Congruence subgroups as abstract groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:21:06Z http://mathoverflow.net/feeds/question/71969 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71969/congruence-subgroups-as-abstract-groups Congruence subgroups as abstract groups Alex Bartel 2011-08-03T06:10:29Z 2011-08-12T06:42:11Z <p>This is probably well know, and maybe even trivial, but not to me. Consider for concreteness the subgroup <code>$$\pm\Gamma_0(3)=\left\{\begin{pmatrix}a &amp; b \\ c &amp; d\end{pmatrix}:\;a,b,c,d\in\mathbb{Z},ad-bc=\pm1, c\equiv 0\pmod 3\right\}$$</code> of $GL_2(\mathbb{Z})$. This has of course index 4 in $GL_2(\mathbb{Z})$. The first (possibly completely ridiculous) question is</p> <blockquote> <p>Does $\pm\Gamma_0(3)$ contain a subgroup isomorphic to $GL_2(\mathbb{Z})$?</p> </blockquote> <p>It's not even obvious to me that the two are not isomorphic as abstract groups. The second question is</p> <blockquote> <p>Does $GL_2(\mathbb{Z})$ contain subgroups that are isomorphic to $\pm\Gamma_0(3)$ with finite index other than 4? If the answer is yes, then what is the least common multiple of all such indices? E.g. is there a subgroup of index 3 (or 5, or 7, or...) in $GL_2(\mathbb{Z})$ isomorphic to $\pm\Gamma_0(3)$? Or will all such indices be multiples of 4?</p> </blockquote> <p>An answer or technique that is applicable to other congruence subgroups and to other values of 2 would be a great bonus, but for now I would happily settle for an answer to this concrete question.</p> http://mathoverflow.net/questions/71969/congruence-subgroups-as-abstract-groups/71975#71975 Answer by Geoff Robinson for Congruence subgroups as abstract groups Geoff Robinson 2011-08-03T07:46:43Z 2011-08-03T13:58:35Z <p>Junkie's comment answers both parts of the first question, since $\pm \Gamma_{0}(3)$ contains no element of order $4$ (its image after reduction (mod 3) would still have order $4$). They also provide a suggestion to deal with other primes. For other $p> 3$, I think you can do something like this. The matrices congruent to the identity (entrywise) (mod p) form a torsion free normal subgroup $H$ of ${\rm GL}(2,\mathbb{Z})$. The image of the congruence subgroup (mod $H$) is solvable, and has a normal Sylow $p$-subgroup with Abelian factor group. However, if $X$ is a subgroup of the congruence subgroup isomorphic to ${\rm GL}(2,\mathbb{Z})$, then $X/X \cap H$ contains a dihedral subgroup of order $8$.</p> http://mathoverflow.net/questions/71969/congruence-subgroups-as-abstract-groups/72744#72744 Answer by Lior Silberman for Congruence subgroups as abstract groups Lior Silberman 2011-08-12T06:42:11Z 2011-08-12T06:42:11Z <p>For $n\geq 3$ (and in higher-rank in general) the question is basically settled by <A href="http://en.wikipedia.org/wiki/Superrigidity" rel="nofollow">Super-rigidity</a>. For example, if you have two lattices of $\mathrm{SL}_n(\mathbb{R})$ which are abstractly isomorphic then they must be obtained from each other by an automorphism of $\mathrm{SL}_n(\mathbb{R})$ (inverse transpose and/or inner conjugation).</p>