48 CHAPTER 2. THE STRUCTURE OF GROUPS 2.5 The Sylow Theorems In this section, let p denote a prime, and G a finite group. Lem 2.28. If a group H of order pn acts on a finite set S and if SH := fx 2 S j hx = x for all h 2 Hg; then jSj ≡ jSH j mod p. Proof. Let Ox denote the H-orbit of x 2 S. Then S can be partitioned into a disjoint union of H-orbits H S = S [Ox1 [···[Oxk ; H where p divides jOxi j = [H : Hxi ] for i = 1; ··· ; k. Therefore, jSj ≡ jS j mod p. Thm 2.29 (Cauchy). If G is a finite group whose order is divisible by a prime p, then G contains an element of order p. Proof. (J. H. McKay) Let S be the set of p-tuples of group elements f(a1; a2; ··· ; ap) j ai 2 −1 p−1 G and a1a2 ··· ap = eg. Then ap = (a1a2 ··· ap−1) . Therefore jSj = jGj and thus jSj ≡ 0 mod p. Let Zp act on S by k · ((a1; a2; ··· ; ap)) := (ak+1; ··· ; ap; a1; ··· ; ak). This p p action is well-defined. By Lemma 2.28, jSZ j ≡ jSj ≡ 0 mod p. Note that (a1; ··· ; ap) 2 SZ p p if and only if a1 = a2 = ··· = ap, and (e; e; ··· ; e) 2 SZ . There are at least p elements in SZ . If (a; a; ··· ; a) 2 SZp and a 6= e, then ap = e implies that a is an element of G of order p. A group in which every element has the order a power of p is called a p-group. n Cor 2.30. A finite group G is a p-group if and only if jGj = p for some n 2 N. Cor 2.31. The center C(G) of a nontrivial finite p-group G contains more than one element. P Proof. By the class equation: jGj = jC(G)j + [G : CG(xi)] and by p j [G : CG(xi)]. Lem 2.32. If H is a p-subgroup of a finite group G. Then [NG(H): H] ≡ [G : H] mod p. Proof. Let S be the set of left cosets of H in G. Then jSj = [G : H]. Let H act on S by left translation. xH 2 SH if and only if hxH = xH for all h 2 H, if and only if x−1hx 2 H for all H h 2 H, if and only if x 2 NG(H), if and only if xH 2 NG(H)=H. By Lemma 2.28, jS j ≡ jSj mod p, that is, [NG(H): H] ≡ [G : H] mod p. Thm 2.33 (First Sylow Theorem). Let G be a group of order pnm, with n ≥ 1, p prime, and gcd(p; m) = 1. Then 1. G contains a subgroup of order pi for every 1 ≤ i ≤ n. In particular, a subgroup of G of order pn is called a Sylow p-subgroup of G. 2.5. THE SYLOW THEOREMS 49 2. Every subgroup of order pi (i < n) is normal in some subgroup of order pi+1. Proof. G contains a subgroup hai of order p by Cauchy's Theorem. Proceeding by induction assume H is a subgroup of G of order pi (1 ≤ i < n). Then [NG(H): H] ≡ [G : H] ≡ 0 mod p by Lemma 2.32. Notice that H C NG(H). So p j jNG(H)=Hj. Applying Cauchy's Theorem, i+1 there is bH 2 NG(H)=H of order p. Then H C hb; Hi where jhb; Hij = p . This proves the First Sylow Theorem. As a corollary, every p-subgroup of G is contained in a Sylow p-subgroup of G. Thm 2.34 (Second Sylow Theorem). Any two Sylow p-subgroups of G are conjugate in G. Proof. Let H and P be two Sylow p-subgroups of G. Let S be the set of left cosets of P in G and let H act on S by left translation. By Lemma 2.28, jSH j ≡ jSj ≡ [G : P ] 6≡ 0 mod p. Hence there is xP 2 SH . Then hxP = xP for all h 2 H. In other words, x−1Hx ≤ P . This forces x−1Hx = P . Thus H and P are conjugate in G. Thm 2.35 (Third Sylow Theorem). The number of Sylow p-subgroups of G divides jGj and is of the form kp + 1 for some k ≥ 0. Proof. Let S be the set of all Sylow p-subgroups of G. Then jSj = [G : NG(P )], a divisor of jGj. Let P be any Sylow p-subgroup of G. Let P act on S by conjugation. An element P −1 H 2 S iff bHb = H for all b 2 P , iff P ≤ NG(H). In such case, both P and H are Sylow p-subgroups of NG(H). So they are conjugate in NG(H). This forces P = H since P P H C NG(H). Therefore, S = fP g. Lemma 2.28 implies that jSj ≡ jS j ≡ 1 mod p. 3 Ex. S4 is a group of order 24 = 2 · 3. D4 = h(1234); (12)(34)i is of order 8 and it is a Sylow 2-group of S4. How many Sylow 2-subgroups in S4? How many Sylow 3-subgroups? Prop 2.36. If P is a Sylow p-subgroup of a finite group G, then NG(NG(P )) = NG(P ). See [Hungerford, Algebra, GTM073]. Thm 2.37 (Frattini Argument). Let K be a normal subgroup of a finite group G. If P is a Sylow p-subgroup of K (for some prime p), then G = KNG(P ). Proof. If g 2 G, then gP g−1 ≤ gKg−1 = K. So gP g−1 is a Sylow p-group of K. Hence there −1 −1 −1 −1 −1 −1 is k 2 K such that gP g = kP k . Then (k g)P (k g) = P . Then k g 2 NG(P ). So g 2 KNG(P )..