Prove that a proper subgroup of a $p$-group never equals its normalizer.

Let $G$ be a group of order $p^{n}$, where $p$ is prime. Let $H$ be a proper subgroup. We define its normalizer $$N = \lbrace g \in G, gHg^{-1} = H\rbrace$$ I want to show that $H \neq N$.

For that, I'm trying to use the multiplication action of $H$ on $G/H$.

If $|G.xH| = 1$, then $x \in N$. So I'm looking for $x \not \in H$ such that $|G.xH| = 1$. What can I do from here ?

I may have an idea but I'm not sure how to conclude :

Let $F = \lbrace xH \in G/H, G.xH = \lbrace xH \rbrace \rbrace$, so $|F|$ is the number of orbits that have cardinality $1$. If I can show that |F| > 1 then it's over. $H$ lies in $F$, so let's suppose $|F| = 1$ and we should find a contradiction.

I'ts not hard to prove that $$|F| \equiv |G/H| \pmod p$$ so with our hypothesis $$|G/H| \equiv 1 \pmod p$$ But how do we conclude from here ?

Solution 1:

I give an alternating proof without using the group action. Let $T$ be a subgroup of $G$ containing $H$ with $[T:H]=p$ (you have to prove that such a subgroup exists). Then $H$ is maximal subgroups of the $p$-group $T$. Therefore $H \triangleleft T$ (this also needs a short proof). Hence $T \subseteq N$.