Lowering metrics to finite index subgroups

Question

Let $G$ be an infinite, countable, finitely generated group. Let $H$ be a finite index subgroup of $G$. Let $S$ be a finite, symmetric set of generators of $G$, and let $d(\cdot,\cdot)$ be the word length metric on $G$ induced by $S$. Then $d$ is also a metric on $H$. The question is the following: is $d$ in the same bilipschitz equivalence class as all the word length metrics on $H$? That is, if $d'(\cdot,\cdot)$ is a word length metric on $H$, does there exist a constant $K$ such that $\frac{1}{k}d'(g,h) \leq d(g,h) \leq K d'(g,h)$ for all $g,h \in H$?

Answer 1

Even more is true. A metric space is called D-separated if the distance between any distinct points is at least D.

Claim. Let $X, X'$ be $1$-separated metric spaces. Let $f: X\to X'$ be a bijective quasi-isometry. Then $f$ is bilipschitz.

Proof. Inequality $d(f(x), f(y))\le L d(x,y)+A$ implies that $d(f(x), f(y))\le M d(x,y) + M\le (M+1)d(x, y)$. Here $M$ is the maximum of $L, A$. The same argument applies to the inverse of $f$. Thus, $f$ is bilipschitz. Qed

In your case the identity map is a quasi-isometry (by Milnor-Schwarz Lemma), so the above claim applies. Of course, one can give a more direct argument as I explained in my comments: let $S$ and $S'$ be finite generating sets for $G$ and $H$ respectively. Without loss of generality we can assume that $S'\subset S$. We can also assume that representatives of all the cosets in $G/H$ are in $S$. Then the embedding of $H$ in $G$ is 1-lipschitz. Conversely, let $p: G\to H$ be nearest point projection with respect to the metric $d$. Then, since $H$ is 1-dense in $G$, the map $p$ is 3-lipschitz. This gives you the inequality $$ d\le d'\le 3d $$ on the group $H$.