 Rational conjugation of finite-index modular subgroups

For any finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ and rational positive-determinant $\gamma$, the intersection $S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}\Gamma \gamma$ has finite index. Multiply $\gamma$ by a positive rational scalar to obtain an integral matrix $A$ with positive integer determinant $D$. For $h=I+DB\in \Gamma (D)$, $Ah{A}^{-1}=I+AB\mathrm{adj}(A)$ is integral, so $\Gamma (D)$ lies in $S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. This intersection has finite index. Its further intersection with ${\gamma }^{-1}\Gamma \gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. No assumption that $\Gamma$ itself is a congruence subgroup is needed.