Exercise 1.2.11 (Group of rigid motions of an octahedron)

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Outline of a proof.

Proof. Let $𝒪 = {A, B, C, D, E, F}$ a regular octahedron (for instance the points of coordinates $(1, 0, 0), (0, 1, 0), (- 1, 0, 0), (0, - 1, 0), (0, 0, 1), (0, 0, - 1)$ ), and $G$ be the group of rigid motions of $𝒪$ .

$G$ acts on the set $𝒪$ by $g \cdot M = g (M)$ $(g \in G, M \in 𝒪)$ . Using rotations of angles $\pm π ∕ 3$ and axe $0 I$ , where $I$ is a centre of a face, we can send $A$ on any vertex. So the action is transitive, there is a unique orbit $O_{A}$ , and $| O_{A} | = | 𝒪 | = 6$ .

The stabilizer $G_{A}$ of $A$ has four elements ${e, ρ, ρ^{2}, ρ^{3}}$ where $ρ$ is the rotation of angle $π ∕ 2$ and axe $OA$ . By the fundamental theorem of group actions, $(G : G_{A}) = | O_{A} |$ , so

| G | = | G_{A} | \cdot | O_{A} | = 4 \cdot 6 = 24 .

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Note: If we associate to each face of the cube its center, we obtain an octahedron, and conversely, the same operation with an octahedron gives a cube (these two Platonic solids are dual). This explains why the cube and the octahedron have the same group of rigid motions.

richardganaye

2025-09-09 08:24

Exercise 1.2.11 (Group of rigid motions of an octahedron)

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