Free metabelian group of rank 2 I was trying to find the free metabelian group of rank 2, and I realized that the wreath product of Z and Z is metabelian and has only 2 generators in its minimal generating set. I could not find a way to extend this to another metabelian group without adding another generator. So my question is: Is the free metabelian group of rank 2 isomorphic to the wreath product of Z and Z?
 A: Let  me turn my comment into an answer.  Let $C$ be an infinite cyclic group generated by $t$ and $\mathbb ZC$ be the group ring.  Then $\mathbb Z\wr\mathbb Z=\mathbb ZC\rtimes C$ where $C$ acts on its group ring via the regular representation. The abelianization is $\mathbb Z\times \mathbb Z$ via the map $(x,t^m)\mapsto (\epsilon(x),m)$ where $\epsilon$ is the augmentation map.
The commutator subgoup of $\mathbb Z\wr \mathbb Z$ is then the augmentation ideal $I$.  As a module over the group ring of $\mathbb Z\times \mathbb Z$ we have that the first factor acts trivially and it is one-generated over the second factor (i.e., it is really a $\mathbb ZC$-module).  On the other hand, if $M_2$ is the free metabelian group of rank $2$ generated by $x,y$, then it is well known that the commutator subgroup is freely generated as a module over the group ring of $\mathbb Z\times \mathbb Z$ by $[x,y]$.  Thus $M_2\not\cong \mathbb Z\wr \mathbb Z$.
The word problem is very easy for $M_2$.  It follows from a general result of Jorge Almeida on free objects in semidirect products of varieties and was rediscovered by Vershik many years later.  Two words $w_1,w_2$ represent the same element of $M_2$ iff they represent the same element of $\mathbb Z\times \mathbb Z$ and the number of times the path labeled by $w_1$  in the Cayley graph of $\mathbb Z\times \mathbb Z$ starting at the origin traverses each edge $e$ is the same as the number of times as the path labeled by $w_2$ traverses $e$, where backward traversals are counted negatively. 
Edit. Alternatively, let $E$ be the edge set of the Cayley graph of $\mathbb Z\times \mathbb Z$. Then $\mathbb Z^2$ acts on $E$ so we can form the semidirect product $\mathbb ZE\rtimes \mathbb Z^2$. The free metabelian group on $x,y$ is the subgroup generated by $(1\rightarrow x,x)$ and $(1\rightarrow y,y)$ where I view $x,y$ as generators of $\mathbb Z^2$. 
