# Index of a subgroup containing a Sylow normalizer.

It is the exercise 2.12 on Paolo Aluffi's Algebra: Chapter 0.

Let $$P$$ be a p-Sylow subgroup of $$G$$ and $$H$$ a subgroup containing $$N_G(P)$$. It claims that $$p|([G:H]-1)$$.

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.

• The order of $\;H\;$ is divided by the order of $\;N_G(P)\;$ , but the index of this last group is the number $\;n_p\;$ of Sylow $\;p\,-$ subgroups, and we know $\;n_p=1\pmod p\;$ , so... – DonAntonio Nov 6 '18 at 13:37

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 ...