I'm trying to prove that all hyperbolic cyclic subgroups of $PSL(2, \mathbb{R})$ are Fuchsian.

Let $\Gamma$ be a hyperbolic subgroup of $PSL(2, \mathbb{R})$. It is enough to show that $\Gamma$ is discrete, since all discrete subgroups of $PSL(2, \mathbb{R})$ are Fuchsian.

My aim was to do this by contradiction.

Assume $\Gamma$ is not Fuchsian. This implies there exists a sequence $\{T_k\}$ with $T_k \in \Gamma$ and $T_k \rightarrow Id$ as $k \rightarrow \infty$.

From here I don't know where to go, but surely I will have to use the fact that every hyperbolic element $T \in\Gamma$ has two fixedpoints, say $z_1, z_2$.

  • $\begingroup$ I don't see where your proof has used the hypothesis that $\Gamma$ is cyclic. And you have to use that hypothesis, because it is false that every hyperbolic subgroup is Fuchsian: take, for example, the subgroup $\left\{\begin{pmatrix} t & 0 \\ 0 & 1/t \end{pmatrix} \mid t > 0\right\}$. $\endgroup$ – Lee Mosher Aug 6 '19 at 22:41
  • $\begingroup$ thanks, I know that cyclic is necessary, but I don't know how to use this fact. should I use the fixed points of the $T \in \Gamma$ which generates the whole $\Gamma$? $\endgroup$ – Livpez. Aug 7 '19 at 11:44
  • $\begingroup$ It's easy, but the difficulty might be sensitive to your choice of definition of "hyperbolic cyclic subgroup". $\endgroup$ – YCor Aug 7 '19 at 13:50
  • $\begingroup$ @YCor which definition would you suggest? $\endgroup$ – Livpez. Aug 7 '19 at 17:13
  • $\begingroup$ It's your own question, so I expect you have one in mind. $\endgroup$ – YCor Aug 7 '19 at 19:13

You have the right idea by finding some sequence $\{T_k\}$ which converges to the identity. How to do this, you have to recall that we can conjugate any hyperbolic elements to $z\mapsto \lambda z$ for $\lambda≠1$.

For more detail see my answer here.

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