Quasi-isometry of finitely generated group

Let $$\Gamma$$ be a finitely generated group, with two generating sets $$S_1,S_2$$. Deduce, from the Milnor - Svarc lemma, that $$Cay(\Gamma, S_1)$$ and $$Cay(\Gamma,S_2)$$ are quasi isometric, where $$Cay(\Gamma,S_i)$$ is the Cayley graph of $$\Gamma$$ w.r.t the generating set $$S_i$$.

Now my idea is as follows - we know that $$(\Gamma, d_{S_1})$$ and $$Cay(\Gamma,S_1$$) are quasi isometric. The same goes for $$(\Gamma, d_{S_2})$$ and $$Cay(\Gamma,S_2)$$. So I'm trying to show that $$(\Gamma, d_{S_1})$$ and $$(\Gamma, d_{S_2})$$ are quasi-isometric. But how do I get this from Milnor-Svarc? I understand that $$\Gamma$$ acts on $$(\Gamma, d_{S_2})$$ geometrically and all, so $$(\Gamma, d_{S_2})$$ and $$\Gamma$$ are quasi-isometric, but I'm not sure w.r.t to what metric (on $$\Gamma$$ - the one that acts). As I understand it, Milnor - Svarc states that if Γ acts geometrically on a geodesic proper metric space (X,d) then Γ is finitely generated by a subset S⊆Γ and (Γ,dS) and (X,d) are quasi-isometric

Any help would be appreciated.

• If all you need to show is that $(\Gamma,d_{S_1})$ and $(\Gamma,d_{S_2})$ are quasi-isometric, see this answer for a direct proof that the identity map is a quasi-isometry, in fact it is bi-Lipschitz. Commented Dec 1, 2021 at 3:27
• @Lee Mosher Yes this proof Im aware of. Thing is I want to prove this theorem as a consequence of Milnor Svarc so I think a direct proof is not good enough Commented Dec 1, 2021 at 7:36
• It sounds like your post could be improved by including what you understand to be the statement of the Milnor-Svarc lemma. Without that, it's quite hard for me to guess exactly where your misunderstanding lies. At the moment, my best guess is that the statement you have in mind might be incomplete. Commented Dec 1, 2021 at 14:53
• @LeeMosher As I understand it, if $\Gamma$ acts geometrically on a geodesic proper metric space $(X,d)$ then $\Gamma$ is finitely generated by a subset $S\subseteq\Gamma$ and $(\Gamma,d_S)$ and $(X,d)$ are quasi-isometric Commented Dec 1, 2021 at 14:55

The statement of the Milnor Svarc lemma that you included in your post is somewhat weaker than the statement one usually sees, e.g. the statement on the wikipedia page. Here's the stronger statement, which obviously implies the statement in your post:

If $$\Gamma$$ acts geometrically on a geodesic proper metric space $$(X,d)$$ then $$\Gamma$$ is finitely generated, and for any finite generating set $$S$$ of $$\Gamma$$, $$(\Gamma, d_S)$$ and $$(X,d)$$ are quasi-isometric.

(Emphasis added by me on that universal quantifier).

In particular this statement directly implies the statement that you have asked about, namely that $$(\Gamma,d_{S_1})$$ and $$(\Gamma,d_{S_2})$$ are quasi-isometric for any two finite generating sets $$S_1,S_2 \subset \Gamma$$.

One can also use your weaker statement to prove this stronger one, by directly proving that $$(\Gamma,d_{S_1})$$ and $$(\Gamma,d_{S_2})$$ are quasi-isometric for any two finite generating sets $$S_1,S_2 \subset \Gamma$$ (see the link in my first comment).

However, my best guess is that if you look at whatever proof you know for the weaker version of the lemma included in your post, you will discover that the proof of that statement does not depend on the choice of finite generating set $$S$$. So what that proof is really doing is to prove the stronger statement. Furthermore, if you compare that proof with the proof of the statement in the link of my first comment, you will see that those two proofs are really the same proof.

So, all-in-all, the stronger statement of the Milnor-Svarc Lemma is really the preferred one.