# Proof for group actions by conjugation?

In a proof on the theorem: A finite subgroup of SO$$_3$$ is isomorphic either to a cyclic group, a dihedral group, or the rotational symmetry group of one of the regular solids, they use some logic to determine that there exists an action of $$G$$ on $$x$$. Some proof set up:

"Let $$G$$ be a finite subgroup of $$SO_3$$. The two points where the axis of a rotation $$g\in G$$ meets the unit sphere are called the poles of $$g$$ (which don't move when $$g$$ is applied to another element of $$G$$). Let $$X$$ denote the set of all poles of all elements of $$G - \{e\}$$. Suppose $$x\in X$$ and $$g\in G$$. Let $$x$$ be a pole of the element $$h\in G$$. Then $$(ghg^{-1})(g(x))=g(h(x))=g(x)$$, which shows that $$g(x)$$ is a pole of $$ghg^{-1}$$ and hence $$g(x)\in X$$. Therefore, we have an action of $$G$$ on $$X$$."

I do not understand the conclusion in bold. (I looked up group actions on Wikipedia, but I don't understand how this proof/lemma matches the two axioms "identy & compatibility"). Therefore I also do not understand why they go to show that $$g(x)\in X$$ when $$x\in X$$ and $$g\in G$$.

• A comment containing a partial answer would also be appreciated.. – The Coding Wombat Oct 31 '18 at 11:37