Free abelian groups of rank at least two are not hyperbolic

If the basis is infinite, every finite set of group elements uses only finitely many basis coordinates and cannot generate $A$. Thus [L0] excludes hyperbolicity. Otherwise identify $A$ with ${\mathbb{Z}}^n$, $n\ge 2$, and fix any finite generating set. Replace it by its nonzero symmetric closure $S$, which leaves the geometric Cayley graph unchanged by [L1]. It spans ${\mathbb{R}}^n$.

Choose $s\in S$ of maximal Euclidean norm. The linear functional $f(x)=⟨s,x⟩\mathrm{/}\mathrm{\parallel }s{\mathrm{\parallel }}^2$ satisfies $f(s)=1$ and $\mathrm{\mid }f(a)\mathrm{\mid }\le 1$ for every $a\in S$, by Cauchy–Schwarz and maximality. Since $S$ spans a space of dimension at least two, choose $v\in S$ independent of $s$ and put $w=v-⟨v,s⟩s\mathrm{/}\mathrm{\parallel }s{\mathrm{\parallel }}^2$. Then $w\mathrm{\ne }0$, $⟨w,s⟩=0$ and $⟨w,v⟩>0$. Choose $t\in S$ maximizing $⟨w,t⟩$. Symmetry gives a positive maximum and $\mathrm{\mid }⟨w,a⟩\mathrm{\mid }\le ⟨w,t⟩$ on $S$. Hence $g(x)=⟨w,x⟩\mathrm{/}⟨w,t⟩$ has $g(t)=1$, $g(s)=0$ and $\mathrm{\mid }g(a)\mathrm{\mid }\le 1$ on $S$. In particular $s,t$ are independent.

Extend these linear functions from graph vertices affinely over each edge. Their slopes have absolute value at most one, so they are 1-Lipschitz for the graph path metric. Thus a path of $m$ successive $s$-edges, or $m$ successive $t$-edges, has endpoints at distance exactly $m$: the path supplies the upper bound and $f$, respectively $g$, supplies the lower bound. Translates and reversals are likewise geodesics. Therefore the four such paths through $0,ms,ms+mt,mt$ form a geodesic quadrilateral.

Choose linear functionals $\alpha ,\beta$ on ${\mathbb{R}}^n$ with $\alpha (s)=1$, $\alpha (t)=0$, $\beta (s)=0$ and $\beta (t)=1$; solving the nonsingular two-vector Gram system constructs them. Let $C={\max }_{a\in S}\max \{\mathrm{\mid }\alpha (a)\mathrm{\mid },\mathrm{\mid }\beta (a)\mathrm{\mid }\}\ge 1$. Their affine extensions to graph edges are $C$-Lipschitz. At the midpoint of the side $0$ to $ms$, their values are $(m\mathrm{/}2,0)$. On each of the other three sides, either $\alpha =0$, $\alpha =m$, or $\beta =m$. Consequently every point on those sides is at distance at least $m\mathrm{/}(2C)$ from that midpoint. Taking arbitrarily large $m$ contradicts [L2] for every proposed hyperbolicity constant. Since the finite generating set was arbitrary, [L0] excludes hyperbolicity of $A$.