There is a fact that a subgroup of linear fractional transformation $$\Omega := \Omega(\mathbb{H}) = \{M \in GL_2(\mathbb{R}) : \det(M) > 0\}$$ is a self-homeomorphism of the upper half plane $\mathbb{H} := \{x+iy \in \mathbb{C} : y>0\}.$
Any subgroup $A$ of $\Omega$ which acts on $\mathbb{H}$ nice enough (say properly discontinuous and fixed-point free), the the orbit space $$\mathbb{H}/A \text{ is a Riemann surface}$$ with charts created by pre-image under the natural projection $p : \mathbb{H} \rightarrow \mathbb{H}/A.$
Let $G$ be a subgroup of $\Omega$ which is $$G := \Big<\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \Big>$$ ($G$ is a commutator subgroup of the modular group; information https://projecteuclid.org/download/pdf_1/euclid.ijm/1255632506).
I would like to see and verify what is the Riemann surface created by orbit of the action $G$ acts on $\mathbb{H}$ by linear fractional map (what is the surface $\mathbb{H}/G$) ?
I am not quite sure how to start and deal with this kind of questions. Maybe first how to visualize the picture of the resulting Riemann surface. Then the standard technique to construct the homeomorphism of the surface that homeomorphic to $\mathbb{H}/G$.