Normal p-Subgroup contained in All Sylow p-Subgroups
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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: $\S 59 \gamma$