0708-1300/fact

If [math]\displaystyle{ n\neq m }[/math] then [math]\displaystyle{ \mathbb{Z}^n\not\cong\mathbb{Z}^m }[/math].

Proof

Assume that [math]\displaystyle{ f:\mathbb{Z}^n\rightarrow\mathbb{Z}^m }[/math] is an isomorphism. Let [math]\displaystyle{ A }[/math] be the matrix of [math]\displaystyle{ f }[/math] in the canonical basis and [math]\displaystyle{ M }[/math] the maximum of the absolute values of the entries of [math]\displaystyle{ A }[/math]. If we evaluate [math]\displaystyle{ f }[/math] in all the vectors of [math]\displaystyle{ \mathbb{Z}^n }[/math] who's entries have absolute values less than or equal to [math]\displaystyle{ r }[/math] (there are [math]\displaystyle{ (2r)^n }[/math] of such elements) then we get elements who's entries have absolute value less than or equal to [math]\displaystyle{ nrM }[/math] (there are [math]\displaystyle{ (2rnM)^m }[/math] of such elements in [math]\displaystyle{ \mathbb{Z}^m }[/math]). Since [math]\displaystyle{ f }[/math] is injective we must have [math]\displaystyle{ (2r)^n\leq(2rnM)^m }[/math] for every [math]\displaystyle{ r }[/math]. Replacing [math]\displaystyle{ f }[/math] by its inverse if necessary we can assume that [math]\displaystyle{ n\gt m }[/math] but if this is the case the inequality above can not be true for arbitrarily large values of [math]\displaystyle{ r }[/math].