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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!

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

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  • $\begingroup$ Is $N_G(N_G (H)) \neq \emptyset$??? If $N_G(N_G (H)) = \emptyset$ we can't choose $g \in N_G(N_G (H)) \neq \emptyset$. $\endgroup$ – Minh Apr 5 at 4:58
  • $\begingroup$ No subgroup of any group can be the empty set. $\endgroup$ – Derek Holt Apr 5 at 5:20
  • $\begingroup$ $N_G(N_G (H))$ is a subgroup of $G$, but is $N_G(N_G (H)) \setminus N_G (H)$ be a subgroup of $G$??? $\endgroup$ – Minh Apr 5 at 5:34
  • $\begingroup$ No, I never said it was. It could not possibly be a subgroup because it does not contain the identity element. I said that it is nonempty. $\endgroup$ – Derek Holt Apr 5 at 5:41
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    $\begingroup$ Please read what I write. As I said in the first paragraph, it is nonempty because $G$ is a $p$-group. $\endgroup$ – Derek Holt Apr 5 at 5:49

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