If $H$ is a proper subgroup of a $p$-group $G$, then $H$ is proper in $N_G(H)$. 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.
 A: You are on the right track. Now look at the subgroup $H/Z$ of $G/Z$. By induction, its normalizer is strictly larger than $H/Z$. Say it contains the residue class $\overline x$ of $x \in G$ where $\overline x \not\in H/Z$. Now show that $x$ also normalizes $H$ in $G$ to get a contradiction.
A: Let me write a complete answer with the hint proposed by @marlu .

Let $G$ be a $p$-group. So, $|G|=p^r$ for some $r\ge 1$.
Let us look the base case that $|G|=p$. If $H<G$, then $H=\{e\}<G=N_G (\{e\})$.
Assume that, for all $s<r$, if $|G|=p^s$ and $H<G$, then we $H< N_G (H)$.
Let us now consider the case that $|G|=p^r$. Note that $H\leq N_G(H)$. If $H<N_G(H)$, then we are done. So we may assume that $H=N_G(H)$, which forces that $Z\unlhd H$, where $Z:=Z(G)$.
By our inductive assumption and the fact that $|G/Z|=p^t$ with $t<r$ (where we have used the fact that every finite $p$-group has a non-trivial center), we know the normalizer (group) of $H/Z$ properly contains $H/Z$. Therefore, there exists $\bar{x} \in G/Z$ such that $\bar{x} \notin H/Z$ and $\bar{x}$ normalizes $H/Z$, i.e. $\bar{x} (H/Z)\bar{x}^{-1}=H/Z$.
Let $h\in H$. We then have $\bar{x} (hZ)\bar{x}^{-1}=\tilde{h}Z$ for some $\tilde{h}\in H$. Therefore, $ (xhx^{-1})Z=\tilde{h}Z$, where we have used the fact that $\bar{x}^{-1}=\overline{x^{-1}}$. So, there exist $z,\tilde{z}\in Z$ such that $x hx^{-1}z=\tilde{h}\tilde{z}$, which implies that $xhx^{-1}=\tilde{h}\tilde{z}z^{-1}$. Recall that $Z\leq H$, which implies $\tilde{h}\tilde{z}z^{-1}\in H$ and so $x$ does normalize $H$. Since $\bar{x}\notin H/Z$, it is clear that $x\notin H$, which contradicts with $H=N_G(H)$, which implies that we can only have the case that $H<N_G(H)$.
$\tag*{$\blacksquare$}$
