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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.

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

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