# Direct product of free groups in $\mathrm{SL}_3(\mathbb{Z})$

Let $$\mathbb{F}_2$$ be the free group on two generators. Does $$\mathbb{F}_2 \times \mathbb{F}_2$$ embed as a subgroup of $$\mathrm{SL}_3(\mathbb{Z})$$?

• No; more generally, let $G$ be a non-virtually-solvable subgroup of $\mathrm{GL}_3(\mathbf{C})$. Then the centralizer of $G$ is abelian. Indeed its Zariski closure contains a Zariski-closed copy of $\mathrm{(P)SL}_2(\mathbf{C})$. There are two such subgroups up to conjugation: the irreducible $\mathrm{PSL}_2(\mathbf{C})(=\mathrm{SO}_3)$ and the upper-left block. The first has a trivial centralizer, and the second has centralizer equal to the diagonal matrices $(a,a,b)$. [I'm pretty sure this argument already exists somewhere on this site.] – YCor Jan 8 at 9:26

Here is a brief elementary argument only relying on Burnside's theorem that proper unital subalgebras of matrix algebras are non-irreducible (over an algebraically closed field).

Proposition: Let $$K$$ be an algebraically closed field. Let $$A,B$$ be noncommuting matrices in $$\mathrm{M}_3(K)$$. Then the centralizer of $$\{A,B\}$$ in $$\mathrm{M}_3(K)$$ is triangulable, i.e., stabilizes a flag (i.e., is conjugate into the subalgebra of upper triangular matrices).

Proof: there are essentially 6 types of $$3\times 3$$ matrices: (a) 3 eigenvalues (b) 2 eigenvalues, diagonalizable (c) 2 eigenvalues, not diagonalizable (d) scalar (=1 eigenvalue, diagonalizable) (e) 1 eigenvalue, scalar+(nilpotent of rank 1) (f) 1 eigenvalue, scalar+(nilpotent of rank 2).

Matrices of type (a),(c),(f) have abelian centralizer. Matrices of type (e) have triangulable centralizer (hint: compute the centralizer of the matrix $$E_{13}$$). By assumption, $$A$$ is not central and not of type (d). If $$A$$ has type (acef) then it has triangulable centralizer. So the only case to consider is when $$A$$ has type (b): we can suppose that $$A$$ is the diagonal matrix $$(0,0,1)$$. The centralizer $$C$$ of $$A$$ is the set of matrices diagonal by blocks $$2+1$$, and the double centralizer $$C'$$ is reduced to $$K+KA$$ (so $$C'\subset C$$). Hence $$B\notin C$$. Therefore, the intersection $$I$$ of the centralizers of $$A$$ and $$B$$ is properly contained in $$C$$, and hence by the contraposite of Burnside's theorem (in dimension 2, in which it is an elementary exercise) implies that $$I$$ is triangulable.

Corollary: for every field $$K$$, every non-abelian subgroup of $$\mathrm{GL}_3(K)$$ has a 3-step solvable centralizer. In particular, no group of the form $$H_1\times H_2$$ with $$H$$ non-abelian and $$H_2$$ non-solvable, is embeddable into $$\mathrm{GL}_3(K)$$ for any field $$K$$.

Remarks:

1) a slight refinement shows that a 3-step solvable subgroup has an abelian centralizer, so "3-step" can be replaced with "2-step").

2) Here are two commuting non-abelian subgroups of $$\mathrm{SL}_3(\mathbf{Q})$$, each isomorphic to the Baumslag-Solitar $$\mathrm{BS}(1,p^3)$$ (here $$p\in\mathbf{Z}\smallsetminus\{0,1\}$$), with trivial intersection, thus generating their direct product: $$\Gamma_1=\left\langle\begin{pmatrix}p & 0 & 0\\0&p^{-2}&0\\0&0&p\end{pmatrix},\begin{pmatrix}1 & 1 & 0\\0&1&0\\0&0&1\end{pmatrix}\right\rangle,\;\Gamma_2=\left\langle\begin{pmatrix}p & 0 & 0\\0&p&0\\0&0&p^{-2}\end{pmatrix},\begin{pmatrix}1 & 0 & 1\\0&1&0\\0&0&1\end{pmatrix}\right\rangle.$$

This cannot exist. See Misha Kapovich's answer to this question: https://mathoverflow.net/a/163754/1345

In particular, part (2) of his answer implies that there can be no semisimple $$\mathbb{F}_2\times \mathbb{F}_2$$ subgroup of $$SL_3(\mathbb{Z})$$. That is, there is no such subgroup in which every element is diagonalizable.

Now one sees that if there is an arbitrary $$\mathbb{F}_2\times \mathbb{F}_2$$ subgroup of $$SL_3(\mathbb{Z})$$, then each $$\mathbb{F}_2$$ factor contains a semisimple $$\mathbb{F}_2$$ subgroup. Passing to these subgroups, we obtain a semisimple such subgroup, a contradiction.

• What's a reference for the fact that any $F_2$ contains a semisimple $F_2$? – YCor Jan 9 at 0:36
• @YCor: This follows from some version of the proof of Tits' alternative, but I don't have a specific reference. Actually, it suffices just to see that there is a single element with 3 distinct eigenvalues (this is semisimple). If not, one can see that each matrix is quasi-unipotent (with eigenvalues $\pm1$). Then one can see that the subgroup is virtually nilpotent, contradicting that it is free. Once one has such an element with 3 distinct eigenvalues in one factor, then all the elements in the other factor must have the same eigenspaces in order to commute. But this is a contradiction. – Ian Agol Jan 9 at 2:28
• OK. Well if I can put it otherwise, you prove that if $G$ is a non-virtually-unipotent subgroup of $\mathrm{GL}_3(\mathbf{Z})$ then its centralizer is abelian semisimple [this conclusion fails in $\mathrm{GL}_3(\mathbf{Q})$ as I have noticed in my answer]. Indeed, considering the Zariski closure of $G$, one sees that $G$ has an element $g$ with eigenvalues not contained in $\{\pm 1\}$, and hence (using that $g$ is integral) $g$ has 3 distinct eigenvalues, hence its centralizer is abelian. – YCor Jan 9 at 9:39