$H\lhd G$, $|H|=p^k$ then $H$ is included in every $p$-Sylow subgroup of $G$

Question

Let $G$ be a finite group and let $H$ be a normal subgroup of G, with $|H|=p^k$. Then you have to prove that $H \subset P$, for every $P\in\mathrm{Syl}_p(G)$. What I thought is that, as $p$-Sylow are maximal $p$-subgroups then, as $H$ is a $p$-subgroup then it is included in $P$. But I'm not using the fact that $H$ is a normal subgroup, thus I guess I'm missing something, what it is?

Answer 1

Fistly $H\subset P$ for some $P\in \mathrm{Syl}_p(G)$ because a Sylow $p$-subgroup is a maximal $p$-subgroup of $G$. Note that Sylow $p$-subgroups are conjugate. We have $H = gHg^{-1}\subset gPg^{-1}$ because $H$ is normal. This gives the result.

Or argue by contradiction. If $H$ is not contained in $P$ for some $P\in\mathrm{Syl}_p(G)$, then $HP$ is a $p$-subgroup of $G$ (because $H$ is normal) and that $P\subsetneq HP$, a contradicition for $P$ to be a maximal $p$-subgroup.