Let $N=\mathbb Z^2$ and let $t$ generate the other factor of $\Gamma_B=N\rtimes_B\mathbb Z$. For $n\in N$,
$$
[t,n]=tnt^{-1}n^{-1}=(B-I)n.
$$
After <abelianization>, the image of $N$ is therefore
$$
N/(B-I)N.
$$
If $1$ is not an <eigenvalue> of $B$, then $\det(B-I)\ne0$. 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
$$
\Gamma_B^{\mathrm{ab}}
\cong\mathbb Z\oplus N/(B-I)N.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 133 / 2 / b / Solution

Let $B\in\operatorname{GL}_d(\mathbb Z)$ and $\Gamma_B=\mathbb Z^d\rtimes_B\mathbb Z$. If $t$ generates the second factor, then the elements $tnt^{-1}n^{-1}$ fill $(B-I)\mathbb Z^d$, and hence
$$
\Gamma_B^{\mathrm{ab}}\cong
\mathbb Z\oplus\mathbb Z^d/(B-I)\mathbb Z^d.
$$
If $1$ is not an <eigenvalue> of $B$, the second summand is finite.

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 Incoming links: Abelianization of a semidirect product by the integers