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!

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