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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!

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You should look at the answers to mathoverflow.net/questions/2757/… –  Richard Kent Aug 22 '12 at 15:01
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3 Answers 3

up vote 7 down vote accepted

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.

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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 $$ |PSL(2,Z_n)|= \frac{n^3}{2} ~~~\prod_i (1- p_i^{-2}). $$

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

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