Let $\Gamma$ a virtual cyclic subgroup of $PSL(2,\mathbb{R})$, i.e. there are $\Lambda \leq \Gamma$ with $\Lambda$ cyclic of finite index. ¿Is $\Gamma$ elementary?, where elementary means that there are $z\in \mathbb{H}\cup \partial \mathbb{H}$ such that $|\Gamma \cdot z|<\infty$.
I think the answer is not, but I don't know how to construct virtual cyclic subgroups of $PSL(2,\mathbb{R})$. So, I tried to prove it. Suppose that $\Lambda = \big<h \big>$, with $ h \in PSL(2, \mathbb{R})$. Since the index of $\Lambda$ is finite we can choose the representatives of the lateral classes as $g_1, \ldots, g_k$, then for all $g \in \Gamma$ there are $i \in \{1,\ldots,k\}$ such that $g^{-1}\circ g_i\in \Lambda$, and since $\Lambda $ is cyclic $g^{-1}\circ g_i= h^{n_i}$. I think that the only possible $z\in \mathbb{H} \cup \partial\mathbb{H}$ with finite $\Gamma$-orbit must be a $z_0 \in Fix(h)$.