sylow subgroup of a subgroup 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.
 A: 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 $H \cap gPg^{-1} \in {\rm Syl}_2(H)$.
A: 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 $Q\cap H = K$. 
A: 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 $G$)?
A: Let $P'$ be any Sylow $p$-subgroup of $H$. Then there is a maximal $p$-subgroup of $G$ containing it. What is it?
