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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 $P=(P\cap H)(P\cap K)$.

I looked at the proof here, but I can't understand step "(3) It is clear in this situation that $P=(P\cap H)(P\cap K)$." Help.

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  • $\begingroup$ This should be a comment there, not a new question. $\endgroup$ – user641 Apr 1 '12 at 18:41
  • $\begingroup$ I tried to put it as a comment there but I don't know how? $\endgroup$ – user28083 Apr 1 '12 at 18:46
  • $\begingroup$ @user28083: You don't have enough reputation yet to make a comment in someone else's question. $\endgroup$ – Arturo Magidin Apr 1 '12 at 20:44
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    $\begingroup$ @user28083: The comment by Jack Schmidt suggests one way of doing it: note that $|G|=|HK|=|H||K|/|H\cap K|$, and $|(P\cap H)(P\cap K)|=|P\cap H||P\cap K|/|P\cap H\cap K|$. Show that this equals the largest power of $p$ that divides $|G|$. $\endgroup$ – Arturo Magidin Apr 1 '12 at 20:52
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    $\begingroup$ |G|/ |(P∩H)(P∩K)|=(|H||K||P∩H∩K|)/(|P∩H||P∩K||H∩K|) is not divisible by p ==>|(P∩H)(P∩K)| is the largest p power that divides the order of G. Thank you very much. $\endgroup$ – user28083 Apr 1 '12 at 22:38

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