 Solution

Use the standard coordinate generators of ${\mathbb{Z}}^d$. Its word metric is the ${\ell }^1$ metric, so its radius-$r$ ball consists of integer points satisfying ${\sum }_i|x_i|\le r$. For $d\ge 1$ it contains the cube $|x_i|\le \lfloor r/d\rfloor$ and is contained in the cube $|x_i|\le r$. Thusand its growth is bounded above and below by constant multiples of $r^d$ for large $r$.

If ${\mathbb{Z}}^n$ and ${\mathbb{Z}}^m$ are quasi-isometric, part (i) gives equivalent growth functions of a finitely generated group. For positive $n>m$, the required comparison $r^n\lesssim r^m$ would bound a positive constant times $r^n$ by $O(r^m+r)$, impossible as $r\to \infty$. Interchanging $n,m$ rules out $m>n$. Hence $n=m$. Conversely, for equal dimensions the identity is an isometry with standard generators; changing finite generators preserves quasi-isometry by Q2(ii). ThereforeIf dimension zero is allowed, ${\mathbb{Z}}^0$ is bounded and cannot be quasi-isometric to an unbounded positive-dimensional lattice. This boundedness argument is needed because the additive linear term in ${\sim }_e$ makes constant and linear growth functions of a finitely generated group equivalent; growth equivalence alone would not distinguish dimensions zero and one.