I know the following: If $P$ is a $p$-Sylow subgroup of $G$ then its normalizer $N_G(P)$ equals $N_G(N_G(P))$.

But I don't know how to go about proving the following stronger version:

With finite group $G$ and Sylow subgroup $P$ of $G$ and $K=N_G(P)$, for any subgroup $H$ of $G$ with $H\ge K$, $H = N_G(H)$.

Any advice?


A nice exercise (for me at least, the people who have studied groups seriously may be bored). One way is to work out the following steps. Leaving a few details out for now.

  1. If a $p$-subgroup $Q\le G$ normalizes a Sylow $p$-subgroup $P$ of $G$ then $Q\le P$ (this is something you have surely seen).
  2. If $P$ is a $p$-subgroup of $N_G(H)$ then $P$ normalizes at least one of the Sylow $p$-subgroups of $H$ (use orbit-stabilizer).
  3. If $P$ is a $p$-subgroup of $N_G(H)$ then $P$ is contained in one of the Sylow $p$-subgroups of $H$. In particular, the groups $H$ and $N_G(H)$ have the same number of Sylow $p$-subgroups.
  4. For any intermediate subgroup $M$, $K\le M\le G$, the number of Sylow $p$-subgroups of $M$ is $n_p(M)=[M:K]$ (orbit-stabilizer again).
  5. $[N_G(H):H]=1$.
  • $\begingroup$ This is very helpful. Thanks so much! $\endgroup$ – TheExchanger Dec 6 '17 at 8:07

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