6
$\begingroup$

Are the congruence subgroups of the modular group $\Gamma\equiv\mathrm{PSL}\left(2,\mathbb{Z}\right)$ (e.g. $\Gamma\left(n\right)$, $\Gamma_{0}\left(n\right)$, $\Gamma_{1}\left(n\right)$ etc.) finitely presented? If so, is there a proof of this? Assuming they are finitely presented, are the presentations of e.g. the principal congruence subgroups $\Gamma\left(n\right)$ (for small $n$) documented anywhere? I know they could be computed using the Reidemeister-Schreier process, but it would be nice to have some independent confirmation of the presentations.

Many thanks!

$\endgroup$
7
$\begingroup$

A subgroup of finite index in a finitely presented group is finitely presented (see Exercise 6.1.6 in Robinson: A course in the theory of groups), so all congruence subgroups of the modular group are finitely presented. I cannot answer your second question.

$\endgroup$
18
$\begingroup$

Not only these subgroups are finitely presented, they are all finite free products of cyclic groups; most of them (for sufficiently large $n$) are actually free of finite rank (once congruence subgroup contains no elements of order 2 and 3). For instance, you can easily check that $\Gamma(n)$ is torsion-free for all $n\ge 2$ by looking at traces for $n\ge 3$ (since $1\ne 2$ mod $n\ge 2$ and $2\ne 0$ mod $n\ge 3$) and by looking at matrix coefficients for $n=2$. Rank is easily computable if you know index of the congruence subgroup in the modular group. The magic formula is multiplicativity of the Euler characteristic: For the modular group $\Gamma$, $\chi=-1+\frac{1}{2} + \frac{1}{3}=-\frac{1}{6}$. If $\Gamma'\subset \Gamma$ is a subgroup of index $i$ then $\chi(\Gamma')=i \chi(\Gamma)$. If $\Gamma$ is free of rank $r$ then $\chi(\Gamma)= 1-r$. For instance, to find index $i$ for $\Gamma(n)$, compute the order of the quotient group $SL(2, Z_n)/\pm I$. There is a closed formula for the order of this group (in terms of prime factors of $n$) which will tell you what the index is:

If $n$ is the product of powers of primes $\prod_i p_i^{k_i}$ then $$ |PSL(2,Z_n)|= \frac{n^3}{2} ~~~\prod_i (1- p_i^{-2}). $$

$\endgroup$
  • 1
    $\begingroup$ Like this post except that I don't find the index formula ugly at all! $\endgroup$ – GH from MO Aug 22 '12 at 17:34
  • $\begingroup$ I should note that the formula above can be found in Shimura's book (I am not sure what other sources there are...) $\endgroup$ – Igor Rivin Aug 27 '12 at 4:39
4
$\begingroup$

Well, the congruence subgroups are finite index, the group $PSL(2,\mathbb{Z})$ is finitely presented, so they are too. The magic words are: "Reidemeister-Schreier" and Magnus-Karras-Solitar.

$\endgroup$
  • 1
    $\begingroup$ You repeated my earlier response (except for the magic words). $\endgroup$ – GH from MO Aug 22 '12 at 14:42
  • 4
    $\begingroup$ I think the OP knows the magic words already. $\endgroup$ – Autumn Kent Aug 22 '12 at 15:02
  • 2
    $\begingroup$ @GH I didn't see your response when I posted... $\endgroup$ – Igor Rivin Aug 22 '12 at 15:14
  • $\begingroup$ @Igor: That's what I thought. Thanks for the note! $\endgroup$ – GH from MO Aug 22 '12 at 17:35
  • $\begingroup$ @GH: not at all... $\endgroup$ – Igor Rivin Aug 22 '12 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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