TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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$Aut (Z^{n}) ≅ G L_{n} (Z)$ for every finite rank $n$

Statement

For every finite rank $n \geq 0$ ,

$Aut (Z^{n}) ≅ G L_{n} (Z),$

where $G L_{n} (Z)$ denotes the group of invertible $n$ -by- $n$ arrays of integers under the product

$(A B)_{i j} = \sum_{k = 1}^{n} A_{i k} B_{k j} .$

Facts & Assumptions

Given: The free abelian group $Z^{n}$ with its standard basis $e_{1}, \dots, e_{n}$ .

[L1]

A homomorphism from a free abelian group is determined uniquely by the images of a free basis (Free abelian group on a set).

[L2]

An isomorphism is a bijective group homomorphism, and an automorphism of $G$ is an isomorphism from $G$ to itself (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Proof

technique · direct

1.1L1

For an endomorphism $f$ , write $f (e_{j}) = \sum_{i} a_{i j} e_{i}$ . By [L1], the integer array $A_{f} = (a_{i j})$ determines $f$ , and every integer array arises from a unique endomorphism.

2.1step 1.1algebra

A finite-sum calculation on each basis vector gives $A_{f \circ g} = A_{f} A_{g}$ with the product displayed in the Statement. Thus $f \mapsto A_{f}$ is an isomorphism between the endomorphism monoid and the monoid of integer arrays.

3.1step 2.1L2algebra

An endomorphism $f$ is an automorphism exactly when some endomorphism $g$ satisfies $f g = g f = id$ . If $f$ is an automorphism it is bijective by [L2], so its set-theoretic inverse $g$ exists, and $g$ is a homomorphism because $f (g (x) + g (y)) = f g (x) + f g (y) = x + y = f (g (x + y))$ and $f$ is injective; conversely such a $g$ is a two-sided set inverse, so $f$ is bijective and hence an automorphism by [L2]. By step 2.1 this is equivalent to an integer array $A_{g}$ satisfying $A_{f} A_{g} = A_{g} A_{f} = I$ . These are exactly the elements of $G L_{n} (Z)$ .

4.1step 2.1step 3.1∎

Restricting the correspondence in step 2.1 to the invertible elements proves the isomorphism. When $n = 0$ , both sides are the one-element group.

Depends on

Used by

A shear and a quarter-turn exhibit noncommuting automorphisms of ℤ² Example

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Martin R. Bridson and Karen Vogtmann, Automorphism groups of free groups, surface groups and free abelian groups (standard reference, not scraped)

Aut⁡(Zn)≅GLn(Z)Aut(Zn)≅GLn​(Z) for every finite rank nn