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If a group acts properly discontinuously, cocompactly, and isometrically on a proper geodesic metric space, then the group is finitely generated and every orbit map from a word metric is a quasi-isometry.

The infinite dihedral group is $D_{\infty }\cong C_2\ast C_2$, with factors $A=⟨a⟩$ and $B=⟨b⟩$. Its Bass-Serre tree has vertex setand one edge indexed by each $g\in D_{\infty }$, joining $gA$ to $gB$. Since both factors have order two, every vertex has degree two. The connected tree is therefore a bi-infinite line.

The action is cocompact, and its vertex stabilizers are the finite conjugates of $A$ and $B$, so it is proper. By the Milnor–Švarc lemma, an orbit map from $D_{\infty }$ with a word metric to this line is a quasi-isometry. A simplicial bi-infinite line is quasi-isometric to $\mathbb{R}$, hence so is $D_{\infty }$.

Solved by gpt-5.6-sol high.

Let $f:X\to T$ be a $(\lambda ,\epsilon )$-quasi-isometry to a tree. The image under $f$ of every geodesic segment in $X$ is a $(\lambda ,\epsilon )$-quasigeodesic in $T$. By the Morse lemma for quasi-geodesics, it lies within a constant $R=R(\lambda ,\epsilon )$ of the tree geodesic with the same endpoints.

Consider a geodesic triangle in $X$. A point $p$ on one side maps within $R$ of the corresponding side of the comparison triangle in $T$. Every geodesic triangle in a tree is $0$-thin, so that comparison side is contained in the other two sides. Those two tree sides are in turn within $R$ of the images of the other two sides of the original triangle. Hence some point $q$ on one of those sides satisfiesThe lower quasi-isometry inequality givesThus $X$ is Gromov-hyperbolic metric space with $\delta =\lambda (2R+\epsilon )$.

Solved by gpt-5.6-sol high.

Fix a hyperbolicity constant ${\delta }_0>0$ for ${\mathbb{H}}^2$. For a prescribed $\delta >0$, scale its distance byAll distances in every geodesic triangle, including its thinness constant, scale by $\delta /{\delta }_0$. Henceis $\delta$-hyperbolic. Multiplication of a metric by a fixed positive constant is a bilipschitz equivalence and hence a quasi-isometry. Therefore $X_{\delta }$ is quasi-isometric to ${\mathbb{H}}^2$ and cannot be a quasi-tree. This supplies an example for every $\delta >0$.

Solved by gpt-5.6-sol high.

A quasi-isometric embedding satisfies the two-sided coarse distance inequality in the definition of quasi-isometry, without requiring its image to be coarsely dense.