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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 $p\nmid|G:H|.$

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up vote 8 down vote accepted

Let $P$ be a $p$-subgroup of $H$ whose normalizer is completely contained in $H$. Write $|P|=p^n$. If $p$ divides $[G:H]$, then $G$ contains a $p$-subgroup of order $p^{n+1}$ that contains $P$. Call this $Q$. Since $[Q:P]=p$, $P$ is maximal in $Q$ hence normal in $Q$, so $Q\subseteq N_G(P)\subseteq H$. But this means that $H$ contains a $p$-subgroup of order strictly larger than that of $P$, so $P$ is not a $p$-Sylow subgroup of $H$.

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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 $$|G:H| \equiv 1 \pmod{p}$$

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