To check that a subgroup of $SL(n, \mathbb{Z})$ for $n>2$ is profinitely dense, one need only check that it surjects under all the projections mod $m\in \mathbb{Z}$ (which, while in appearance infinite, is usually a finite computation). What about $SL(2, \mathbb{Z})$ is there anything resembling a decision procedure?



profinite density in $SL(2, \mathbb{Z})$
This group is virtually free so subgroup separable so a fg subgroup is dense iff it is the whole thing. For non-fg subgroups an algorithm probably doesn't make sense. 


