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Let $G$ be a finite group, $H$ and $K$ subgroups of $G$ such that $G=HK$. Show that there exists a $p$-Sylow subgroup $P$ of $G$ such that $P = (P \cap H)(P \cap K)$.

It is not hard to prove that there exist $p$-Sylow subgroups of $H$ and $K$, $P_1$ and $P_2$ respectively, such that $P_1 \cap P_2$ is a $p$-Sylow subgroup of $H \cap K$, but I cannot guarantee that $P_1P_2$ is a subgroup of $G$ (and thus the $p$-Sylow of $G$ that we are looking for). Is there any way to complete the proof or it is not the right way?

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It seems like your assumptions aren't quite right. Wouldn't you need that H,K are proper subgroups to avoid the trivial case where you take $H:=\{e\}$ and $K:=G$. –  JSchlather Jun 1 '11 at 4:34
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@Jacob Schlather In your example, the result is trivial. For example, if $P$ is any Sylow $p$-subgroup of $G$, then $(P\cap H)(P\cap K)=\{e\}P=P$. –  Amitesh Datta Jun 1 '11 at 4:47
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Ah, so it is I wasn't reading the second condition correctly. –  JSchlather Jun 1 '11 at 4:54

1 Answer 1

up vote 5 down vote accepted

The following steps lead to a solution:

(1) Let us first find a Sylow $p$-subgroup $P$ of $G$ such that $P\cap H$ is a Sylow $p$-subgroup of $H$ and $P\cap K$ is a Sylow $p$-subgroup of $K$. Let $P_1$ be a Sylow $p$-subgroup of $H$ and let $P_2$ be a Sylow $p$-subgroup of $K$. Choose a Sylow $p$-subgroup $Q$ of $G$ such that $P_1\subseteq Q$. We know that there exists $g\in G$ such that $P_2\subseteq Q^g$. In particular, $Q\cap H=P_1$ and $Q^g\cap K=P_2$.

(2) We can write $g=hk$ for some $h\in H$ and $k\in K$. Use this to find a Sylow $p$-subgroup $P$ of $G$ such that $P\cap H$ and $P\cap K$ are Sylow $p$-subgroups of $H$ and $K$, respectively.

(3) It is clear in this situation that $P=(P\cap H)(P\cap K)$.

Thank you for asking this question! It encouraged me to think about finite groups (a beautiful subject) which I have not done for a few months. If you have further questions, please do not hesitate to ask them.

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Oh thank you for the answer, I was missing step 2. –  Clokr_ Jun 1 '11 at 12:02
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For (3) I had to count elements. Is it easy to show how to factor an element of P into a product of elements of P∩H and P∩K? If K was normal, then you could conjugate "h" into P∩H without disturbing K, but I don't see it in general. –  Jack Schmidt Jun 1 '11 at 15:02

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