Index of a subgroup containing a Sylow normalizer. 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.
 A: 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 ...
A: *

*$P\subseteq N_G(P)\subseteq H\subseteq G$ [conditions]

*$P$ is a $p$-Sylow subgroup of $H$ [1]

*The number $n_p(H)$ of $p$-Sylow subgroups in $H$ is $[H:N_H(P)]$ [Sylow-II]

*The number $n_p(G)$ of $p$-Sylow subgroups in $G$ is $[G:N_G(P)]$ [Sylow-II]

*$N_H(P)=N_H(P)\cap H$ [definition]

*$N_H(P)=N_G(P)$ [1,5]

*$n_p(H)=[H:N_G(P)]$ [3,6]

*$n_p(H)=1\text{ (mod }p) $ [Sylow-III]

*$n_p(G)=1\text{ (mod }p) $ [Sylow-III]

*$[H:N_G(P)]=1\text{ (mod }p) $ [7,8]

*$[G:N_G(P)]=1\text{ (mod }p) $ [4,9]

*$|H|/|N_G(P)|=pa+1$ [10]

*$|G|/|N_G(P)|=pb+1$ [11]

*$|G|/|H|=(pb+1)/(pa+1)=pc+d$ [12,13]

*$(pc+d)(pa+1)=p^2ac+pc+pad+d=pb+1$ [14]

*$(pc+d)(pa+1)=d\text{ (mod }p)$ [15]

*$(pc+d)(pa+1)=1\text{ (mod }p)$ [15]

*$d=1\text{ (mod }p)$ [16,17]

*$|G|/|H|=1\text{ (mod }p)$ [14,18]

*$[G:H]=1\text{ (mod }p)$ [19]
