Infinite index subgroups of $SL(3,\mathbf{Z})$

Question

Finite index subgroups of $SL(3,\mathbf{Z})$ are well-know (at least we know that they are congruence subgroups).

But I wasn't able to find reference on infinite index subgroups. Does someone knows how to find such a documentation ?

More precisely, I would like to know : 1) Does $SL(3,\mathbf{Z})$ has a infinite index normal subgroup ? 2) If no, what are, up to conjugacy, the infinite index subgroups of $SL(3,\mathbf{Z})$ ?

Answer 1

Excluding the trivial group, the answer to 1) is that there are none. This is because the center of $SL(3,\mathbf{Z})$ is trivial combined with the Margulis normal subgroup theorem:

Let $\Gamma$ be an irreducible lattice in a connected semi-simple Lie group $G$ with finite center and of real rank at least two. The normal subgroups of $\Gamma$ are contained in the center (hence finite) or have finite index.

I don't know if there is a substantially simpler proof of this fact for $\Gamma = SL(3,\mathbf{Z})$ and $G = SL(3,\mathbf{R})$ than specializing the proof of the general case --- I doubt that there is a way around proving that for a non-central normal subgroup the quotient $\Gamma/N$ is amenable and has property (T) and hence is finite. A nice exposition of the fundamental ideas leading to Margulis's normal subgroup theorem can be found in these notes by Clara Löh.

The resources available on the homepage of Dave Morris (books/lecture notes/slides) could also be helpful.

I am pessimistic that there is a reasonable description of all infinite index subgroups of $SL(3,\mathbf{Z})$ up to conjugacy, since this includes in particular all subgroups of $SL(2,\mathbf{Z})$.