# all hyperbolic cyclic subgroup of P$SL(2, \mathbb{R})$ are Fuchsian.

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 $$z_1, z_2$$.

• 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 $\left\{\begin{pmatrix} t & 0 \\ 0 & 1/t \end{pmatrix} \mid t > 0\right\}$. – Lee Mosher Aug 6 '19 at 22:41
• 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 $\Gamma$? – Livpez. Aug 7 '19 at 11:44
• It's easy, but the difficulty might be sensitive to your choice of definition of "hyperbolic cyclic subgroup". – YCor Aug 7 '19 at 13:50
• @YCor which definition would you suggest? – Livpez. Aug 7 '19 at 17:13
• It's your own question, so I expect you have one in mind. – 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 $$\lambda≠1$$.