Some background (see e.g. the books by Ghys & de la Harpe or Bridson & Haefliger for more information):

Let $\Gamma$ be a group with a finite symmetric generating set $S$. Recall that $\Gamma$ is called a *(word) hyperbolic group* if the Cayley graph of $(\Gamma, S)$ is hyperbolic (in the sense of Gromov). This notion is independent of the choice of the generating set $S$.

Assuming $\Gamma$ to be hyperbolic, there is a finite directed graph canonically associated to $(\Gamma,S)$ called the *geodesic automaton* (GA), whose arrows are labelled by elements of $S$. Paths in the GA graph starting from a special basepoint correspond to segments of geodesics in the Cayley graph.

Now consider the *recurrent* vertices of the GA graph, i.e., the set of vertices that belong to loops. Following this paper by Haissinsky, Mathieu, and Mueller, let us say that the GA graph is *strongly connected* if given any two recurrent vertices, there is one path from one to the other.

For example, if $\Gamma = \mathbb{Z}$ and $S = \{1,-1\}$ then the GA graph is: which is obviously not strongly connected.

Question.If $\Gamma$ is a nonelementary hyperbolic group, is the geodesic automaton graph strongly connected for some (all?) choice of $S$?