Subgroups of $SL(2, \mathbb{Z}/p \mathbb{Z})$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:46:17Z http://mathoverflow.net/feeds/question/108585 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108585/subgroups-of-sl2-mathbbz-p-mathbbz Subgroups of $SL(2, \mathbb{Z}/p \mathbb{Z})$ Igor Rivin 2012-10-02T00:37:51Z 2012-10-02T02:03:31Z <p>This must be known to everyone who might remotely be considered an expert (but not to me...): What are the subgroups of index three of $SL(2, \mathbb{Z}/p \mathbb{Z})?$</p> <p><strong>EDIT</strong> Following up on Qiaochu's answer, the situation for $SL(2, 3)$ is described <a href="http://groupprops.subwiki.org/wiki/Subgroup_structure_of_special_linear_group%3ASL%282,3%29" rel="nofollow">here.</a></p> http://mathoverflow.net/questions/108585/subgroups-of-sl2-mathbbz-p-mathbbz/108586#108586 Answer by Mark Sapir for Subgroups of $SL(2, \mathbb{Z}/p \mathbb{Z})$ Mark Sapir 2012-10-02T00:52:53Z 2012-10-02T00:52:53Z <p>@Igor: If a group has subgroup of index 3, it has a homomorphism into $S_3$. </p> http://mathoverflow.net/questions/108585/subgroups-of-sl2-mathbbz-p-mathbbz/108587#108587 Answer by Qiaochu Yuan for Subgroups of $SL(2, \mathbb{Z}/p \mathbb{Z})$ Qiaochu Yuan 2012-10-02T00:58:53Z 2012-10-02T02:03:31Z <p>A subgroup $H$ of index $3$ determines a homomorphism $\text{SL}_2(\mathbb{F}_p) \to S_3$ whose kernel $N = \bigcap_{g \in \text{SL}_2(\mathbb{F}_p)} gHg^{-1}$ is a normal subgroup of index either $3$ or $6$. The image of $N$ in $\text{PSL}_2(\mathbb{F}_p)$ is therefore also a normal subgroup of index $3$ or $6$. For $p \ge 5$ this group is simple and has size larger than $6$, so no such $N$ exists, hence no such $H$ exists.</p> <p>This leaves two cases $p = 2, 3$. When $p = 2$ we have $\text{SL}_2(\mathbb{F}_2) \cong S_3$ which has $3$ subgroups of index $3$ generated by each transposition. When $p = 3$ we have $|\text{SL}_2(\mathbb{F}_3)| = 24$, so its subgroups of index $3$ are precisely its Sylow $2$-subgroups. I think there are $3$ of them and that they are isomorphic to the <a href="http://en.wikipedia.org/wiki/Quaternion_group#Matrix_representations" rel="nofollow">quaternion group</a> $Q_8$. (Edit: apparently there is only one.) </p>