Index of a subgroup containing a Sylow normalizer.

Question

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.

Answer 1

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 ...

Answer 2

1. $P\subseteq N_G(P)\subseteq H\subseteq G$ [conditions]
2. $P$ is a $p$-Sylow subgroup of $H$ [1]
3. The number $n_p(H)$ of $p$-Sylow subgroups in $H$ is $[H:N_H(P)]$ [Sylow-II]
4. The number $n_p(G)$ of $p$-Sylow subgroups in $G$ is $[G:N_G(P)]$ [Sylow-II]
5. $N_H(P)=N_H(P)\cap H$ [definition]
6. $N_H(P)=N_G(P)$ [1,5]
7. $n_p(H)=[H:N_G(P)]$ [3,6]
8. $n_p(H)=1\text{ (mod }p) $ [Sylow-III]
9. $n_p(G)=1\text{ (mod }p) $ [Sylow-III]
10. $[H:N_G(P)]=1\text{ (mod }p) $ [7,8]
11. $[G:N_G(P)]=1\text{ (mod }p) $ [4,9]
12. $|H|/|N_G(P)|=pa+1$ [10]
13. $|G|/|N_G(P)|=pb+1$ [11]
14. $|G|/|H|=(pb+1)/(pa+1)=pc+d$ [12,13]
15. $(pc+d)(pa+1)=p^2ac+pc+pad+d=pb+1$ [14]
16. $(pc+d)(pa+1)=d\text{ (mod }p)$ [15]
17. $(pc+d)(pa+1)=1\text{ (mod }p)$ [15]
18. $d=1\text{ (mod }p)$ [16,17]
19. $|G|/|H|=1\text{ (mod }p)$ [14,18]
20. $[G:H]=1\text{ (mod }p)$ [19]