# Normal p-Subgroup contained in All Sylow p-Subgroups

## Theorem

Let $G$ be a finite group.

Let $p$ be a prime number.

Let $H$ be a normal subgroup of $G$ which is a $p$-group.

Then $H$ is a subset of every Sylow $p$-subgroup of $G$.

## Proof

Let $P$ be a Sylow $p$-subgroup of $G$.

By Second Sylow Theorem, $H$ is a subset of a conjugate of $P$.

Then:

$\exists g \in G: H \subseteq g P g^{-1}$.

This implies:

 $\ds H$ $=$ $\ds g^{-1} H g$ Subgroup equals Conjugate iff Normal $\ds$ $\subseteq$ $\ds g^{-1} \paren {g P g^{-1} } g$ Subset Relation is Compatible with Subset Product/Corollary 2 $\ds$ $=$ $\ds P$

Since $P$ is arbitrary, $H$ is a subset of every Sylow $p$-subgroup of $G$.

$\blacksquare$