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Let $p$ be a prime and $H$ a subgroup of a finite group $G$. Let $P$ be a p-sylow subgroup of G. Prove that there exists $g\in G$ such that $H\cap gPg^{-1}$ is sylow subgroup of $H$.

I have no idea how to do this, any hints?

Note: Originally it was unclear if the problem was for possibly infinite groups or just finite ones. However, since the definition of $p$-Sylow subgroup being used is that it is a $p$-subgroup such that the index and the order are relatively prime, the definition only applies to finite groups.

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    edit the question and add the information that $G$ need not be finite *there*.2010-12-29

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Let P' be any Sylow $p$-subgroup of $H$. Then there is a maximal $p$-subgroup of $G$ containing it. What is it?

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    @Chu: The trivial subgroup is a p-subgroup.2010-12-29
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Let $G$ be the direct product of countably many copies of the dihedral group $D$ of order 6 (or, if you prefer, $D$ is the symmetric group $S_3$).

We can construct a Sylow $2$-subgroup of $G$ by choosing Sylow $2$-subgroups of each of the direct factors of $G$, and taking their direct product. Since $D$ has three Sylow $2$-subgroups, $G$ has uncountably many Sylow $2$-subgroups, so they cannot all be conjugate in the countable group $G$.

If we let $P$ and $H$ be non-conjugate Sylow $2$-subgroups of $G$, then there is no $g \in G$ such that $H \cap gPg^{-1} \in {\rm Syl}_2(H)$.

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    +1 - I like this better than the counterexample I found. Thank you.2010-12-30
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Note: The following works if $G$ is finite, but may fail in the infinite case; there are infinite groups in which there are $p$-Sylow subgroups $P$ and P' for which no automorphism (inner or outer) of $G$ maps $P$ to P'.

Let $K$ be a $p$-Sylow subgroup of $H$ (we know it exists, though it may be trivial). Then $K$ is a $p$-subgroup of $G$ (even if $K={e}$) and by the Sylow Theorems, is contained in some Sylow $p$-subgroup $Q$ of $G$. By the Sylow Theorems, $Q$ and $P$ are conjugate. Now just verify that $Q\cap H = K$.

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    @Chu: Yes, if you defined it as a subgroup in which the order and index are coprime, then your definition, *a fortiori* assumes that the group $G$ is finite (otherwise, the definition does not make sense since at least one of the order and index will necessarily be infinite). One can instead define Sylow subgroups as "maximal $p$-subgroups", but then they don't always exist (in the infinite case), and as B.H. Neumann's paper shows, we can have weird things happening in the infinite case.2010-12-29
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Well, we have $\vert P \vert =p^n$ where $\vert G \vert = p^n m$ and $(p,m)=1$. What can be said about $\vert H \cap P \vert$ (or, for that matter, $\vert H \cap Q \vert$ for any Sylow p-subgroup $Q$ of $G$)?

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    Ah, in that case, we know that for all $x\in P$, $\vert x \vert=p^n$ for some $n\geq 0$. What does that tell us about elements of $H\cap P$?2010-12-29