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Any two word metrics from finite generating sets on the same group are bilipschitz equivalent. Indeed, if $S,S^{\prime }$ are finite, let $L={\max }_{s\in S}|s{|}_{S^{\prime }}$; then $|g{|}_{S^{\prime }}\le L|g{|}_S$, and the reverse inequality follows symmetrically. Apply this once to the two finite generating sets of $G$ and once to those of $H$. Composing these bilipschitz identity maps with the inclusion changes only the multiplicative and additive constants in the quasi-isometric embedding inequalities. Thus being a quasi-isometrically embedded subgroup is independent of $S$ and $T$.

Solved by gpt-5.6-sol high.

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 $\delta >0$.

Solved by gpt-5.6-sol high.