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!

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $H^g \ne H$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.