Let $G$ be a group acting on itself.
Prove $A=\{g\in G: |g^G|< \infty \}$ is a characteristic subgroup of $G$
I've found the following question: Does the subgroup $\{g\in G\,|\,o(\operatorname{Cl}(g))<\infty\}$ of $G$ have a name? In that question the group action is specified as conjugation, but the exercise in the text I'm reading does not specify this.
This leads me to ask whether this would be valid in a more general setting as well, when the group action is not specified.
Idea
I have an idea, but I think it flawed. Is it? And is this theorem true in the general case?
- Notice how $A \not = \varnothing$, since $1\in A$
Choose $g,h \in A$, does $gh^{-1}\in A$?
Consider the function $\psi: g^G\to (gh^{-1})^G; g^a \mapsto (gh^{-1})^a$ which maps the orbit $g^G$ to the orbit $(gh^{-1})^G$. This function is surjective, which implies $\infty > |g^G|\geqslant |(gh^{-1})^G|$. Which would imply $gh^{-1} \in A$.
Flaw 1: I think the function $\psi$ is not well defined (?)
Characteristic subgroup?
Choose a $\theta \in \mathrm{Aut}(G)$, now is $A^\theta = A$?
Choose a $g\in A$, I have to prove $g\in A^\theta = \{g\in G: |g^G|<\infty\}^\theta = \{g^\theta\in G: |g^G|< \infty\}$
Notice how $g=h^\theta$ for a unique $h\in G$, namely $h=g^{\theta^{-1}}$, does now $|h^G|< \infty$?
Consider $\psi: g^G\to h^G; g^a\mapsto (g^{\theta^{-1}})^a$ which is once again surjective. Then $|h^G|\leqslant |g^G|< \infty$.
Once again I think the function $\psi$ is not well defined (?)