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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?

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Not true, $\Gamma_2$ has an element of order 2. –  user641 Aug 3 '12 at 12:25
Group-theoretically, though, you can proceed as follows: $SL(2,\mathbb{Z})\cong C_4\ast_{C_2} C_6$, and the subgroup will be free iff it is torsion-free, that is, avoids all elements of orders $2$, $3$, $4$, and $6$. All of these have a handful of standard forms in $SL(2,\mathbb{Z})$, so just check if any are congruent to the identity mod $p$. –  user641 Aug 3 '12 at 12:26
The criterion given by Steve D is for example proved in Serre's Trees, paragraph 4.3. –  PseudoNeo Aug 3 '12 at 12:48
Thank you, you completely answered my question. –  Seirios Aug 3 '12 at 14:51
@SteveD I suggest you post your comment as an answer, then. –  M Turgeon Aug 3 '12 at 15:30

1 Answer 1

up vote 4 down vote accepted

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 $H/\Gamma_p$.)

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Another way to find the number of generators is with Euler characteristic: $\chi(SL(2,\mathbb{Z}))= 1/4 + 1/6 - 1/2 = -1/6$. Now $\Gamma_p$ has index $|SL(2,p)|=p(p-1)(p+1)$, and so $\chi(\Gamma_p)=-p(p-1)(p+1)/6$. Thus the rank of $\Gamma_p$ is $1+p(p-1)(p+1)/6$. –  user641 Aug 3 '12 at 16:47

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