Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
1  
Thank you, Arturo Magidin. –  Waldott Apr 1 '11 at 1:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.