Cusp holomorphy under rational slash operators

If $f$ is a modular form on a finite-index subgroup and $\gamma$ is rational with positive determinant, then $f{|}_k\gamma$ is bounded near infinity. Choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =\gamma \infty$. The matrix ${\sigma }^{-1}\gamma$ is upper triangular and sends $z$ to $Az+B$ with $A>0$. Thus the existing cusp expansion of $f{|}_k\sigma$ stays bounded under this substitution. Rational conjugation of finite-index modular subgroups gives a positive period for the translate, so boundedness gives a removable singularity in its cusp parameter. Apply this to $\gamma \rho$ for every $\rho \in S{L}_2(\mathbb{Z})$ to obtain holomorphy at every cusp of the new subgroup. Vanishing at the cusps is preserved too.