Exercise 1.7.23 (Kernel of the action of $G$ on the set of three pairs of opposite faces)

Proof. Let

be the set of pairs of opposite faces of a cube (labelled as for a dice). Each face is considered as a set of $4$ vertices. The group $G$ of rigid motions of a cube acts on $A$ by

Let $r_1$ be the rotation with axis $\mathit{IJ}$, where $I,J$ are the centers of the faces $F_1,F_6$ of $\pi$ radians. We define similarly $r_2$ and $r_3$.

Then $r_1(F_1)=F_1,r_1(F_6)=F_6,r_1(F_2)=F_5,r_1(F_5)=F_2,r_1(F_3)=F_4,r_1(F_4)=F_3$, so $r_1$ fixes every pair in $A$. Therefore $r_1$ is in the kernel of the action.

This shows that the kernel of the action in not trivial, so the action is not faithful.

Similarly, $r_2$ and $r_3$ are in the kernel of the permutation action $\phi$, and also $e$, but no other rigid motion among the $24$ elements of $G$ (The axial rotations of order $3$ map each face on an adjacent face, and also the rotations of order $4$). So

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