Abstract Commensurator Group of $\mathbb{Z}^n$ $Comm(\mathbb{Z}^n)\cong GL(n,\mathbb{Q})$? Hello! In a paper I read that $\mathrm{Comm}(\mathbb{Z}^n)\cong \mathrm{GL}(n,\mathbb{Q})$. Why is that true? How can I find an isomorphism of this groups?
I know that $\mathrm{Aut}(\mathbb{Z}^n)\cong\mathrm{GL}(n,\mathrm{Z})$.
Definition of the abstract commensurator of a group $G$:
Let $G$ be a group. Consider the set 
$\Omega(G)$ of all isomorphisms between subgroups
of finite index of $G$. Two such isomorphisms $\phi_1:H_1\to H_1'$ and $\phi_2:H_2\to H_2'$ are called equivalent, written $\phi_1\sim\phi_2$, if there exists a subgroup $H$ of finite index in $G$ such that both $\phi_1$ and $\phi_2$ are defined on $H$ and $\phi_1\mid_{H}=\phi_2\mid_{H}$.
For any two isomorphisms $\alpha:G_1\to G_1'$ and $\beta:G_2\to G_2'$ in
$\Omega(G)$, we define their product $\alpha\beta:\alpha^{-1}(G_1'\cap G_2)\to \beta(G_1'\cap G_2)$ in $\Omega(G)$. The factor-set $\Omega(G)/\sim$ inherts the multiplication $[\alpha][\beta]=[\alpha\beta]$ and is a group, called the abstract commensurator of
G and denoted by $\mathrm{Comm}(G)$.
Thanks for help!
 A: First, it is easy to see that $\mathrm{Comm}(\mathbb{Z}^n)$ must be isomorphic to a subgroup of $\mathrm{GL}(n,\mathbb{Q})$.  In particular, every finite-index subgroup of $\mathbb{Z}^n$ is an $n$-dimensional lattice, so any isomorphism between two such subgroups extends uniquely to an isomorphism $\mathbb{Q}^n\to\mathbb{Q}^n$.  Note that two isomorphisms of $\mathbb{Q}^n$ that agree on an $n$-dimensional lattice must be equal, so it really does work to think of elements of $\mathrm{Comm}(\mathbb{Z}^n)$ as matrices.
All that remains is to show that every element of $\mathrm{GL}(n,\mathbb{Q})$ corresponds to some element of $\mathrm{Comm}(\mathbb{Z}^n)$.  This is fairly easy: if $A\in\mathrm{GL}(n,\mathbb{Q})$, then there exists a positive integer $k$ so that the $n\times n$ matrix for $kA$ has integer entries.  In this case, $A$ maps the finite-index subgroup $k\\,\mathbb{Z}^n$ (of index $k^n$) isomorphically to another finite-index subgroup of $\mathbb{Z}^n$.  The image subgroup has finite index since it spans $\mathbb{Q}^n$, and is therefore an $n$-dimensional lattice.
Edit: As Mark points out, this is essentially the same answer given in the comments above.
A: This is shown (with no claim to originality) by Jonathan Hillman in "Commensurators and Deficiency" (Theorem 7)
