# Why does a fundamental domain have a freely acted upon element?

I want to understand the proof of Theorem 1.58 in "Groups, Graphs and Trees" by John Meier. I have some problems with the second line (see screenshot).

With every connected simple graph $$\Gamma$$ we can associate a 1-complex $$X_\Gamma$$. An automorphism of $$\Gamma$$ is an autohomeomorphism of $$X_\Gamma$$ which maps vertices to vertices. Let $$\sigma\colon G \to \mathsf{Aut}(\Gamma)$$ be an action of a group $$G$$ on a graph $$\Gamma$$. A closed subset $$\mathcal F \subset X_\Gamma$$ is called fundamental domain if the set $$\{\sigma(g)(\mathcal F)\mid g \in G\}$$ covers $$X_\Gamma$$ and the subset $$\mathcal F$$ is minimal closed subset with this property.

Lemma. Assume that if $$\sigma(g)(\mathcal F) = \mathcal F$$ for $$g \in G$$ then $$g = 1$$. Then there is a point $$x \in \mathcal F$$ that is moved freely under the action $$\sigma$$ (that is, $$\sigma(g)(x) = \sigma(h)(x) \Rightarrow g = h$$ for any $$g, h \in G$$).

I can't figure out why would this statement be true. In this book there is no explanation (perhaps, author assumes that this is obvious and maybe it really is).

I would very appreciate any help! • Please do not rely on pictures of text. Aug 5 at 19:29
• What does Theorem 1.33 say? Aug 6 at 8:43
• @MoisheKohan Theorem 1.33 states that there is a bijection between the orbit of an element and the left cosets of its stabilizer. It's not related to the question though. yesterday