1
$\begingroup$

Suppose that $G$ is a finitely generated group, $X$ is a $G$-set, and $x \in X$ is a point. Are there any sorts of conditions on $X$ and $G$ that would let me conclude that $\operatorname{Stab}(x)$ is finitely generated as well?

Improve this question
$\endgroup$
1
  • $\begingroup$ Note that for any subgroup $H$ you can take $X=G/H$, so what you are asking is for a characterization of all finitely generated subgroups of a given group. $\endgroup$
    – user1688
    Commented Jul 14, 2016 at 8:08

2 Answers 2

Reset to default
6
$\begingroup$

"Stabilizer of $x$" is not any more specific than "subgroup of $G$" (consider the regular action of $G$ on $G/H$). In other words, you are asking for conditions on $G$ such that its subgroups are finitely generated.

This is false, in general. It's true in nilpotent (and virtually nilpotent) groups. It's false in solvable groups (consider the lamplighter).

Improve this answer
$\endgroup$
2
$\begingroup$

Lior Silberman is absolutely correct, but I wanted to add one other natural and simple condition in which you can conclude that the stabilizer of $x$ is finitely generated, namely, when the orbit of $x$ is finite. Equivalently, subgroups of finite index in finitely generated groups are finitely generated. This follows from Schreier generators (see "Schreier's lemma" or "Schreier's subgroup lemma").

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .