most recent 30 from http://mathoverflow.net 2013-05-25T22:08:47Z http://mathoverflow.net/feeds/question/113797 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113797/2-generated-arithmetic-groups Aakumadula 2012-11-19T05:19:34Z 2013-03-03T03:25:09Z

<p>Suppose $G({\mathbb Z})$ is a higher rank non-cocompact arithmetic group (e.g. $SL_n({\mathbb Z})$ with $n\geq 3$, or $Sp_{2g}({\mathbb Z})$ with $g\geq 2$). I have seen a result (http://arxiv.org/abs/math/0409345) which says that every finite index subgroup $\Gamma $ of $G({\mathbb Z})$ contains a smaller finite index subgroup generated by three elements. </p> <p>Does anyone know ANY example of $G({\mathbb Z})$, where three can be replaced by two? I believe Alan Reid has some result in this direction. </p> <p>[Edit] That 2 should suffice is a conjecture, attributed to Alex Lubotzky. That $3$ DO suffice for non-uniform higher rank lattices in the result mentioned in the link. What I am asking is just ONE example where 2 generators suffice. </p>

http://mathoverflow.net/questions/113797/2-generated-arithmetic-groups/113798#113798 Igor Rivin 2012-11-19T05:34:58Z 2012-11-19T05:34:58Z

<p>The result to which you refer is not a result but a conjecture of A. Lubotzky. Long and Reid have constructed some examples. -- the relevant preprints can be found on <a href="http://www.ma.utexas.edu/users/areid/publications.html" rel="nofollow">Alan Reid's web page.</a> I assume that Lubotzky's conjecture is about three and not two generators because he did not want to be too ambitious -- nobody knows anything concrete, to the best of my knowledge.</p>