Let $A$ be a finitely generated abelian subgroup of $\operatorname{Aff}(\mathbb{Z^n}) = \mathbb{Z}^n \rtimes \operatorname{GL}_n(\mathbb{Z})$. Then $A$ acts naturally on $\mathbb{Z}^n$: $$A \times \mathbb{Z}^n \to \mathbb{Z}^n \colon ((x,M),y) \mapsto x+My.$$ Given a generating set $ \{ a_1, \ldots, a_k \} $ of $A$, is there a way to determine whether this action has finitely many orbits, and if so, determine representatives of these orbits?
Some thoughts:
If $n = 1$, this is relatively straightforward. Each $a_i$ is of the form $(x_i,\pm 1)$ with $x_i \in \mathbb{Z}$. Using that $A$ is abelian, we (eventually) end up in one of three cases:
- $A = \langle (x,-1) \rangle$ for some $x \in \mathbb{Z}$. Then $A \cong \mathbb{Z}_2$ and there are infinitely many orbits.
- $A = \langle (x,1) \rangle$ for some $x \in \mathbb{Z}$ with $x \neq 0$. Then $A \cong \mathbb{Z}$ and there are finitely many orbits, with representatives $0, 1, \ldots, x-1$.
- $A = \langle (0,1) \rangle$. Then $A$ is trivial and there are infinitely many orbits.
However, the situation for $n \geq 2$ does not seem to be so straightforward. Intuitively, I expect there to be finitely many orbits if and only if $A \cong \mathbb{Z}^n$. If $A \subseteq \mathbb{Z}^n$, then this is certainly the case and we can easily find representatives of the orbits. However, we can have finitely many orbits when $a_i \notin \mathbb{Z}^n$: the group $A$ given by
$$ A := \left\langle \left( \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \right), \left( \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right)\right\rangle$$ is isomorphic to $\mathbb{Z}^2$ and it acts transitively.