Does $p$ always divide $[N_G(H) : H]$ when $G$ is a $p$-group?

Question

On this question, $G$ is a finite $p$-group and $H$ is a proper subgroup of $G$.

The accepted answer says

Note that p divides $|N_G[H]/H|$ so $N_G[H]/H$ has a subgroup of order $p$.

But what about if $H$ were equal to its own normalizer, $N_G(H)$? In that case, $|N_G[H]/H|$ would be $1$, hence, not divisible by $p$.

Is there any result that prevents a proper subgroup of being equal to its normalizer considering that they are $p$-groups?

Or do you know how to prove that quoted preposition?

Answer 1

Yes, because then $H$ is in turn a $p$-group, and hence $[G:H]\equiv [N_G(H):H]\pmod p$ (see e.g. here). But in this case, $[G:H]\equiv 0\pmod p$ either, whence the claim follows.