I've been trying to prove the following statement, and believe that I have the correct idea. Below is my attempt at a proof, but I believe there is a mistake or gap in this proof, which I will highlight.
Notational remark: If $S$ is a finite generating set of $G$, we denote by $d_S$ the word metric on $G$ with respect to $S$.
Claim. Let $G$ and $H$ be finitely-generated groups, with subgroups $A \leqslant G$ and $B \leqslant H$ of the same finite index. Suppose that $A$ is bi-Lipschitz equivalent to $B$, then $G$ is bi-Lipschitz equivalent to $H$.
Proof. Let $K_A = \{k_1, \ldots, k_m\}$, $K_B = \{\ell_1, \ldots, \ell_m\}$ be (left) transversals of $A$ and $B$ respectively, both containing the identity. We then have that every element element of $G$ can be uniquely written in the form $k_ia$ where $i \in [m]$, $a \in A$, and similarly every element of $H$ can be uniquely written in the form $\ell_j b$ where $j \in [m]$, $b \in B$.
Let $\phi : A \to B$ be an $L$-bi-Lipschitz map, and let $P$, $Q$ be finite generating sets of $A$ and $B$ respectively. Then $S := K_A \cup P$ and $T:=K_B \cup Q$ are finite generating sets of $G$ and $H$. Define $\psi : G \to H$ via $$ \psi \ : \ k_i a \ \longmapsto \ \ell_i \phi(a). $$ We claim that $\psi$ is bi-Lipschitz. Indeed $\psi$ is clearly bijective. We then compute \begin{align} d_T(\ell_i\phi(a_1), \ell_j\phi(a_2)) &\leq d_T(\phi(a_1), \phi(a_2)) + 2 \\ &\leq d_Q(\phi(a_1), \phi(a_2)) + 2 \\ &\leq L \cdot d_P(a_1, a_2) + 2 \\ &= L \cdot d_S(a_1, a_2) + 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\ast)\\ &\leq (L+2) \cdot d_S(a_1, a_2), \end{align} where the last inequality follows from only considering the non-trivial case where $a_1 \neq a_2$. An identical calculation proves the lower bound, and we are done. //
The issue in this proof is in the equality $$ d_P(a_1, a_2) = d_S(a_1, a_2), $$ which appears subtly at $(\ast)$. This equality says that if we take a generating set $P$ for $A$ and extend it to a generating set $S$ for $G$ by taking a transversal, then these word metrics agree on $A$. I am not convinced such an equality is obvious, let alone true.
I suspect this claim holds but have been unable to prove so - if it does then it would fill the above gap. Conversely if this claim doesn't hold, then I think the above proof needs reworking.
Any help on the above would be appreciated. Directions to where a similar result can be found in the literature would also be welcomed.
Thanks!