Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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 $H/\Gamma_p$.)

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.