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 $g\in G$.

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

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