Sylow’s Theorems

Sylow’s Theorem I

Let $G$ be a group of order $p^{n}m$, where $p$ is a prime and $p\nmid m$. Then $G$ has a subgroup of order $p^n$, we call it $p$-Sylow subgroup.

Proof We use induction on $|G|$. Clearly for $|G|=1$ there is nothing to prove.

Sylow’s Theorem II

Let $G$ be a finite group and $p$ a prime dividing $|G|$. Then

• Any two $p$-Sylow subgroups in a finite group $G$ are conjugate
• Any $p$-subgroup of $G$ is contained in a $p$-Sylow subgroup
  

Sylow’s Theorem III

The number of $p$-Sylow subgroups in a finite group $G$ of order $p^{n}m$ with $(m,p)=1$ is of the form $1+kp$ and divides $m$.

Proof

Corollary

A group $G$ has exactly one $p$-Sylow subgroup $P$ if and only if $P$ is normal.