Let $O$ be the ring of $S$-integers in a real quadratic number field. Let $G$ be an $S$-arithmetic subgroup of $SL\_2(O)$ whose intersection with $SL\_2(\mathbb Z)$ is not of finite index in $SL\_2(\mathbb Z)$.


*Assume the action of $G$ is properly discontinuous for simplicity.*


Can the fundamental domain of the action of $G$ on the complex upper half plane $\mathbb H$ have finite volume?


What if we assume $G$ to be of finite index in $SL\_2(O)$?


What if we remove the hypothesis that $G\cap SL\_2(\mathbb Z)$ be of infinite index in $SL\_2(\mathbb Z)$?


The reason I ask is simply because I wish to understand whether the "quotient" of $\mathbb H$ by $G$ could behave "well" in certain situations.



Subgroup of $SL_2(O)$ with nice fundamental domain in complex upper half-plane
No arithmetic subgroup of $SL\_2(O\_S)$ will ever act properly on the upper half-plane. These groups are simply too big, and the orbits will not be discrete subgroups. 


To see this explicitly, consider the orbit of $z \in \mathbb{H}$ under the subgroup of elements of $G$ of the form $\begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix}$, which is a finite-index subgroup $L$ of the additive group of $O\_S$. Then the orbit of $z$ is just a coset of the image of $L$ in $\mathbb{R}$, which is a dense subgroup of $\mathbb{R}$. 


If you want proper actions you have to make $G$ act on $\mathbb{H} \times \mathbb{H}$ (if $S = \varnothing$), as @MatthiasWendt points out; or the product of this with some non-archimedean spaces too if your set $S$ is nonempty


