Congruence subgroup of $SL_2(\mathbb{Z})$

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 $H/\Gamma_p$.)