Sylow's theorems

A group of order pⁿ, with p a prime number, is called a p-group. We shall examine actions of p-groups on various sets.

Let H be a p-group acting on a set S. Since the length of an orbit divides the order of the group, which is a power of p, it follows that the length of an orbit is either 1 or is divisible by p. Let $S_0=\{x\in S:h(x)=x,$ for all $x\in H\}$ be the set of orbits of length 1. The lengths of the orbits in S but not in S₀ are all divisible by p; their cardinalities are proper powers of p. We thus conclude as follows:

The mod p lemma (MPL)

If a p-group acts on S, then $\vert S\vert =\vert S_0\vert$ (mod p).


Theorem (Cauchy)

If a prime p divides the order of a group, then the group contains elements of order p.

Proof: Denote the group by G. Let $S=\{(g_1,\ldots ,g_p):g_i\in G$ and $\prod_ig_i=1\}.$ Since exactly p-1 elements can be freely selected from G into an n-tuple $(g_1,\ldots ,g_p)$, and the remaining element is uniquely determined as the inverse of the product of the elements already selected, it follows that the cardinality of S is $\vert G\vert^{p-1}.$ Let the cyclic group of order p, Z_p, act on S by cyclic shifts, specifically as follows:

$k(g_1,g_2,\ldots ,g_p)=(g_{k+1},g_{k+2},\ldots ,g_p,g_1,\ldots ,g_k).$

(Use the fact that gh=1 implies hg=1, to verify that this is indeed a group action.)

In this case $S_0=\{(g,\ldots ,g):g^p=1\}.$ Clearly S₀ is nonempty, since $(1,\ldots ,1)\in S_0$. On the other hand, the MPL informs us that

$0=\vert G\vert^{p-1}=\vert S\vert =\vert S_0\vert$ (mod p),

since p divides $\vert G\vert$, by assumption . We thus conclude that $\vert S_0\vert$ is divisible by p. This shows that there exist elements of order p in G. End of proof.

We write (a,b)=1 if integers a and b are relatively prime, that is, when they share no common primes in their prime factorizations.

A Sylow p-subgroup of G is a p-subgroup of G maximal with respect to inclusion on the p-subgroups of G.

Theorem (Sylow)

(a) If G is a group of order pⁿm, with p prime, $n\geq 1,$ and (p,m)=1, then G contains a subgroup of order pⁱ for each $1\leq i\leq n$ and every subgroup of G of order pⁱ, $1\leq i<n,$ is normal in some subgroup of order pⁱ⁺¹. In particular, Sylow p-subgroups exist.

(b) If H is a p-subgroup of G, and P is any Sylow p -subgroup of G, then $H\leq gPg^{-1},$ for some $g\in G.$ In particular, any two Sylow p-subgroups are conjugate in G.

(c) The number of Sylow p-subgroups of G divides the order of G and is equal to 1 modulo p.

Proof: (a) We proceed by induction on i. Since p divides $\vert G\vert ,$ elements of order p exist in G by Cauchy's theorem; such an element generates a subgroup of order p. Assume that H is a subgroup of order pⁱ, $1\leq i<n.$ Then p divides $\vert G:H\vert .$ Let S be the set of right cosets of H in G. Act with H on S by left translation, that is h (gH)=(hg)H. Then S₀ consists of cosets left fixed by H. Coset gH in in S₀ if and only if hgH=gH for all $h\in H;$ this happens if and only if $g\in N_G(H).$ We thus conclude that S₀ consists of exactly those cosets of H that are in N_G(H). By the MPL $\vert N_G(H):H\vert =\vert S_0\vert =\vert S\vert =\vert G: H\vert$ (mod p). Since $\vert G:H\vert$ is divisible by p, we conclude that p divides $\vert N_G(H):H\vert .$ By Cauchy's theorem there is a subgroup $\bar {T}$ of order p in N_G(H)/H. We are therefore assured the existence of a subgroup T in N_G(H), $H<T\leq N_G(H),$ such that $\vert T/H\vert =\vert\bar {T}\vert =p.$ The order of T is clearly $\vert H\vert\vert T/H\vert =$ pⁱp=pⁱ⁺¹. Since $H<T\leq N_G(H),$ H is normal in T.

(b) Let S be the set of right cosets of P in G. Act with H on S by left translation. Clearly $\vert S\vert =\vert G:P\vert$, and p does not divide $\vert G:P\vert$ since P is a Sylow p-subgroup. By the MPL $\vert S_0\vert =\vert S\vert\neq 0$ (mod p), and S₀ is, therefore, nonempty. There exists a coset gP in S₀.

$gP\in S_0$ $\Leftrightarrow$ hgP=gP for all $h\in H$ $\Leftrightarrow$ $H\leq gPg^{-1}.$

If H is a Sylow p-subgroup we necessarily have H=gPg^{- 1}.

(c) Let S be the set of all Sylow p-subgroups of G. By part (b) we know that if G acts by conjugation on S, the action is transitive, that is, there is one orbit only. The number of Sylow p-subgroups is the length of the orbit, which in turn divides the order of G.

Let P be a Sylow p-subgroup of G. Act with P on S by conjugation.

$Q\in S_0$ $\Leftrightarrow$ xQx^-1=Q, for all $x\in P$ $\Leftrightarrow$ $P\leq N_G(Q).$

As Sylow p-subgroups of N_G(Q), P and Q are conjugate in N_G(Q). But Q is normal in N_G(Q) and therefore P=Q. Thus $S_0 =\{P\}.$ By MPL $1=\vert S_0\vert =\vert S\vert$ (mod p). This end the proof.