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.

  • $\begingroup$ 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... $\endgroup$ – 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 ...


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