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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!

proofpic

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  • $\begingroup$ Please do not rely on pictures of text. $\endgroup$
    – Shaun
    Aug 5 at 19:29
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    $\begingroup$ What does Theorem 1.33 say? $\endgroup$ Aug 6 at 8:43
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    $\begingroup$ @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. $\endgroup$ yesterday

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