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.
$\begingroup$
$\endgroup$
4
-
3$\begingroup$ None of the finite-index subgroups are free groups or surface groups. This follows from the congruence subgroup property; see, e.g., people.uleth.ca/~dave.morris/talks/CSP1-Intro.pdf $\endgroup$– Andy PutmanCommented Jan 29, 2014 at 2:40
-
$\begingroup$ Better yet, use Margulis' normal subgroup theorem, that the only normal subgroups are finite-index (or finite), whereas a free group has many infinite-index normal subgroups. ;) $\endgroup$– Ian AgolCommented Jan 29, 2014 at 5:02
-
$\begingroup$ @YvesCornulier: Margulis' theorem applies to any finite-index subgroup. Anyway, it was a joke, since this theorem is overkill - Misha's answer is much more direct. $\endgroup$– Ian AgolCommented Jan 29, 2014 at 20:25
-
$\begingroup$ It has a free abelian subgroup of rank two (generated by $e_{12}(1)$ and $e_{13}(1)$) which is an obvious obstruction to being virtually free. $\endgroup$– YCorCommented Nov 28, 2019 at 14:45
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
5
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.
-
1$\begingroup$ I think you could simplify the argument further. If $SL_n(Z)$ had a finite index, hyperbolic subgroup, it would also be hyperbolic, since hyperbolicity is a quasiisometry invariant. The subgroup of those matrices which all but one basis vector is isomorphic to $Z^{n-1}$. Thus it cannot be hyperbolic for $n\ge 3$ since it contains $\mathbb{Z}^2$ as a subgroup. $\endgroup$ Commented Jan 29, 2014 at 11:03
-
$\begingroup$ @HenrikRüping: True, but I also wanted to exclude all surface groups (not just hyperbolic ones). I am also not sure that OP knows what hyperbolic groups (and their properties) are. $\endgroup$– MishaCommented Jan 29, 2014 at 11:05
-
$\begingroup$ @Misha: a uniform way to discard surface and free groups is that they are subgroups of $PSL_2$, while the integral Heisenberg group is not (virtually) linear in dimension 2. $\endgroup$– YCorCommented Jan 29, 2014 at 20:04
-
$\begingroup$ You may exclude the torus group and Klein bottle because all finite-index subgroups of $SL_3(\mathbb{Z})$ are non-abelian. $\endgroup$– Ian AgolCommented Jan 30, 2014 at 4:35
-
$\begingroup$ @YvesCornulier: Yes, of course, see my remarks about "nonelementary arguments": I tried to give a proof using only the tools that a graduate student knows after taking an algebraic topology class. $\endgroup$– MishaCommented Jan 30, 2014 at 9:37
$\begingroup$
$\endgroup$
5
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).
-
6$\begingroup$ This is a bit of an overkill. Just use the fact that the upper triangular subgroup if SL(n,Z) is solvable and noncyclic if $n>2$ to conclude that this group cannot be virtually free. $\endgroup$– MishaCommented Jan 29, 2014 at 3:19
-
$\begingroup$ Since surface groups are also hyperbolic, this also shows that surface subgroups are never of finite index, as asked in the body. However, as Misha says, there are much more elementary ways to see either of these facts. $\endgroup$– HJRWCommented Jan 29, 2014 at 5:31
-
1$\begingroup$ @Misha: as I said, this is based on some googling! I know essentially nothing about this area of group theory. Can you elaborate on your argument in an answer? $\endgroup$ Commented Jan 29, 2014 at 5:42
-
$\begingroup$ @Misha: you mean "not virtually cyclic". $\endgroup$– YCorCommented Jan 29, 2014 at 9:04
-
$\begingroup$
$\endgroup$
3
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.
-
2$\begingroup$ Wouldn't you say that Misha's approach is 'purely algebraic' and yet doesn't use any 'fairly strong information about special linear groups over $\mathbb{Z}$'? $\endgroup$– HJRWCommented Jan 29, 2014 at 19:34
-
$\begingroup$ @Jim: it's just nonsense to use CSP: if you take any subgroup of finite index $\Gamma\subset SL_n(\mathbf{Z})$, and intersect with the group $U$ of upper triangular matrices with diagonal entries 1, it's clear that $\Gamma\cap U$ is solvable and not abelian (e.g. because it's Zariski dense in $U$, or even more naively arguing that for some $n>0$ it will contain the non-commuting matrices $e_{12}(n)$ and $e_{23}(n)$). $\endgroup$– YCorCommented Jan 29, 2014 at 20:16
-
$\begingroup$ My added comment here was meant only to point to the broader question: What are the subgroups of finite index in the given group (and could any of them be free)? Naturally it's possible to focus just on the narrower parenthetic question, but the general question was answered half a century ago and clarifies the picture. $\endgroup$ Commented Jan 29, 2014 at 21:56