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The group $G$ acts on itself by conjugation. If $H$ is a subgroup of $G$, show that the elements of the orbit of $H$ are subgroups of $G$ of the same order.

I'm a little confused about what I need to prove.

My thought is that:

Let h $\in$ H, and $\theta_h$ = {g(h) | g $\in$ G} where $\theta_h$ is the orbit of h.

Then I can say, for example, $\theta_{h_1}$ = {$g_1 h_1 g_1^{-1}$, $g_2 h_1 g_2^{-1}$, $g_3 h_1 g_3^{-1}$ ..} where $g_1, g_2, g_3, .. \in G$ and $h_1 \in H$.

From here, do I need to prove that | $g_1 h_1 g_1^{-1}$ | = |$g_2 h_1 g_2^{-1}$ | = ... ?

If that's the case, my proof is that if $h_1^m$ = e, then |$g_1 h_1 g_1^{-1}|$ = e. So all the elements of the orbit have the same order.

Am I on the right track?

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  • $\begingroup$ I think that they're asking you to prove that any conjugate of $H$ is also a subgroup and it has the same order (size) as does $H$ itself. $\endgroup$ – Robert Shore Sep 18 '20 at 6:07
  • $\begingroup$ @RobertShore oh so you mean to prove $|gHg^-1|$ = $|H|$? $\endgroup$ – jun Sep 18 '20 at 6:09
  • $\begingroup$ Yes, and also to prove that $g^{-1}Hg$ is a subgroup of $G$ (which is quite easy). $\endgroup$ – Robert Shore Sep 18 '20 at 6:10
  • $\begingroup$ @RobertShore and for that one, I need to prove closure, identity, and inverse, right? $\endgroup$ – jun Sep 18 '20 at 6:11
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    $\begingroup$ @RobertShore Then can I say $gHg^-1$ and $H$ are isomorphic? $\endgroup$ – jun Sep 18 '20 at 6:20
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Hint: Conjugation by an element of the group defines an automorphism known as an inner automorphism. It follows that the image of a subgroup is a subgroup of the same order. For, an automorphism is an isomorphism.

Again, $x\mapsto gxg^{-1}$ defines a bijective homomorphism from $G$ to $G$ for any $g\in G$. The proof is straightforward.

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