Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 133 / 3

Identify $G$ with the finite-index subgroup $f(G)\le H$. If $S$ is a finite generating set for $H$ and $T$ is a finite set of right-coset representatives, Schreier's lemma gives a finite generating set for $f(G)$, and hence for $G$.

Equip both groups with word metrics from finite generating sets. The inclusion is Lipschitz because each generator of $G$ has bounded length in $H$. Conversely, rewriting a word in $S$ by tracking its cosets through the finite set $T$ expresses an element of $G$ as a word of length bounded linearly in its $H$-length. Finally, every element of $H$ lies within the maximum word length of an element of $T$ from $f(G)$. Thus the inclusion is a finite-index subgroup quasi-isometry, and composing it with the isomorphism $f:G\to f(G)$ proves that $G$ and $H$ are quasi-isometric.

Let $S$ be a finite generating set of $H$, choose one lift $\tilde{s}\in G$ for every $s\in S$, and put $K=\ker f$. The finite setgenerates $G$: lift a word representing $f(g)$ and observe that the discrepancy from $g$ lies in $K$.

Use this generating set for $G$. The quotient map does not increase word length, while lifting a shortest word in $H$ leaves only a final element of $K$, whose word length is at most one. HenceThe map is surjective, so it is a finite-kernel quotient quasi-isometry. Therefore $G$ is finitely generated and quasi-isometric to $H$.

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,