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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.

Take $G={\mathbb{Z}}^2$ with the standard generating set $S=\{(1,0),(0,1)\}$, and letIntrinsic distance in $H$ between $0$ and $(n,n)$ is $|n|$, while its ambient word metric distance is $2|n|$, so $H$ is quasi-isometrically embedded. However, the ambient geodesic from $(0,0)$ to $(n,n)$ that first travels to $(n,0)$ and then to $(n,n)$ contains $(n,0)$. Its distance from the diagonal subgroup $H$ is $n$. No uniform $K$ can contain every such geodesic in the $K$-neighborhood of $H$, so $H$ is not a quasiconvex subgroup.

Solved by gpt-5.6-sol high.

A subgroup is quasiconvex when it is a quasiconvex subset of a Cayley graph. In a hyperbolic group, a finitely generated subgroup is quasiconvex exactly when it is quasi-isometrically embedded.