# Sylow Theorems

Let $G$ be a finite group.

## Theorem 1

For every prime factor $p$ with multiplicity $n$ of the order of $G$, there exists a Sylow $p$-subgroup of $G$, of order $p^n$.

## Theorem 2

All Sylow $p$-subgroups of $G$ are conjugate to each other, i.e.\ if $H$ and $K$ are Sylow $p$-subgroups of $G$, then there exists an element $g\in G$ with $g^{-1}Hg=K$.

## Theorem 3

Let $p$ be a prime such that $|G|=p^nm$, where $p\nmid m$. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then:
1) $n_p\mid m$, which is the index of the Sylow $p$-subgroup in $G$.
2) $n_p\equiv 1\pmod p$.

## Theorem 3b (Proof)

We have $n_p=[G:N_G(P)]$, where $P$ is any Sylow $p$-subgroup of $G$ and $N_G$ denotes the normalizer.

### Proof

Let $P$ be a Sylow $p$-subgroup of $G$ and let $G$ act on $\text{Syl}_p(G)$ by conjugation. We have $|\text{Orb}(P)|=n_p$, $\text{Stab}(P)=\{g\in G:gPg^{-1}=P\}=N_G(P)$.

By the Orbit-Stabilizer Theorem, $|\text{Orb}(P)|=[G:\text{Stab}(P)]$, thus $n_p=[G:N_G(P)]$.

## Orbit-Stabilizer Theorem

Let $G$ be a group which acts on a finite set $X$. Then $\displaystyle |\text{Orb}(x)|=[G:\text{Stab}(x)]=\frac{|G|}{|\text{Stab}(x)|}.$ 