Congruence subgroup of $SL_2(\mathbb{Z})$ It is known that the congruence subgroup $\Gamma_p$ of $SL_2(\mathbb{Z})$, that is the kernel of the epimorphism $SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}_p)$ (with $p$ a prime number), is a free group.
Have you a reference for this result?
 A: If $\Gamma_p$ is  torsion free (which will be the case provided $p > 2$), then it acts freely and properly diconstinuously on the upper half-plane $H$, and so is identified with the fundamental group of the quotient
$H/\Gamma_p$.  But this quotient is a punctured Riemann surface, and hence its $\pi_1$ is free.  Thus $\Gamma_p$ is free.  (And it is not difficult to compute the number of generators, since this is just a matter of determining the genus and number of punctures of $H/\Gamma_p$.)
A: We can also argue as follows. As in Matt E's answer we first need to begin by observing that $\Gamma(p)$ is torsion-free once $p \ge 3$; in particular this means its intersection with the center $-1 \in SL_2(\mathbb{Z})$ is trivial, so it injects into the modular group $\Gamma = PSL_2(\mathbb{Z})$.
Next, recall that $\Gamma$ can be written as the free product $C_2 \ast C_3$ (a proof is given here). By the Kurosh subgroup theorem every subgroup of $\Gamma$ must be a free product of copies of $\mathbb{Z}, C_2$, and $C_3$, and among these the ones which are torsion-free are precisely the free products of copies of $\mathbb{Z}$. So we get that every torsion-free subgroup of $\Gamma$, and hence of $SL_2(\mathbb{Z})$, is free.
One can work quite explicitly with finite index subgroups of $\Gamma$ by drawing certain graphs of groups / dessins d'enfant; see this blog post for some nice pictures.
