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



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




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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:
$$\Gamma\_1=\left\langle\begin{pmatrix}p & 0 & 0\\0&p^{-2}&0\\0&0&p\end{pmatrix},\begin{pmatrix}1 & 1 & 0\\0&1&0\\0&0&1\end{pmatrix}\right\rangle,\;\Gamma\_2=\left\langle\begin{pmatrix}p & 0 & 0\\0&p&0\\0&0&p^{-2}\end{pmatrix},\begin{pmatrix}1 & 0 & 1\\0&1&0\\0&0&1\end{pmatrix}\right\rangle.$$




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Edit (Oct 19, 2022) [it is mentioned](https://mathoverflow.net/a/432758/14094) that $\mathrm{SL}\_3(\mathbf{Z})$ doesn't contain a copy of $F\_2\times\mathbf{Z}$. More precisely element $g$ of $\mathrm{GL}\_3(\mathbf{Z})$ with $g^2\neq 1$ has a solvable centralizer in $\mathrm{GL}\_3(\mathbf{C})$.


This indeed follows readily from the discussion in the proposition. If $g$ has type (a), (c), (e) or (f) then it has a solvable centralizer. If $g$ has type (d) then it is scalar but the only scalar matrix in $\mathrm{GL}\_3(\mathbf{Z})$ are $\pm$ identity. Remains when $g$ has type (b), i.e., is $\mathbf{C}$-diagonalizable with a double eigenvalue, say $\lambda$. So the eigenvalues are $\lambda,\lambda,\lambda^{-2}$. Taking the gcd between the characteristic polynomial $P$ and $P'$ shows that $\lambda\in\mathbf{Q}$. Then looking at the coefficients of $P$ shows that $2\lambda+\lambda^{-1}$ and $\lambda+2\lambda^{-1}$ both belong to $\mathbf{Z}$. Hence both $3\lambda$ and $3\lambda^{-1}$ belong to $\mathbf{Z}$, which implies $|\lambda|\in\{1/3,1,3\}$, but the condition $2\lambda+\lambda^{-1}\in\mathbf{Z}$ then forces $|\lambda|=1$, so $g^2=1$.


