Consider the semidirect product $$ G=\mathbb{Z}^n \rtimes \mathbb{Z}^m $$ Is $ G $ always virtually abelian?
Is it the case that the abelianization of $ G $ is $ \mathbb{Z}^{n+m} $ if and only if $ G $ was already $ \mathbb{Z}^{n+m} $ to begin with?
If $ G $ is a nontrivial semi direct product can we conclude that the free rank of the abelianization $ G $ must be strictly less than $ n+m $?
I am especially interested in whether every semidirect product $$ \mathbb{Z}^2 \rtimes \mathbb{Z} $$ is virtually abelian. If not are they all virtually nilpotent? All virtually solvable?
For example I believe that $ \mathbb{Z}^2 \rtimes \mathbb{Z} $ with respect to the semidirect product $$ n \mapsto \begin{bmatrix} 0 & -1 \\ 1 & -d \end{bmatrix}^n $$ is virtually solvable but not virtually nilpotent (or virtually abelian).
No. The semidirect product of ${\mathbb Z}$ acting on ${\mathbb Z}^2$ with action $$\left( \begin{array}{cc}0&1\\1&1\end{array}\right)$$ is solvable but not nilpotent, and it is not virtually abelian.
Its abelianization if just ${\mathbb Z}$.