2.5 the Sylow Theorems

Home , Finite group, Subgroup, Sylow theorems

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 := {x ∈ S | hx = x for all h ∈ H}, then |S| ≡ |SH | mod p.

Proof. Let Ox denote the H-orbit of x ∈ S. Then S can be partitioned into a disjoint union of H-orbits H S = S ∪ Ox1 ∪ · · · ∪ Oxk , H where p divides |Oxi | = [H : Hxi ] for i = 1, ··· , k. Therefore, |S| ≡ |S | 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 {(a1, a2, ··· , ap) | ai ∈ −1 p−1 G and a1a2 ··· ap = e}. Then ap = (a1a2 ··· ap−1) . Therefore |S| = |G| and thus |S| ≡ 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, |SZ | ≡ |S| ≡ 0 mod p. Note that (a1, ··· , ap) ∈ SZ p p if and only if a1 = a2 = ··· = ap, and (e, e, ··· , e) ∈ SZ . There are at least p elements in SZ . If (a, a, ··· , a) ∈ 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 |G| = p for some n ∈ 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: |G| = |C(G)| + [G : CG(xi)] and by p | [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 |S| = [G : H]. Let H act on S by left translation. xH ∈ SH if and only if hxH = xH for all h ∈ H, if and only if x−1hx ∈ H for all H h ∈ H, if and only if x ∈ NG(H), if and only if xH ∈ NG(H)/H. By Lemma 2.28, |S | ≡ |S| 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 | |NG(H)/H|. Applying Cauchy’s Theorem, i+1 there is bH ∈ NG(H)/H of order p. Then H C hb, Hi where |hb, Hi| = 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, |SH | ≡ |S| ≡ [G : P ] 6≡ 0 mod p. Hence there is xP ∈ SH . Then hxP = xP for all h ∈ 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 |G| 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 |S| = [G : NG(P )], a divisor of |G|. Let P be any Sylow p-subgroup of G. Let P act on S by conjugation. An element P −1 H ∈ S iﬀ bHb = H for all b ∈ 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 = {P }. Lemma 2.28 implies that |S| ≡ |S | ≡ 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 ∈ 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 ∈ K such that gP g = kP k . Then (k g)P (k g) = P . Then k g ∈ NG(P ). So g ∈ KNG(P ).