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Let $\Gamma$ be a congruence subgroup, $s$ be a cusp of $\Gamma$, $\sigma \in SL_{2}(\mathbb{Z})$ be the map which sends $\infty$ to $s$, and $\Gamma_{s} = \{\gamma \in \Gamma \mid \gamma(s) = s\}$. If we let $\Gamma$ act on the upper half plane in the usual way, why must each $\rho \in \sigma^{-1}\Gamma_{s}\sigma$ be a translation by some $h_{1}$?

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up vote 4 down vote accepted

Every element of $\sigma^{-1} \Gamma_s \sigma$ is in particular a Möbius transformation $\rho : z \mapsto \frac{az+b}{cz+d}$ which fixes $\infty$. It follows that $c = 0$ (otherwise $\rho(\infty) = \frac{a}{c}$ is finite). Now, since in addition we know that everything lies in $\text{SL}_2(\mathbb{Z})$, we know that $ad = 1$, so $a = d = \pm 1$. Then $\rho : z \mapsto z \pm b$ is a translation as desired.

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