0708-1300/fact

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 $r$.