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

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!

• What do you mean " $H$ be a normal subgroup of $G$ then either $H\triangleleft G$ "? – user549397 Apr 5 at 1:08
• I have edited it. – Minh Apr 5 at 1:13
• I don't think you want to assume that $H$ is a normal subgroup of $G$. I think you just want to assume $H$ is a subgroup of $G$. – Robert Shore Apr 5 at 1:23
• @Minh Well you did not. – user549397 Apr 5 at 1:26
• If $H$ is not normal in $G$ then choose any $g \in N_G(N_G(H)) \setminus N_G(H)$. – Derek Holt Apr 5 at 2:40

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 $$H^g \ne H$$.
• Is $N_G(N_G (H)) \neq \emptyset$??? If $N_G(N_G (H)) = \emptyset$ we can't choose $g \in N_G(N_G (H)) \neq \emptyset$. – Minh Apr 5 at 4:58
• $N_G(N_G (H))$ is a subgroup of $G$, but is $N_G(N_G (H)) \setminus N_G (H)$ be a subgroup of $G$??? – Minh Apr 5 at 5:34
• Please read what I write. As I said in the first paragraph, it is nonempty because $G$ is a $p$-group. – Derek Holt Apr 5 at 5:49