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Is every solvable subgroup of $GL(n,\mathbb{Z})$ polycyclic?

The first solvable group that is not polycyclic is $\mathbb{Z}[1/2]\rtimes \mathbb{Z}$ (where the automorphism is given by multiplication with 2) and I do not see a way of embedding it into $GL_n(\mathbb{Z})$ for some $n$.

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It is a theorem of Mal'cev that all solvable subgroups of $GL(n,\mathbb Z)$ are polycyclic, and a theorem of Auslander that every polycyclic group is isomorphically embeddable in $GL(n,\mathbb Z)$, for some $n$. Auslander's theorem was later reproved by Swan purely algebraically by adapting the proof of Ado's theorem.

A. I. Mal’cev, "On certain classes of infinite solvable groups", Mat. Sb. 28 (1951) 567–588; Amer. Math. Soc. Transl. (2) 2 (1956) 1–21

R.G. Swan, "Representations of polycyclic groups", Proc. Amer. Math. Soc, 1967

L. Auslander, "On a problem of Philip Hall", The Annals of Mathematics, 1967

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    $\begingroup$ You can find both theorems in Kargapolov-Merzljakov, as well as Segal's Polycyclic Groups. $\endgroup$
    – Steve D
    Commented Oct 25, 2011 at 16:28
  • $\begingroup$ Does the same thing holds for arithmetic groups? $\endgroup$
    – Xiaolei Wu
    Commented Feb 12, 2020 at 9:11

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