Let $G$ be a $p$-group and let $H$ be a proper subgroup of $G$. Show that $H$ is a proper subgroup of $N_G(H)$ Let $G$ be a $p$-group and let $H < G$. Show that $H < N_G(H)$.
If $H \trianglelefteq G$, then it should be clear that $H<N_G(H)$. So we suppose, $H$ isn't normal. So we let $H$ act on the set
$$S = \{gHg^{-1}: g \in G, gHg^{-1} \neq H\}$$
via conjugation. If I can show that $p$ doesn't divide $|S|$, then I can use the fixed point theorem to complete the problem. Any advice on how to show that $p$ doesn't divide $|S|$?

Hint

*

*Finite $p$- groups are nilpotent.

*For any proper subgroup $H$ of a nilpotent group $G$, we have $N_G(H)\supsetneq H$.

Both these results are well-known.
The result is false without requiring the $p$- group to be finite, as Tarski monsters show.
