$\DeclareMathOperator\GL{GL}$Let $G \subset M_{n\times n~}(\mathbb Z)$ be a **finitely generated** subgroup of $\GL(n,\mathbb Q)$ (i.e. $g\in G$ is an invertible matrix with entries in $\mathbb Z$). Then $G$ acts on $\mathbb R^n = \mathbb Z^n \otimes_{\mathbb Z} \mathbb R$ through $\GL(n,\mathbb Q)$.

Suppose that there is a **rational affine** subspace $V \subset \mathbb R^n$ (by this, I mean that there is a sub-lattice $L \subset \mathbb Z^n$ and $a \in \mathbb Z^n$ such that $V = a + (L \otimes_{\mathbb Z} \mathbb R)$), and $V$ is invariant under the action of $G$ (i.e. for any $v\in V, g\in G$, we have $g\cdot v \in V$). Moreover, there exists $v \in L$ (in fact, we can take $v=a$) such that
$$G \cdot v = L.$$

**Question**: is there a bounded subset $P \subset V$ such that $$\bigcup_{g \in G}\ g\cdot P = V \quad ? $$

Any suggestion on relevant questions/references is very welcome! Particularly, I don't know which field studies such problems ….

**Edit:**

**Example.** Consider $(0,1)+L:=(\mathbb Z,1) \subset \mathbb Z^2$, and $$G=\{\begin{pmatrix} 1&k\\0&1\end{pmatrix}\mid k\in\mathbb Z\}.$$ For $v=(0,1)$, we have $G \cdot v =(0,1)+L$. In this case, we can take $P$ to be the interval from $(0,1)$ to $(1,1)$.

primitiveinteger vectors? $\endgroup$8more comments