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 $G$.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \gamma$