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.
Best Answer
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 $H \cap gPg^{-1} \in {\rm Syl}_2(H)$.