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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.

Let $h\in H$ and letbe a geodesic in the Cayley graph ${\mathrm{Cay}}_S(G)$. By $K$-quasiconvexity choose $h_i\in H$ with $d_S(x_i,h_i)\le K$, taking $h_0=1$ and $h_N=h$. ThenThe elements $h_i^{-1}{h}_{i+1}$ telescope to $h$, so the finite setgenerates $H$.

Moreover $|h{|}_U\le |h{|}_S$, while $|h{|}_S\le (2K+1)|h{|}_U$. Thus the inclusion $(H,d_U)\to (G,d_S)$ is a quasi-isometric embedding, and $H$ is quasi-isometrically embedded.

Solved by gpt-5.6-sol high.

A finitely generated subgroup $H\le G$ is quasi-isometrically embedded when its inclusion, equipped with word metrics from finite generating sets, is a quasi-isometric embedding. This property is independent of those generating sets.