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 $SL(2, \mathbb{Z}/p \mathbb{Z})?$


**EDIT**
Following up on Qiaochu's answer, the situation for $SL(2, 3)$ is described [here.](http://groupprops.subwiki.org/wiki/Subgroup_structure_of_special_linear_group%3ASL%282,3%29)



Subgroups of  $SL(2, \mathbb{Z}/p \mathbb{Z})$
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.


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 [quaternion group](http://en.wikipedia.org/wiki/Quaternion_group#Matrix_representations) $Q\_8$. (Edit: apparently there is only one.) 


