0708-1300/fact

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If n\neq m then \mathbb{Z}^n\not\cong\mathbb{Z}^m.

Proof

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