Abelian subgroups of $\mathrm{GL}(n,\mathbb{Z})$

Question

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?

Answer 1

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.