 Solution

Let $N={\mathbb{Z}}^2$ and let $t$ generate the other factor of ${\Gamma }_B=N{⋊}_B\mathbb{Z}$. For $n\in N$,After abelianization, the image of $N$ is thereforeIf $1$ is not an eigenvalue of $B$, then $\det (B-I)\ne 0$. Hence $(B-I)N$ is a full-rank sublattice of $N$ with finite index of a subgroup $|\det (B-I)|$. The displayed quotient, and therefore the image of the ${\mathbb{Z}}^2$ factor in ${\Gamma }_B^{\mathrm{ab}}$, is finite. In fact the abelianization of a semidirect product by the integers gives