Let $\textbf{T}:=\bigl(\begin{smallmatrix} 1&1\\ 0&1 \end{smallmatrix} \bigr)$, $\textbf{S}:=\bigl(\begin{smallmatrix} 0&1/\sqrt{N}\\ -\sqrt{N}&0 \end{smallmatrix} \bigr)$ and $H$ the subgroup of $\operatorname{SL}_2(\textbf{R})$ generated by $\textbf{T}$ and $\textbf{S}$.
Is that true that $H$ contains $\Gamma_0(N)$ as a subgroup of index two ? (this statement is true for $N=4$)
Thanks !
For all $N>4$ the group $H$ is discrete, but the area of $U/H$ is infinite, where $U$ is the upper half-plane with the hyperbolic metric. (You can see this by drawing a fundamental polygon, which is bounded by three disjoint hyperbolic geodesics.) On the other hand, for each finite-index subgroup $\Gamma'$ (e.g. a congruence-subgroup) of the modular group $\Gamma$ the area of $U/\Gamma'$ is finite. Therefore, for all $N>4$, $H$ cannot contain a finite index subgroup of $\Gamma$, in particular, it cannot contain a congruence subgroup.