Let $H$ be a proper subgroup of $p$-group $G$. Show that the normalizer of $H$ in $G$, denoted $N_G(H)$, is strictly larger than $H$, and that $H$ is contained in a normal subgroup of index $p$.
Here's what I've got so far:
- If $H$ is normal, $N_G(H)$ is all of $G$ and we are done.
- If $H$ is not normal, then suppose for the sake of contradiction that $N_G(H)=H$. Then there is no element outside of $H$ that fixes $H$ by conjugation. But the center $Z(G)$ of $G$ does fix $H$, so $Z(G)$ must be a subgroup of $H$.
Don't know if I'm going on the right path or not, but either way can't really think my way out of this one... Any help would be appreciated.