# Every p-subgroup is conjugate to a p-subgroup whose normalizer in the Sylow is Sylow in its normalizer

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A related survey article is: finding conjugate subgroups

## Statement

Suppose $G$ is a finite group and $P$ is a $p$-Sylow subgroup of $G$. Suppose $H$ is a subgroup contained inside $P$. Then, there exists a subgroup $K$ of $P$, such that $H$ and $K$ are conjugate subgroups inside $G$, and such that $N_P(H)$ is a Sylow subgroup of $N_G(H)$.

## Proof

Given: A finite group $G$, a $p$-Sylow subgroup $P$, a subgroup $H$ contained inside $P$.

To prove: There exists a subgroup $K$ of $P$ such that $H$ and $K$ are conjugate subgroups inside $G$, and such that ,math>N_P(H)[/itex] is $p$-Sylow inside $N_G(H)$.

Proof:

1. Let $L$ be a $p$-Sylow subgroup of $N_G(H)$. Such a $L$ exists by fact (1) in the subgroup $N_G(H)$.
2. Let $g \in G$ be such that $L \le g^{-1}Pg$. Note that such a $g$ exists by fact (2) in the group $G$.
3. Let $K = gHg^{-1}$. Then, $gLg^{-1}$ is a ,math>p[/itex]-Sylow subgroup of $N_G(K)$. By construction, $gLg^{-1} \le P$, so a $p$-Sylow subgroup of $N_G(K)$ is contained in $P$. Thus, $N_P(K)$ is $p$-Sylow in $N_G(K)$.