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.
 A: 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.
