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

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Jan 9, 2019 at 9:39	comment	added	YCor		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.
Jan 9, 2019 at 2:28	comment	added	Ian Agol		@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.
Jan 9, 2019 at 0:36	comment	added	YCor		What's a reference for the fact that any $F_2$ contains a semisimple $F_2$?
Jan 8, 2019 at 18:13	vote	accept	burtonpeterj
Jan 8, 2019 at 18:13
Jan 8, 2019 at 5:03	history	answered	Ian Agol	CC BY-SA 4.0