# Let $P$ and $P'$ be Sylow $p$-subgps of $G$, $P' \subset N_G(P)$. Show that $P=P'$.

Let $G$ be a finite group and let $P$ and $P'$ be Sylow $p$-subgroups of $G$ such that $P' \subset N_G(P)$, with $N_G(P)$ the normalizer of $P$ in $G$. Show that $P=P'$.

I'm not sure where to start here.

• You can think of P as a sylow subgroup of its normalizer. What do you know about the conjugates of P? – Improve Sep 24 '17 at 17:28

By the Sylow theorems applied to $N_G(P)$, we see that since $P \trianglelefteq N_G(P)$, $P$ is the unique Sylow $p$-subgroup of $G$. But as $P' \subset N_G(P)$, $P'$ is also a Sylow $p$-subgroup. Thus $P = P'$.