TITLE: If $H$ is a subgroup of $G$, and $N_G(P)\subseteq H$ for a $p$-Sylow of $H$, then $P$ is a $p$-Sylow of $G$
QUESTION [5 upvotes]: I'm stuck on this problem that I've found in Isaacs' Finite Group Theory. I've tried thinking about it for a while but didn't came up with a solution, so I'm asking for some hints or even for a full solution if you wish, everything is welcomed since I don't see any way to follow.
The problem is the following.
Let $G$ be a finite group and $P\in \text{Syl}_p(H)$, where $H\subseteq G$, meaning that $P$ is a Sylow $p$ group in $H$. Let $N_G(P)$ be the normalizer of $P$ in $G$, and let $N_G(P)\subseteq H$. Then show that $p\nmid|G:H|.$

REPLY [3 votes]: We actually have $|G:H| \equiv 1 \pmod{p}$ in this case. Indeed, considering the action of $P$ on the set of left cosets $G/P$ we get 
$$   |G:P| \equiv |\mathbf{N}_G(P) : P|\pmod{p}$$
since $P$ is a $p$-group. Now since $\mathbf{N}_G(P) \subseteq H$, $|\mathbf{N}_G(P) : P|$ is not divisible by $p$ hence we can cancel this term in the above equivalence and get
$$ |G : \mathbf{N}_G(P)| \equiv 1 \pmod{p}$$ (This comes for free if you show $P \in \text{Syl}_p(G)$ first, since $|G:\mathbf{N}_G(P)| = n_p(G)$)
Furthermore $|H:\mathbf{N}_G(P)| =n_p(H) \equiv 1 \pmod{p}$ and hence $$|G:H| \equiv 1 \pmod{p}$$