) and divides |G|.Proof. Let P be a Sylow p-subgroup acting on the set of Sylow p-subgroups,P = {P = P 1 , P 2 , . . . , P k },by conjugation. From the proof of the Second Sylow Theorem, the onlyP -conjugate of P is itself and the order of the other P -conjugacy classes is a