# If a Subgroup Contains a Sylow Subgroup, then the Normalizer is the Subgroup itself ## Problem 117

Let $G$ be a finite group and $P$ be a nontrivial Sylow subgroup of $G$.
Let $H$ be a subgroup of $G$ containing the normalizer $N_G(P)$ of $P$ in $G$.

Then show that $N_G(H)=H$. Add to solve later

## Hint.

Use the conjugate part of the Sylow theorem.
See the second statement of the Sylow theorem.

## Proof.

It is clear that $H \subset N_G(H)$.
So we show that $N_G(H) \subset H$.

Take any $a \in N_G(H)$. Since $P \subset N_G(P) \subset H$, we have
$aPa^{-1} \subset aHa^{-1}=H,$ where the last step follows from $a \in N_G(H)$.

It follows that $P$ and $aPa^{-1}$ are both Sylow subgroups in $H$.
By the Sylow theorem, any two $p$-Sylow subgroups are conjugate. Thus there exists $b \in H$ such that $bPb^{-1}=aPa^{-1}$.

This implies that $(b^{-1}a)P(b^{-1}a)^{-1}=P$ and thus $b^{-1}a \in N_G(P) \subset H$. Hence we have $a \in H$ since $b \in H$.
This shows that $N_G(H) \subset H$, hence $N_G(H)=H$ as required.

## Corollary (The Normalizer of the Normalizer of a Sylow subgroup)

We apply the result to the case $H=N_G(P)$, and obtain the following result.

The normalizer of the normalizer of a Sylow subgroup $P$ of a finite group $G$ is the normalizer of $P$.
That is, we have
$N_G(N_G(P))=N_G(P).$ Add to solve later

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