The following is exercise $2.12$ in Paolo Aluffi's Algebra: Chapter $0$.
Let $P$ be a $p$-Sylow subgroup of $G$ and $H$ a subgroup containing $N_G(P)$. Then $p \mid |G:H|-1$.
I tried to use the fact that $H$ is self-normalizing, but could not see any relevance to this exercise. I also tried to imitate the trick which proved Sylow-III but failed.
Best Answer
Hint: $P \subseteq N_G(P) \subseteq H$, so $N_H(P)=H \cap N_G(P)=N_G(P)$. Hence $n_p(H)=|H:N_H(P)|=|H:N_G(P)| \equiv 1$ mod $p$, by Sylow Theory in $H$. But $n_p(G)=|G:N_G(P)| \equiv 1$ mod $p$, by Sylow Theory in $G$. And one gets, $n_p(G)=|G:H|n_p(H)$. So ...