# Generators for congruence subgroups of SL_2

For positive integers $n$ and $L$, denote by $SL_n(Z,L)$ the level $L$ congruence subgroup of $SL_n(Z)$, i.e. the kernel of the homomorphism $SL_n(Z)\rightarrow SL_n(Z/LZ)$.

For $n$ at least $3$, it is known that $SL_n(Z,L)$ is normally generated (as a subgroup of $SL_n(Z)$) by Lth powers of elementary matrices. Indeed, this is essentially equivalent to the congruence subgroup problem for $SL_n(Z)$.

However, this fails for $SL_2(Z,L)$ since $SL_2(Z)$ does not have the congruence subgroup property.

Question : Is there a nice generating set for $SL_2(Z,L)\ ?$ I'm sure this is in the literature somewhere, but I have not been able to find it.

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Hi Andy,

I don't know if you are still interested in this, but I just found the reference:

MR0049937 (14,250d) Grosswald, Emil On the parabolic generators of the principal congruence subgroups of the modular group. Amer. J. Math. 74, (1952). 435--443.

It is based on the previous work of H.Frasch (1933) who gave an explicit set of free generators for principal congruence subgroups Gamma(p) in PSL(2,Z), for prime p's.

-Ignat

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Kulkarni (American J of Math, 113, 6, 1053-1133) gives a method to compute nice fundamental domains for the action of subgroups on $SL_2(\mathbf{Z})$ on the upper half-plane.

"Nice" means that in particular that the subgroup is a free product of the subgroups generated by the edge-pairing transformations. In the case of $\Gamma(N) =\ker SL_2(\mathbf{Z})\to SL_2 (\mathbf{Z} / {N})$ we have a free group and so we get a free system of generators. Kulkarni's approach is based on the observation that the congruence subgroups of $SL_2(\mathbf{Z})$ are in a bijection with "bipartite cuboid graphs", which are unitrivalent graphs with a cyclic order on the edges meeting at a trivalent vertex plus some extra data. However, Kulkarni's method involves "trial and error" and I don't think explicit sets of generators are known for general $N$.

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I was at a workshop on noncongruence modular forms recently (a noncongruence subgroup of $SL_2(\mathbb{Z})$ is a subgroup of finite index not containing any $\Gamma(N)$ ) when this question came up. I believe that the consensus was that although one can, in principle, compute a generating set for $\Gamma(N)$, the algorithm is not terribly effective for $N$ large (where by large I mean greater than $13$ or so). The algorithm involves computing the Farey symbol associated to $\Gamma(N)$ and getting coset representatives. The difficulty is that the index of $\Gamma(N)$ increases very rapidly, making the calculation of the Farey symbol very lengthy.

Ling Long and Chris Kurth have written a paper about the algorithm (it references the paper of Kulkarni that algori mentioned), which is available here.

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