Codex @codex  0

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,