# 0708-1300/fact

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