Does $\mathrm{GL}(n,\mathbb{Z})$ contain torsion-free abelian subgroups that are not isomorphic to $\mathbb{Z}^k$ for some $k$? Said otherwise (as suggested by comment below), does $\mathrm{GL}(n,\mathbb{Z})$ contain torsion-free abelian subgroups that are not finitely generated?

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    $\begingroup$ Your question is equivalent (in view of the structure theorem of finitely generated abelian groups) to «does $GL(n,Z)$ contain non-finitely generated torsion free subgroups?» $\endgroup$ – Mariano Suárez-Álvarez Aug 27 '13 at 15:46
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    $\begingroup$ There is a more general result of Mal'cev that a soluble subgroup of ${\rm GL}(n,{\mathbb Z})$ is polycyclic. It's done in Chapter 2 of Segal's book on Polycyclic Groups. I would guess that the highest rank of a torsion-free abelian subgroup of ${\rm GL}(n,{\mathbb Z})$ is $n-1$, but I haven't found a reference for that. $\endgroup$ – Derek Holt Aug 28 '13 at 7:37

Sorry if this is ridiculous.

Let $A<GL_n(\mathbb{Z})$ be a torsion-free abelian subroup. Let $B$ be the Zariski closure of $A$ in $GL_n(\mathbb{C})$. Then $B$ is a group variety, so is of the form $B_1\times B_2$, with $B_1$ finite and $B_2$ connected. $A$ is torsion-free, so $A\subset B_2$. Since $B_2$ is connected, it remains connected in the complex topology (see A complex algebraic variety which is connected in the usual topology). If we put the complex topology on $B_2$, then forget the complex structure to view it as a real Lie group, $B_2\cong \mathbb{R}^n\times (S^1)^m$ for some integers $n,m$. $A$ is discrete in $B_2$, and one checks that every discrete subgroup of $\mathbb{R}^n\times(S^1)^m$ is finitely generated.

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  • $\begingroup$ why not $B_2 \cong \mathbb{R}^n \times (\mathbb{R}^{\times})^m$, where $\mathbb{R}^{\times}$ is the multiplicative group of $\mathbb{R}$? $\endgroup$ – user68316 Aug 27 '13 at 16:43
  • $\begingroup$ $\mathbb{R}^\times$ is not connected. $\endgroup$ – Julian Rosen Aug 27 '13 at 16:43
  • $\begingroup$ Isn't it connected in Zariski topology? $\endgroup$ – user68316 Aug 27 '13 at 16:46
  • $\begingroup$ @user68316 Oh yes, of course! I overlooked this, so my argument will need to be modified. $\endgroup$ – Julian Rosen Aug 27 '13 at 16:48
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    $\begingroup$ You cannot conclude that $A$ is contained in $B_2$ from knowing that $A$ is torsion-free. $\mathbb R\times\mathbb Z_2$ has plenty of infinite cyclic subgroups which are not contained in the connected component of the unit elemet. $\endgroup$ – Mariano Suárez-Álvarez Aug 27 '13 at 17:25

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