Let $G$ be a finite group and $N$ a normal subgroup of $G$. Let $\pi:G\rightarrow G/N$ be the canonical homomorphism.
Suppose $P$ is a Sylow $p$-subgroup of $G$. Then $\pi[P]\leq G/N$. Note that $G$ operates transitively on $(G/N)/\pi[P]$. Therefore there exists a subgroup $S$ of $G$ such that $[G:S]=\left[G/N:\pi[P]\right]$.
It is claimed that $P\subset S$. I don't know why this holds. Hints?
Let $f:G/S\rightarrow (G/N)/\pi[P]$ be the $G$-isomorphism mentioned. I know that $f(S)=\pi[P]$. This implies that $P=\pi^{-1}[f(S)]$.
Edit:
I think I have to pick $S:=\text{stab}_{G}(\pi[P])$ for this to work, since $P\subset\text{stab}_{G}(\pi[P])$.