Let $G \subset PSL(2, \Bbb R)$ be a discrete subgroup without elliptic or parabolic elements. Does it follow that it is at most countable?
Subgroups as above have the property that the quotients of the hyperbolic upper half-plane $H$ by them are compact smooth surfaces. The fact that $G$ acts properly discontinuously on $H$ makes me suspect that it cannot be uncountable (otherwise, "something" would have a limit point, and this would somehow contradict the proper discontinuous action), but since I cannot visualize these objects, I do not know how to encode my intuition into a formal reasoning.
(For me, $G$ acts properly discontinuously on $H$ if and only if for every $z \in H$ there exist an open subset $U_z \subset H$ such that $U_z \cap g U_z = \emptyset, \ \forall g \in G \setminus \{1\}$. This is equivalent to $G$ being a discrete subgroup with no elliptic elements.)
A posteriori, the fact that, for $f \in L^1 (H)$ and $D$ a fundamental domain for $G \backslash H$, one may write $\int _H f = \sum _{g \in G} \int _{gD} f$, is a further argument that $G$ must be at most countable, otherwise $\sum _{g \in G}$ wouldn't make sense. This, on the other hand, is not a proof.
