Sylow subgroup of $G/H$ is image of Sylow subgroup of $G$ under projection

Question

I already know that if $H$ is a normal subgroup of $G$ and $S$ a $p$-Sylow subgroup of $G$, and let $\pi:G\to G/H:g\mapsto gH$, then $\pi(S)$ is a $p$-Sylow subgroup of $G/H$.

Now I'm wondering how to show that the converse is also true:

If $S'$ is a $p$-Sylow subgroup of $G/H$, then there exists a $p$-Sylow subgroup of $G$ (say $S$) such that $\pi(S)=S'$?

Could I generate $S$ directly using $S'$ and a $p$-Sylow subgroup of $H$? Please give me some hint.

Answer 1

First of all, if $S \in Syl_p(G)$, then $SH/H \in Syl_p(G/H)$. Why? $SH/H \cong S/(H \cap S)$ and the latter is a $p$-subgroup, since $S$ is a $p$-subgroup. In addition, $|G/H : SH/H|=|G:SH| \mid |G:S|$, hence $p \nmid |G/H : SH/H|$. I leave it to you to prove (similarly) that $H \cap S \in Syl_p(H)$.

Now, let $S' \in Syl_p(G/H)$. Take an $S \in Syl_p(G)$, then by the above, $SH/H \in Syl_p(G/H)$. Hence by the Sylow theorems, $S'$ must be conjugate to $SH/H$, so for some coset $gH \in G/H$ we have $S'=(g^{-1}H) SH/H (gH)=g^{-1}SgH/H \cong (g^{-1}Sg)/(H \cap g^{-1}Sg)$. By the remarks above, $H \cap g^{-1}Sg \in Syl_p(H)$, so your question Could I generate $S$ directly using $S'$ and a Sylow $p$-subgroup of $H$? is therefore answered. All you need is an extra $g \in G$.

Answer 2

The argument I had in mind was the following.

Fix a Sylow $p$-subgroup $S'$ of $G/H$. Let's consider $K=\pi^{-1}(S')\le G$. By the correspondence principle $[G:K]=[G/H:S']$ is coprime to $p$. So if $S$ is a Sylows $p$-subgroup of $K$, then $S$ is also a Sylow $p$-subgroup of $G$. I claim that any such $S$ works. The reason is that $\pi(S)\simeq S/(S\cap H)$ has the correct cardinality for a Sylow $p$-subgroup of $\pi(K)=S'$. But that Sylow $p$-subgroup must be $S'$ itself. Hence $S'=\pi(S)$.