Computing congruence subgroups indeces in $\mathrm{PSL}_2(\mathbb{Z})$

Question

Let $N$ be an integer, it is easy compute the indices in $\mathrm{SL}_2(\mathbb{Z})$ of the congruence subgroups $\Gamma(N),\Gamma_1(N),\Gamma_0(N)$.

My question is: can I compute the indices $$[\mathrm{PSL}_2(\mathbb{Z}):\bar{ \Gamma}(N)],[\mathrm{PSL}_2(\mathbb{Z}):\bar{ \Gamma}_1(N)],[\mathrm{PSL}_2(\mathbb{Z}):\bar{ \Gamma}_0(N)]$$

just knowing their indices in $\mathrm{SL}_2(\mathbb{Z})$?

Answer 1

If $G$ is a subgroup of $\mathrm{SL}_(\Bbb Z)$ and $\bar G$ is a the image of $G$ in $\mathrm{PSL}_2(\Bbb Z)$, then $$|\mathrm{PSL}_2(\Bbb Z):\bar G|=|\mathrm{SL}_2(\Bbb Z):\pm G|.$$ This is either $|\mathrm{SL}_2(\Bbb Z):G|$ or $\frac12|\mathrm{SL}_2(\Bbb Z):G|$ according whether or not $-I\in G$.

Answer 2

Explicitly the group index is given as follows:

$$ [PSL(2,\mathbb{Z}):\Gamma (m)]=\frac{m^3}{2}\prod_{p\mid m}\left(1-\frac{1}{p^2}, \right),\; m\ge 3 $$

$$ [PSL(2,\mathbb{Z}):\Gamma_1 (m)]=\frac{m^2}{2}\prod_{p\mid m}\left(1-\frac{1}{p^2}, \right),\; m\ge 3 $$

$$ [PSL(2,\mathbb{Z}):\Gamma_0 (m)]=m\prod_{p\mid m}\left(1+\frac{1}{p}, \right),\; m\ge 1. $$

Here $$ \Gamma (m) = \{A \in PSL(2,\mathbb{Z}) \mid A\equiv \pm I \bmod m\}/\{\pm I\}. $$