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Let $G \subset GL(n, \Bbb Z)$ be a f.g. linear group. The Tits alternative says that $G$ is either virtually solvable (i.e. has a solvable subgroup of finite index), or contains a free group $F_2$.

I am interested in linear groups $G$ which do not contain $F_2 \times F_2$ and are not virtually solvable. Do these groups have a characterization of some sort? In particular, is this true for $SL(3,\Bbb Z)$?

For the motivation, let me mention that if $G$ contains $F_2 \times F_2$, thеn the group membership problem (decision problem whether a group generated by a finite set of elements contains a given element) is undecidable.
K. A. Mihailova, The occurrence problem for direct products of groups, Mat. Sb. 70 (1966), 241-251.

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    $\begingroup$ In $SL_3$ the centralizer of any non-virtually-solvable subgroup $G$ is abelian, hence there's no $F_2\times F_2$. To prove it we can suppose $G$ Zariski-closed and the field algebraically closed of char. 0, and in turn that $G$ is a copy of $(P)SL_2$, and up to conjugation there are only two pssibilities. $\endgroup$
    – YCor
    Commented Feb 20, 2015 at 18:26
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    $\begingroup$ I don't think ${\rm SL}(3,K)$ can contain $F_2 \times F_2$ for any field $K$. Assuming $K$ is algebraically closed, the dimensions of the irreducible constituents of a subgroup $F_2$ could not all be $1$, so they would be a single $3$ or $1+2$,and then Schur's Lemma implies this is not centralized by any $F_2$. $\endgroup$
    – Derek Holt
    Commented Feb 20, 2015 at 18:27
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    $\begingroup$ For an arbitrary $G$ in a larger group, it depends on $G$ (and not only on its Zariski closure); for instance for a Zariski-dense subgroup in $SL_4$ it depends whether $G$ has large enough intersections with subgroups with large centralizers (i.e. pointwise stabilizers of lines for the action in $P^3$). $\endgroup$
    – YCor
    Commented Feb 20, 2015 at 18:30

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