= Solution
The relations $aka^{-1}k=1$ and $bkb^{-1}k=1$ say that conjugation by either $a$ or $b$ sends $k$ to $k^{-1}$. Thus $K=\langle k\rangle$ is a normal subgroup of order at most $3$. Quotienting by it gives
$$
\Gamma/K
\cong\langle a,b\mid a^4=b^4=(ab)^4=1\rangle,
$$
the 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>,
$$
\boxed{\Gamma\simeq_{\mathrm{qi}}H^2.}
$$
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