Either $H \triangleleft G$ or exist a conjugated subgroup $H^g \subseteq N_G (H)$, in which $g \in G$, with $H^g \neq H$.

Question

Let $G$ be a $p-$group. If $H$ be a subgroup of $G$, prove that either $H \triangleleft G$ or exist a conjugated subgroup $H^g \subseteq N_G (H)$, in which $g \in G$, with $H^g \neq H$.

In my opinion, to solve this problem we must use the formula of orbit $$|\mathcal{O} (x)| = |G : S_G (x)|$$

Could you give me some hint to solve this problem! Thank all!

Answer 1

I will make my comment into an answer. If $H$ is not normal in $G$, then $N_G(H) \ne G$ and so, by a standard property of $p$-groups, $N_G(H)$ is properly contained in its normalizer $N_G(N_G(H))$.

Now choose any $g \in N_G(N_G(H)) \setminus N_G(H)$, and we have $H^g \le N_G(H)$ with $H^g \ne H$.