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The Milnor–Švarc lemma applies to an isometric properly discontinuous group action that is also a cocompact group action on a nonempty proper metric space that is a geodesic metric space. It concludes that the group is finitely generated and, for a basepoint $x$ and any finite generating set of a group $S$, the orbit mapHere $d_S$ is the word metric, and a quasi-isometry means a map satisfying two-sided coarse distance bounds with a uniformly coarsely dense image. A free action is not required: finite stabilizers are compatible with proper discontinuity. The properness and cocompactness hypotheses ensure that only finitely many group elements move a chosen basepoint by the bounded amount used to generate the group. Replacing a finite generating set by another does not change the conclusion, by equivalence of finite word metrics.

Retain the metric geodesic subdivision and orbit-density parts of the usual Milnor–Švarc lemma proof. Drop the step using proper discontinuity and compactness to prove that the displacement-bounded generating set is finite.

More explicitly, let $L=d_X(x,gx)$ and $m=\max \{1,\lceil L\rceil \}$. Subdivide a metric geodesic from $x$ to $gx$ into $m$ segments of length at most one. Choose orbit points $g_{i}x$ within distance $R$ of the subdivision points, choosing the endpoints exactly: $g_0=1$, $g_m=g$. Thenso every nonidentity increment $g_{i-1}^{-1}{g}_i$ lies in $S$. Their telescoping product is $g$, and henceThis also covers $L=0$, when the one increment may be a nontrivial stabilizer element. Conversely, for any word $g=s_1\cdots s_k$ in $S$, the triangle inequality and invariance under the isometric action giveTaking the shortest word gives the other inequality. Thus the proof works unchanged except that the resulting generating set may be infinite. Neither local compactness nor a finite stabilizer is used in this weaker argument.

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 $D_{\infty }$.