The stability group of the action of fuchsian group on the upper half plane of is finite?

Let $$\Gamma$$ be a fuchsian group acting on the upper half plane $$\mathbb H$$, I wonder if for any $$z\in \mathbb H$$, the stability group of the action of fuchsian group on the upper half plane $$\Gamma_z:=\{\gamma\in \Gamma:\gamma.z=z\}$$ is necessarily finite.

I need this result to show that the subgroup of $$\Gamma$$ generated by an elliptic element is finite. If the above assertion is wrong I wish someone could show me how to prove this one. Thanks in advance!

• The stabilizer in $SL_2(\Bbb{R})$ of $i$ is $SO_2(\Bbb{R})$ which is compact isomorphic to $\Bbb{R/Z}$ a discrete subgroup of it is finite. The stabilizer of $z = g.i$ is $g SO_2(\Bbb{R}) g^{-1}$. Oct 1, 2019 at 12:48
• If you only know that $\Gamma\le SO_2(\Bbb{R})$ acts discretely on $\Bbb{H}$ then take $s\ne g.i \in \Bbb{H}$ such that $g SO_2(\Bbb{R}) g^{-1}$ is not contained in the stabilizer of $s$ thus $g SO_2(\Bbb{R})g^{-1} .s$ contains a curve $C$ and if $\Gamma \cap g SO_2(\Bbb{R}) g^{-1}$ is not finite then it is dense in $g SO_2(\Bbb{R}) g^{-1}$ and hence $\Gamma.s$ is dense in $C$ contradicting the discrete action. Oct 1, 2019 at 13:03
• It's called "stabilizer", and there is not one stabilizer per action, but a stabilizer $\Gamma_z$ for each point $z$.
– YCor
Oct 1, 2019 at 22:15

Yes, this is true, here's a proof.

By definition, a Fuchsian group $$\Gamma$$ is a discrete subgroup of $$\operatorname{PSL(2,\mathbb R)} = \operatorname{SL(2,\mathbb R)} / \pm \text{Id}$$ which is the full group of orientation preserving isometries of $$\mathbb H$$ (where a matrix representing an element of $$\operatorname{PSL(2,\mathbb R)}$$ acts on $$\mathbb H$$ by a fractional linear transformation. The topology on the group $$\operatorname{PSL(2,\mathbb R)}$$ can be described either as the quotient of the matrix topology on $$\operatorname{SL(2,\mathbb R)}$$.

Consider $$z \in \mathbb H$$. I'll use your notation $$\Gamma_z$$ for the stabilizer subgroup of $$z$$ in $$\Gamma$$. I also need a notation $$Stab(z)$$ for the stabilizer group of $$z$$ in the full isometry group $$\operatorname{PSL(2,\mathbb R)}$$.

The two key facts needed are:

• $$Stab(z)$$ is a compact group, in fact it is isomorphic to the circle group $$S^1$$ under complex multiplication.
• $$\Gamma_z$$ is a discrete subgroup of $$Stab(z)$$, because the intersection of a discrete subgroup of $$Stab(z)$$ and an arbitrary subgroup $$H < Stab(z)$$ is a discrete subgroup of $$H$$.
• Every discrete subgroup of a compact group is finite, and easy exercise in topological groups.

So yes, $$\Gamma_z$$ is finite. Furthermore, $$\Gamma_z$$ is a finite subgroup of the circle group, which implies that $$\Gamma_z$$ is a finite cyclic group.