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!

  • $\begingroup$ 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 $g SO_2(\Bbb{R}) g^{-1}$. $\endgroup$
    – reuns
    Oct 1, 2019 at 12:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – reuns
    Oct 1, 2019 at 13:03
  • $\begingroup$ It's called "stabilizer", and there is not one stabilizer per action, but a stabilizer $\Gamma_z$ for each point $z$. $\endgroup$
    – YCor
    Oct 1, 2019 at 22:15

1 Answer 1


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.


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