A Criterion for being Sylow p-group

Question

Show that if $H$ is a $p$-group of finite group $G$ and $N_G(H)=H$ then $H$ is a Sylow $p$-group of $G$? Or prove the following more general property,$$[G:H]\equiv1\ (\mod\ p)$$

Answer 1

Let $H$ be a $p$-subgroup of the group $G$, and assume $H=N_G(H)$. By one of the Sylow Theorems, $H \subseteq P$, for some $P \in Syl_p(G)$. Assume that $H \neq P$. It is well-known that in $p$-groups (and in general nilpotent groups) normalizers grow, that is $H \subsetneq N_P(H)$. But $N_P(H) \subseteq N_G(H)=H$, a contradiction. Hence $H=P$. Another Sylow Theorem, dealing with the number of Sylow $p$-subgroups, asserts that $[G:N_G(P)] \equiv 1$ mod $p$, and you are done.

By the way, in general, if $H$ is a $p$-subgroup of a group $G$, then $[G:H] \equiv [N_G(H):H]$ mod $p$. This can be proved by applying the orbit-stabilizer theorem on the left multiplication action on the left cosets of $H$.

Answer 2

Since this is a homework problem I will give you a few things to think about to get you started:

The normalizer $N_G(H)$ is the stabilizer of $H$ under the conjugation action of $G$. So you should be thinking about how $G$ acts by conjugation on the set of $p$-subgroups of $G$.

The index $[G:H] = [G:N_G(H)]$ is equal to the size of the orbit of $H$ under this action. This is the "orbit-stabilizer theorem."