If $H$ is a subgroup of $G$ and $S$ is a $p$-Sylow subgroup of $G$, then there’s a $g\in G$ such that $H\cap gSg^{-1}$ is a $p$-Sylow subgroup of $H$

Question

I’m reading a proof of the above statement and having difficulties understanding the first step. It says that the stabilizer of any element $gS\in X=G/S$ is $H\cap gSg^{-1}$. I guess I should be considering here $H$’s action on $X$ by left traslation, but I don’t see how the result is obtained.

Also, the proof seems to use this fact to conclude that $H\cap gSg^{-1}$ is a $p$-subgroup but this seems obvious to me (the conjugated subgroup of a $p$-subgroup is always a $p$-subgroup, and its intersection with another subgroup must be a $p$-subgroup too, right?).

So I don’t understand why the stabilizer was needed at all.

Answer 1

Let me give you a different proof: Let $P$ be a Sylow subgroup of $H$. By Sylow's theorem $P$ lies in a Sylow subgroup $Q$ of $G$ and $Q$ is conjugate to $S$. Hence, there exists $g\in G$ such that $P\le gSg^{-1}$. Now $H\cap gSg^{-1}$ is a $p$-subgroup of $H$ which cannot be bigger than the Sylow subgroup $P$. Therefore $H\cap gSg^{-1}=P$.