Finite index free subgroups of $\mathrm{SL}(3,\mathbb{Z})$ Does $\mathrm{SL}(n,\mathbb{Z})$ have a free subgroup of finite index for some $n \geq 3$? I know that $\mathrm{SL}(3,\mathbb{Z})$ has many free subgroups and that in the case of $\mathrm{SL}(2,\mathbb{Z})$ the principal congruence subgroups are free for levels $\geq 3$. I believe that $\mathrm{SL}(3,\mathbb{Z})$ also has surface groups as subgroups, and I would be interested to know if any of those have finite index.
 A: Apparently the answer (to the title question, anyway) is no. Based on some googling, it looks like (all groups below are finitely generated)


*

*virtually free groups are hyperbolic,

*hyperbolic groups have linear Dehn functions, and

*$\text{SL}_n(\mathbb{Z})$ is known not to have a linear Dehn function for $n = 3$ and $n \ge 5$ (the case $n = 4$ apparently remains conjectural). 

A: Let's show that $SL(n,Z)$ ($n\ge 3$) contains no free groups (and surface groups) of finite index. The same argument shows that it contains no finite index hyperbolic subgroups; in other words, each $SL(n,Z)$ (for $n\ge 3$) has nonlinear Dehn function. Consider the subgroup $N$ of strictly upper triangular matrices in $SL(3,Z)$ (clearly, $N$ is also a subgroup of $SL(n,Z)$, $n\ge 3$). It is an elementary exercise to see that $N$ is 2-step nilpotent and not virtually abelian. Hence, intersection of every finite index subgroup of $SL(n,Z)$ with $N$ cannot be a free group or a surface group. (The former is immediate, the latter requires going through the list of surface groups and thinking a little bit about the fact that infinite index noncyclic subgroups in surface groups cannot be abelian.) 
The above argument is completely elementary, there are many nonelementary arguments, e.g., using cohomological dimension, Property T, Margulis normal subgroups theorem or congruence subgroup property (see comments by Andy Putman and Ian Agol above). 
On the other hand, there are several cute constructions of closed hyperbolic surface subgroups in $SL(3,Z)$, some are Zariski dense, some are not. It is an interesting open problem to determine if $SL(3,Z)$ contains Zariski dense torsion-free finitely-generated subgroups of infinite index which are neither free nor surface groups. 
A: The fact that $\mathrm{SL}(n, \mathbb{Z})$ for $n \geq 3$ has no free subgroup of finite index goes back a while, but one standard way to see this is via the work of Bass-Lazard-Serre on the Congruence Subgroup Problem here, or more comprehensive later work by Bass-Milnor-Serre and others.
Since every subgroup of finite index contains a congruence subgroup and hence the kernel of reduction modulo some prime $p$, and since subgroups of free groups are also free, it's enough to locate a non-free subgroup of this kernel.   Take the upper triangular matrices with diagonal entries 1 and entries above the diagonal divisible by $p$.   Since $n \geq 3$ this group is solvable and nonabelian, thus non-free.  (As noted in the question, such an argument won't work for $n=2$.)
This purely algebraic approach, like all others, is indirect and requires some fairly strong information about the special linear groups over $\mathbb{Z}$.  Whether you like a more geometric or more algebraic approach is mostly a matter of taste, influenced by what you plan to do with the information. 
