In a 1985 paper named "Dimension function for discrete groups" Lubotzky conjectured that:

For any integer $n \geq 3$ the group $\mathrm{SL}_n(\mathbb{Z})$ contains infinitely many finite index subgroup of rank $2$.

He has probably also conjectured that the rank $2$ subgroups form a basis for the profinite topology (see Problem 1 of Section 4 in his work).

A 2010 work of Long and Reid ("Small subgroups of $\mathrm{SL}_3(\mathbb{Z})$") confirmed the conjecture for $n=3$.

I am interested in the current status of this conjecture.

More concretely, denote by $q(n)$ the least integer for which there are infinitely many finite index subgroups of $\mathrm{SL}_n(\mathbb{Z})$ of rank at most $q(n)$.

What is the best known bound for $q(n)$ ?

Sury and Venkataramana ("GENERATORS FOR ALL PRINCIPAL CONGRUENCE SUBGROUPS OF $\mathrm{SL}_n(\mathbb{Z})$ WITH $n \geq 3$") show that $q(n) \leq n^4$ by presenting generating sets for principal congruence subgroups.

Both proofs and references to related works will be highly appreciated.

Here the rank of a group is the smallest cardinality of a generating set.

A related work is that of Sharma and Venkataramana ("Generations for arithmetic groups") where it is shown that any noncocompact irreducible lattice in a higher rank real semi-simple Lie group contains a subgroup of finite index with rank at most $3$.

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    $\begingroup$ as far as I know, this conjecture is open; the only progress on two generators is by Reid and Long, but it does not confirm the conjecture; they produce infinitely many two generated finite index subgroups, whose intersection is identity, but they do not prove that every finite index subgroup contains one of these.. $\endgroup$ – Venkataramana Oct 4 '15 at 10:16
  • $\begingroup$ @Venkataramana thank you so much for clarifying this point! Do you know what happens if we allow more than $2$ generators (but still a finite number) even for the special case of $\mathrm{SL}_3(\mathbb{Z})$? $\endgroup$ – Pablo Oct 4 '15 at 10:21
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    $\begingroup$ @ pablo. No problem. But you have already answered your question: since $SL_3(\mathbb Z)$ is a lattice in $SL_3(\mathbb R)$, your $q(3)$ is no larger than $3$; it could be two, if you can prove the full Lubotzky question. $\endgroup$ – Venkataramana Oct 4 '15 at 14:39
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    $\begingroup$ There is a recent pre-print by Chen Meiri where he proves the conjecture of Lubotzky that you have stated $\endgroup$ – Venkataramana Dec 13 '15 at 3:44
  • $\begingroup$ @Venkataramana Thanks for informing me! I have already seen it, and myself written a paper on the analogous problem for boundedly generated groups. You can see it in arxiv.org/abs/1510.05904 $\endgroup$ – Pablo Dec 13 '15 at 17:37

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