I am trying to get through the proof of the following theorem:
Let $G = (V,E)$ be a locally finite, connected, infinite graph with linear growth and infinite motion. Then $G$ is 2-distinguishable.
In the proof of this theorem we claim that the stabilizer of every vertex $v\in V$ is finite and since the vertex set is countable, the stabilizer can only have countably many conjugacy classes and hence the automorphism group itself must be countable.
I don't quite understand that. Why can the stabilizer have only countably many conjugacy classes and how does that imply that the automorphism group must be countable?
Maybe I am missing some definitions here. The stabilizer of the element $s$ in the group $\Gamma$ is the set: $$\Gamma_s = \{ \gamma \in \Gamma : \gamma s = s \}$$ And the conjguacy class of an element $a$ in the group $G$ is a set, such that: $$Cl(a) = \{b \in G : \exists g \in G: b = gag^{-1}\}$$
We would like to get assumptions to use a following statement to prove our thesis.
Let $\Gamma$ be a group of permutations of a set $S$ and assume that $\Gamma$ has infinite motion. If $\Gamma$ is countable then there is $\Gamma$-distinguishing 2-colouring of $S$.
Could anyone explain that to me, please?