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

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.

-

2 Answers

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 $Q_8$. (Edit: apparently there is only one.)

-
 An even better answer than @Mark's! Thanks... The situation for $SL(2, 3)$ is described in nauseating detail on the appropriate groupprops page. – Igor Rivin Oct 2 at 1:01 $S_3$ has 3 normal subgroups of index 3?? – algori Oct 2 at 1:59 Whoops. Fixed. Thank you. – Qiaochu Yuan Oct 2 at 2:03

@Igor: If a group has subgroup of index 3, it has a homomorphism into $S_3$.

-
 Hence a normal subgroup of index at most six, which $SL(2, p)$ usually does not. Duh, that was easy, thanks! – Igor Rivin Oct 2 at 1:00