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this question was posted on aops about 3 weeks ago, but never received an answer. Maybe it will fare better here?

Let $G$ be a finite group and $H$ a subgroup. Let $P_H$ be a $p$-Sylow subgroup of $H$. Prove that there exists a $p$-Sylow subgroup $P$ of $G$ such that $P_H=P\cap H$.

Source: The exercise comes from the first chapter of Lang. Thanks.

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up vote 3 down vote accepted

By the Sylow Theorems, every maximal $p$-subgroup of a finite group is a Sylow $p$-subgroup. $P_H$ is a $p$-subgroup of $G$, hence it is contained in a maximal $p$-subgroup of $G$, and this subgroup $P$ must be a $p$-Sylow subgroup of $G$.

So $P$ is a Sylow $p$-subgroup of $G$ that contains $P_H$. In particular, $P_H\subseteq P\cap H$. We just need to show that $P_H=P\cap H$.

For the converse inclusion, let $h\in H\cap P$. Then $\langle P_H,h\rangle\subseteq P$, hence is a $p$-group, and $P_H\subseteq \langle h,P_H\rangle\subseteq H$. Since $P_H$ is a maximal $p$-subgroup of $H$, $P_H=\langle h,P_H\rangle$, hence $h\in P_H$. This proves that $P\cap H\subseteq P_H$, giving equality.

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Thank you, Arturo Magidin. – Waldott Apr 1 '11 at 1:01

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