Degree and ramification index of a natural projection Let $\Gamma$ a subcongruence group of $\text{SL}_2(\mathbb{Z})$, $\mathbb{H}^* = \mathbb{H} \cup \{\infty\} \cup \mathbb{Q}$ and $\overline\Gamma = \Gamma / (\Gamma\cap\{\pm 1\})$. Given the natural projection
$$\pi:\Gamma/\mathbb{H}^* \to {\text{SL}_2(\mathbb{Z})}/\mathbb{H}^*,$$
why does there hold for the degree of the mapping $\pi$:
$$\deg \pi = (\overline{\text{SL}_2(\mathbb{Z})}:\overline{\Gamma})$$
and for the ramification index:
$$e_y = (\overline{\text{SL}_2(\mathbb{Z})}_y:\overline{\Gamma}_y)$$
This is one exercise that has been "left to the reader". Unfortunately, by starting from the definitions $\deg \pi = \sum_{y\in\pi^{-1}(x)} e_y$ where $\pi(z)=\sum_{n=e_y} a_n z^n, a_{e_y}\neq 0$, I failed to give a proof.
 A: More generally, let $\Gamma_1\subseteq\Gamma_2$ be congruence subgroups, and let $f:X(\Gamma_1)\to X(\Gamma_2)$ be the usla projection $\Gamma_1\tau\mapsto \Gamma_2\tau$. I claim that $\text{deg}(f)=[\Gamma_2:\Gamma_1]$ (where everything is taking place in $\text{PSL}_2(\mathbb{Z})$).
So, let's suppose that $\Gamma_2\tau_0\in X(\Gamma_2)$. Note then that by definition an element $\Gamma_1\tau$ is in the fiber $ f^{-1}(\Gamma_2\tau_0)$ if and only if $\tau=\gamma\tau_0$ for some $\gamma\in\Gamma_2$. Thus, we can deduce that the fiber $f^{-1}(\Gamma_2\tau_0)$ is nothing but $\{\Gamma_1\gamma\tau_0:\gamma\in\Gamma_2\}$. Now, $\Gamma_1\gamma\tau_0=\Gamma_1\gamma'\tau_0$ if and only if there exists some $\delta\in\Gamma_1$ s.t. $\delta\gamma\tau_0=\gamma'\tau_0$, or $(\gamma')^{-1}\delta\tau_0=\tau_0$. Now, if $\tau_0$ is not an elliptic point of $\Gamma_2$, then this implies that $(\gamma')^{-1}\delta\gamma=e$. Thus, if $\tau_0$ is not an elliptic point $\Gamma_1\gamma\tau_0=\Gamma_1\gamma'\tau_0$  if and only if $\gamma\Gamma_1=\gamma'\Gamma_1$. So, $|f^{-1}(\Gamma_2\tau_0)|$ is just $[\Gamma_2:\Gamma_1]$. Since there always exists a non-elliptic point, we may conclude, obviously that $\text{deg}(f)=[\Gamma_2:\Gamma_1]$ as desired.
Your phrasing of the problem comes merely from the go-between between thinking of subgroups of $\text{PSL}_2(\mathbb{Z})$ and $\text{SL}_2(\mathbb{Z})$. 
The computation of degrees can be found in Diamond and Shurman, page 67.
