Let $G$ be a non-amenable group (or perhaps more generally, a group with exponential growth). For any $\epsilon>0$, define the shell of radius r, $S_\epsilon(r)$, as the set of points that lie at a distance $[r-\epsilon,r+\epsilon]$ from the identity: $$S_\epsilon(r) = \{ g \in G \, : \, r-\epsilon \leq |g| \leq r+\epsilon\},$$ where $|g| = d(g,e)$ is the geodesic length of the element $g$.
Now regard $S_\epsilon(r)$ as a subgraph. My question is the following: how does the largest distance between two points in $S_{\epsilon}(r)$ scale with $r$? That is, what is the value of $${\rm diam}(S_\epsilon(r)) \equiv \max_{u,v\in S_\epsilon(r)} d_{S_\epsilon(r)}(u,v),$$ where $d_{S_\epsilon(r)}(u,v)$ is the length of the shortest path between $u,v$ which stays entirely in $S_\epsilon(r)$? (in the case where $S_\epsilon(r)$ is disconnected, define ${\rm diam}(S_\epsilon(r)) = \infty$).
Since $G$ is assumed non-amenable, $|S_\epsilon(r)| \sim e^{\alpha r}$ grows exponentially with $r$ (here $\alpha$ is some $O(1)$ constant). This implies that ${\rm diam}(S_\epsilon(r))$ is at least linear in $r$ (edit: as R W points out, this fact does not rely on non-amenability). Can we say anything stronger? In particular, does ${\rm diam}(S_\epsilon(r))$ always grow exponentially with $r$? This is true if $G$ is a hyperbolic group, but I am curious if it holds more generally.
