$\exists g\in G: H\cap gPg^{-1}$ is a Sylow $p-$subgroup of $H$ Question:
If $P$ is a Sylow $p-$subgroup of $G$ and $H\leq G$ with $p||H|$ then $\exists g\in G: H\cap gPg^{-1}$ is a Sylow $p-$subgroup of $H$.
Attempt:
We consider the action $H\times G/P\to G/P$ with $(h,xP)\to hxP$ and we know that 
$$|G/P|=\sum_{x\in S}|[xP]_{H}|=\sum_{x\in S}|H:H\cap xPx^{-1}|$$
since $|[xP]_{H}|=|H:Stab_H(xP)|=|H\cap xPx^{-1}|$.
Now, $P$ is a Sylow $p-$subgroup of $G$ so $p\nmid |G/P|=m$ so from above $\exists g\in G:p\nmid |H:H\cap gPg^{-1}|$


*

*If $|H:H\cap gPg^{-1}|\not=1$: we have that $p\nmid |H:H\cap gPg^{-1}|$ and $H\cap gPg^{-1}$is a $p-$subgroup of $H$(please explain this statement) so $H\cap gPg^{-1}$ is a Sylow $p-$subgroup of $H$.

*If $|H:H\cap gPg^{-1}|=1$ then $H=H\cap gPg^{-1}\Rightarrow H\leq gPg^{-1}(= $$p-$group) and therefore $H\cap gPg^{-1}$ is a Sylow $p-$subgroup of $H$.


Is this proof correct? Is there a quicker way to prove the initial statement?
 A: This somewhat depends on what you are allowed to use. If you can use Sylow's Theorem, then there's an easier way:
Let $Q$ be a Sylow-$p$ sugroup of $H$. Since $Q$ is a $p$-subgroup of $G$, it is contained in a Sylow-$p$ subgroup $P^\ast$ of $G$. By Sylow's Theorem, $P$ and $P^\ast$ are conjugate in $G$, so there exists a $g\in G$ such that $P^\ast=gPg^{-1}$. Therefore $Q\le H\cap gPg^{-1}$.
On the other hand, $H\cap gPg^{-1}$ is a $p$-subgroup of $H$ and is therefore contained in a Sylow-$p$ subgroup $Q^\ast$ of $H$, and thus
$$Q\le H\cap gPg^{-1}\le Q^\ast$$
But since $Q$ and $Q^\ast$ are both Sylow-$p$ subgroups of $H$, their orders must be the same, so we have equality in the above formula. In particular, $Q=H\cap gPg^{-1}$, i.e. $H\cap gPg^{-1}$ is a Sylow-$p$ subgroup of $H$.
A: For the "please explain this statement" part: The conjugate $gPg^{-1}$ has the same order with $P$, hence it is a power of $p$. Therefore, $H\cap gPg^{-1}$, being a subgroup of $gPg^{-1}$, has order dividing $|gPg^{-1}|$, therefore, it a power of $p$. Hence, it is a group of order $p^k$ for some $k\neq 0$, hence a $p$-subgroup of $H$.
Now, in the first bullet, the only "rough" part is why $H\cap gPg^{-1}$ is a Sylow $p$-subgroup. That is because it a non-trivial $p$-subgroup with $p$ not dividing $[H:H\cap gPg^{-1}]$, hence a maximal $p$-subgroup (aka Sylow).
In the second bullet, it is clear that $H\cap gPg^{-1}=H$ hence it is trivially a $p$-Sylow subgroup.
Keep in mind, η Ταλέλλη δεν κάνει λάθος...
