# $G$ 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.

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?

• 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. Commented Sep 18, 2020 at 6:07
• @RobertShore oh so you mean to prove $|gHg^-1|$ = $|H|$?
– jun
Commented Sep 18, 2020 at 6:09
• Yes, and also to prove that $g^{-1}Hg$ is a subgroup of $G$ (which is quite easy). Commented Sep 18, 2020 at 6:10
• @RobertShore and for that one, I need to prove closure, identity, and inverse, right?
– jun
Commented Sep 18, 2020 at 6:11
• @RobertShore Then can I say $gHg^-1$ and $H$ are isomorphic?
– jun
Commented Sep 18, 2020 at 6:20

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