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Subgroups of $SL_2(\mathbb R)$ which contain $SL_2(\mathbb Z)$ as a finite index subgroup

$G = SL_2(\mathbb{Z})$ is indeed the only possibility. This follows from the structure of the quotient orbifold $\mathcal{O} = \mathbb{H}^2 / SL_2(\mathbb{Z})$, which is a once-punctured sphere with a …
Lee Mosher's user avatar
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