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A map $f:X\to Y$ is a quasi-isometry if there are $\lambda \ge 1$ and $\epsilon ,R\ge 0$ such thatfor all $x,x^{\prime }\in X$, and every point of $Y$ lies within distance $R$ of $f(X)$. The Milnor–Švarc lemma says that a group acting properly discontinuously, cocompactly and isometrically on a proper geodesic metric space is finitely generated, and each orbit map from a word metric is a quasi-isometry.

Now let $S,T$ be finite generating sets of $G$. PutReplacing each letter in an $S$-word by a $T$-word and conversely givesThus the identity map is a bilipschitz equivalence, hence a quasi-isometry. All finite generating sets of a finitely generated group give quasi-isometric Cayley graphs.

The relations $ak{a}^{-1}k=1$ and $bk{b}^{-1}k=1$ say that conjugation by either $a$ or $b$ sends $k$ to $k^{-1}$. Thus $K=⟨k⟩$ is a normal subgroup of order at most $3$. Quotienting by it givesthe orientation-preserving hyperbolic triangle group $\Delta (4,4,4)$. This group acts properly discontinuously and cocompactly by isometries on the hyperbolic plane. The Milnor–Švarc lemma therefore makes $\Delta (4,4,4)$ quasi-isometric to $H^2$.

The quotient map $\Gamma \to \Gamma /K$ has finite kernel, so part (b), equivalently the finite-kernel quotient quasi-isometry, makes $\Gamma$ quasi-isometric to $\Delta (4,4,4)$. By transitivity of 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 $D_{\infty }$.

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 $\delta =\lambda (2R+\epsilon )$.

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 $\delta >0$.

For a quasi-isometry $f:X\to Y$ with multiplicative constant $\lambda$ and additive/coarse-surjectivity constant $ϵ$, choose $f^{\prime }(y)$ with $d_Y(f(f^{\prime }(y)),y)\le ϵ$. The triangle inequality and the two distortion inequalities giveAlso $d_X(f^{\prime }(f(x)),x)\le 2\lambda ϵ$. Hence $f^{\prime }$ is a quasi-isometry and both composites stay uniformly close to their respective identities. No continuity is required.

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.

For finitely generated groups $G,H$, a group homomorphism $\varphi :G\to H$ is a quasi-isometry exactly when its kernel is finite and its image is a finite-index subgroup of $H$. The finite kernel controls collapse of distances, and finite index is exactly the coarse-surjectivity condition.