# Intersection of finite index subgroups of $\mathbb{Z}^2$

Suppose you have a subgroup $\{g_0,g_1,\ldots,\}=:G < (\mathrm{GL}_2(\mathbb{Q}),\cdot)$ of invertible 2 by 2 matrices over $\mathbb{Q}$, given by all matrices with determinant in a given subgroup of $\mathbb{Q}^+_0$ (i.e. positive rational numbers under multiplication). Assume the enumeration is fixed. Is there anything you can say about $Z_k:= \cap_{i=1}^k \mathbb{Z}^2 \cdot g_i$ (where we act on $\mathbb{Z}^2$ by matrix multiplication,in terms of the subgroup $G$ you started from (and not just 'it's a finite index subgroup'). If not, does anyone have any suggestions on how to put extra conditions on $G$ such that more would be known?

Thanks

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How does $GL_2(\mathbb Q)$ act on $\mathbb Z^2$? I think I know what you mean, though... – Mariano Suárez-Alvarez Jan 10 '13 at 18:54
Is your group $G$ finite? (It looks like you want $k$ to be the number of its elements; or is $Z_k$ the intersection of the first $k$ of the $\mathbb Z^2\cdot g_i$ for some enumeration of the elements of $G$?) – Mariano Suárez-Alvarez Jan 10 '13 at 18:58
$G$ is infinite, in fact, you can suppose that $G$ is the subgroup of $\mathrm{GL}_2(\mathbb{Q})$ consisting of all matrices with determinants in a given subgroup of $(\mathbb{Q}^+_0,\cdot)$. I was hoping to be able to write these $Z_k$ as $\mathbb{Z}^2 \cdot g$ for some element $g$ closely related (or even inside) $G$. So it is as you say the intersection of the first $k$ of the $\mathbb{Z}^2\cdot g_i$ for some enumeration of the elements of $G$. Lets consider the enumeration fixed. – Jan Keersmaekers Jan 10 '13 at 19:02
Please add all useful information to the question itself :-) – Mariano Suárez-Alvarez Jan 10 '13 at 19:04
What is $\mathbb{Q}_0^+$? Positive rational numbers under multiplication? – Alexander Gruber Jan 10 '13 at 20:41