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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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up vote 9 down vote accepted

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.


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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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My first stab at this would be to think of SL_2(Z,L) acting on the upper half plane. You can see what the cusps are and what conjugacy classes in SL_2(Z,L) they correspond to; mod out by the subgroup normally generated by those, and you get the fundamental group of the closed Riemann surface X(L); I guess I would try drawing in the upper half plane the explicit paths that you know generate the homology ("modular symbols") and see to what extent these generate the whole fundamental group. (But I have been careless in many places here and doubtless someone will just provide a reference so you don't have to do anything...)

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I knew your question rang a bell. If L is an odd prime, then it should be the normal closure of a single element, I think. I don't know if the element is "nice" enough for you, though.

See MR0565476.

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