Quasi-isometry of Schreier coset graphs Let $G$ be a finitely generated group and $H$ a subgroup. For any choice of a generating set $X$ we can form the Schreier coset graph. Is this graph independent of the generating set up to quasi-isometry?
I know the corresponding question for the Cayley graph has positive answer, but the proof I know relies on the Svarc-Milnor Lemma; this lemma requires in particular a proper action of $G$ on the spaces involved, but here the stabilizer of a vertex $Hg$ is the whole $H$, so the action is not proper when $H$ is infinite. Anyway the lemma implies that $G$ is quasi-isometric to any of its Cayley graphs, but surely $G$ is not quasi-isometric to any of its Schreier graphs (for $H\neq 1$). So this is really not the way to try to prove this thing (assuming it's true).
 A: Yes.  This is essentially the proof that studiosus suggests:
Let $X_1$ and $X_2$ be two finite generating sets of $G$, and $S_1$ and $S_2$ the associated Schrier graphs.  Map the vertices of $S_1$ to the vertices of $S_2$ via the identity map (the vertices in both cases are just cosets of $H$).  Note that any graph is quasi-isometric to its set of vertices (leaving the same metric on the vertices), so it suffices to show this vertex mapping is a quasi-isometry.
Express each $x_i \in X_1$ as a word $w_i$ in elements of $X_2$.  Let $K_1$ be the maximum length among the $w_i$.  Similarly, let $K_2$ be the maximum length needed to express an element of $X_2$ as a word in elements of $X_1$.
Then for any two adjacent vertices $Hg$ and $Hg'$ in $S_1$, we have $Hg = Hg'x$ for some $x \in X_1$.  But we can express $x$ as a product $y_1...y_n$, where each $y_i \in X_2$ and $n \leq K_1$.  Thus $Hg = Hg'x_1...x_n$, which means the distance between $Hg$ and $Hg'$ in $S_2$ is no more than $K_1$, since in $S_2$ the cosets $Hg'$, $Hg'x_1$, $Hg'x_1x_2$, etc. are a sequence of adjacent vertices.
A similar argument shows that vertices which are adjacent in $S_2$ are of distance no more than $K_2$ in $S_1$.
Thus, for any two cosets $Hg, Hg'$, we have
$$
\frac{1}{K_2}d_{S_1}(Hg, Hg') \leq d_{S_2}(Hg, Hg') \leq K_1d_{S_1}(Hg, Hg') 
$$
Which shows that mapping cosets under the $S_1$ metric to the same cosets under the $S_2$ metric is a quasi-isometry, thus $S_1$ and $S_2$ are isometric.
