= Solution
Any two <word metric>[word metrics] from finite generating sets on the same group are <bilipschitz equivalence>[bilipschitz equivalent]. Indeed, if $S,S'$ are finite, let $L=\max_{s\in S}|s|_{S'}$; then $|g|_{S'}\leq L|g|_S$, 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 $T$.
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