Discrete subgroups are lattices
The additive group $\Gamma/\Gamma_0$ has order $q$, so for every $\gamma+\Gamma_0\in\Gamma/\Gamma_0$, $q(\gamma+\Gamma_0)=0+\Gamma_0=\Gamma_0$. But $q(\gamma+\Gamma_0)=q\gamma+\Gamma_0$, so $q\gamma+\Gamma_0=\Gamma_0$, and therefore $q\gamma\in\Gamma_0$. Since $\gamma\in\Gamma$ was arbitrary, $q\Gamma\subseteq\Gamma_0$.