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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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    $\begingroup$ Does abstract commensurator have a universal property? If it does, you may want use it to prove the claim. $\endgroup$
    – user23860
    Commented May 23, 2012 at 13:27
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    $\begingroup$ A subgroup of ${\mathbb Z}^n$ of finite index contains a basis of ${\mathbb Q}^n$. So an isomorphism of two such subgroups maps a basis to a basis. Being additive, it is ${\mathbb Z}$-linear, so after tensoring with $\mathbb Q$, it is $\mathbb Q$-linear so it defines an element of $P{\rm GL}(n,{\mathbb Q})$. $\endgroup$
    – user1688
    Commented May 23, 2012 at 13:30
  • $\begingroup$ I mean ${\rm GL}(n,{\mathbb Q})$. $\endgroup$
    – user1688
    Commented May 23, 2012 at 13:31
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    $\begingroup$ Conversely, if the denominators of a matrix $A$ are $\le D$, then the matrix takes the subgroup $D!\mathbb{Z}^n$ to another subgroup of finite index of $\mathbb{Z}^n$, hence belongs to the commensurator. $\endgroup$
    – user6976
    Commented May 23, 2012 at 13:44
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    $\begingroup$ @Peter: Any basis of a subgroup of finite index of $\mathbb{Z}^n$ is a basis of $\mathbb{Q^n}$ (because it contains $n$ elements by the strocture theorem of finitely generated Abelian groups) and is linearly independent over $\mathbb{Z}$, hence over $\mathbb{Q}$. In fact the structure theorem (or the theorem about the structure of modules over PIDs) gives that for every subgroup $H$ of finite index there exists a basis $e_1,...,e_n$ of $\mathbb{Z}^n$ and numbers $k_1 >0,...,k_n>0$ such that $H$ is generated by $k_1e_1,...,k_ne_n$. This is the basis of $H$ you are looking for. $\endgroup$
    – user6976
    Commented May 23, 2012 at 14:57

2 Answers 2

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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.

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    $\begingroup$ @Jim: All that has been explained in the comments above. $\endgroup$
    – user6976
    Commented May 23, 2012 at 15:48
  • $\begingroup$ @Mark: Oh, I hadn't noticed! $\endgroup$
    – Jim Belk
    Commented May 24, 2012 at 18:35
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This is shown (with no claim to originality) by Jonathan Hillman in "Commensurators and Deficiency" (Theorem 7)

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  • $\begingroup$ Thanks. In the proof, what is meant with the $\alpha m=\alpha\mid_{m\mathbb{Z}^n}\circ m$ What does the $\circ$ mean? Is it composition? If so, which map is meant by $m$? $\endgroup$
    – Peter
    Commented May 23, 2012 at 14:30
  • $\begingroup$ $m$ is multiplication by $m.$ $\endgroup$
    – Igor Rivin
    Commented May 23, 2012 at 15:18
  • $\begingroup$ Sorry. But I don't see, why this map is an injectic endomorphism. I know that $m\mathbb{Z}^n\leq H$ because $[G:H]=m$ and so the we can restrict $\alpha$ to this subgroup. But how is this map defined on the other elements $\mathbb{Z}^n\backslash m\mathbb{Z}^n$. And if we multiply with $m$, we imagine $\alpha$ as matrix with integers in $\mathbb{Q}$, right? $\endgroup$
    – Peter
    Commented May 23, 2012 at 17:37
  • $\begingroup$ I am puzzled by your question. The map $\alpha$ is only defined on $H.$ The map $\alpha \circ m$ is given by $\alpha \circ m (x) = \alpha(m x).$ Since $m x \in H$ you are good. $\endgroup$
    – Igor Rivin
    Commented May 23, 2012 at 17:46
  • $\begingroup$ Oh. hehe. Sorry. I was a little bit confused. I thought that I have to multiply $\alpha$ with $m$. So I didn't know how to do that. But know I'm fine. Thanks. $\endgroup$
    – Peter
    Commented May 23, 2012 at 18:32

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