# 0708-1300/fact

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