I'm reading "Groups, Graphs and Trees" by Meier.
In pages 31-34 (which could be found here) He defined fundumental domain for group action on graphs as
Defintion Let $G\curvearrowright \Gamma$ be group acting on a connected graph. a set $F\subseteq \Gamma$ is called a fundamental domain if:
- $F$ is closed
- The set $\{g\cdot F | g\in G \}$ covers the graph $\Gamma$.
- No proper subset of $F$ satisfies properties (1) and (2)
Then, a few pages after, there is the Theorem:
Theorem 1.55 Let $G$ act on a connected graph $\Gamma$ with connected fundamental domain $F$. Then the set of elements
$$S = \{g\in G|g\ne e \ and \ g\cdot F\cap F\ne \emptyset\}$$
is a generating set for $G$.
I don't understand the proof of this theorem (the proof can be found in the link above) - the part that I don't understand is when the autor say:
"let $\{g_0F,...,g_nF\}$ be a finite sequence of (distinct) images of the fundamental domain such that $g_iF \cap g_{i+1}F \ne \emptyset $."
Why can we take such a sequence?
How can we be sure that there are two distinct $g,g'\in G$ s.t. $gF\cap g'F \ne \emptyset$?
In other words, how do we even know that the set $S$ is not empty?