A normal subgroup $H$ with $[G:H]$ coprime to $p$ contains every Sylow $p$-subgroup of $G$.

I was working on the following question:

Let $p$ be a prime factor of the order of a finite group $G$. If $H$ is a normal subgroup $G$ whose index is not a multiple of $p$, show that H must contain every Sylow $p$-subgroup of G.

Since $G$ is a finite group, the index of H in G is defined as $\left|{G}\right|$/$\left|{H}\right|$. If the index of $H$ is not a multiple of p, then $\left|{H}\right|$ must be a multiple of p. Also, $H$ is a normal subgroup of $G$ so afaik it is enough to show that H contains one Sylow $p$-subgroup. However I am not making much progress in proving this.

• If $p^n$ divides |G| then it also divides |H|. So the Sylow subgroups of H are those of G. – awllower Jan 13 '13 at 21:51

Every $p$-group in $G$ is contained in some Sylow $p$-subgroup of $G$, so any $P\in \text{Syl}_p(H)$ is contained in some $Q\in \text{Syl}_p(G)$. $[G:H]$ is coprime to $p$, however, so $P$ and $Q$ have the same order, and are thus equal.
Because $H$ is normal, we have that $P^g\leqslant H$ for every $g\in G$ as $H$. Therefore, since $G$ acts transitively on $\text{Syl}_p(G)$, it follows that $\text{Syl}_p(G)\subseteq \text{Syl}_p(H)$.
Hint: consider the Sylow $p$-subgroups of $H$.