Are all metabelian groups linear Are all metabelian groups linear? (i.e. isomorphic to a subgroup of
invertible matrices over a field)
 A: Every finitely generated metabelian group $\Gamma$ has a faithful representation over a finite product of fields (I think it's due to Remeslennikov). If $\Gamma$ is (virtually) torsion-free, one field of characteristic zero is enough. But for instance, denoting $C_n$ the cyclic group of order $n$, the wreath product $C_6\wr C_\infty$ is not linear over a single field (you need a field of char. 2 and a field of char. 3).
Note that the result is false for 3-solvable finitely generated groups, which can be far from linear or even residually finite: here are 2 examples (one residually finite and not the other):
1) the wreath product $\Lambda=\mathbf{Z}\wr\Gamma$, where $\Gamma$ is f.g. metabelian and not virtually abelian. Then $\Lambda$ is not linear over any commutative ring, although it is 3-solvable and residually finite.
2) (Hall) fix a prime $p$ and consider the group $A_3$ of matrices $M(k;x_{12},x_{13},x_{23})=\begin{pmatrix} 1 & x_{12} & x_{13}\\ 0 & p^k &  x_{23}\\ 0 & 0 & 1\end{pmatrix}$ with $x_{ij}\in\mathbf{Z}[1/p]$ and $k\in\mathbf{Z}$. Then it is 3-solvable and finitely generated (namely by $M(1;0,0,0)$, $M(0;1,0,0)$ and $M(0;0,0,1)$); the element $M(0;0,1,0)$ is central and generates an infinite cyclic subgroup $Z$; the quotient $A_3/Z$ fails to be residually finite because it contains a (central) copy of the abelian group $\mathbf{Z}[1/p]/\mathbf{Z}$. Hence $A_3/Z$ is not linear over any commutative ring.
