Finite symmetry group/Cube from numerical condition/Permutation group/Fact/Proof

Proof

We consider the class of semiaxes ${}K_{2}$ of order ${}3$, which contains ${}8$ semiaxes equivalent to each other. For such a semiaxis ${}H$, the opposite semiaxis belongs also to a class of semiaxes with the same order. Therefore, ${}-H$ belongs to ${}K_{2}$, so that semiaxes of ${}K_{2}$ belong to four axes. We denote the set of these axes by ${}{\mathfrak {A}}$. We consider the group homomorphism

$G\longrightarrow \operatorname {Perm} ({\mathfrak {A}}),g\longmapsto {\left(\sigma _{g}:A\mapsto g(A)\right)}.$

Here, we look at the action on the axes, not at the action on the semiaxes. We claim that not three of these four axes lie in a plane. For if ${}A_{1},A_{2},A_{3}\subseteq E$, then a rotation about ${}120$ degree around ${}A_{1}$, call it ${}f$, would produce the equivalent axes ${}f(A_{2})$ and ${}f(A_{3})$, but these can not lie in the plane ${}E$, and they can not both equal ${}A_{4}$. Suppose that the element ${}g\in G$ has the property, that ${}\sigma _{g}$ is the identity. This means that all lines ${}A\in {\mathfrak {A}}$ are mapped to themselves. Due to exercise, ${}g$ is the identity. Therefore, according to the kernel criterion, the group homomorphism is injective. Hence, we have an isomorphism. In particular, we can apply this reasoning for the cube group; this gives the isomorphism between the cube group and the permutation group ${}S_{4}$.

To fact