Group Action and Orbit Stabilizer on a Polyhedron
In an algebra problem, you are given the group G of rotational symmetries of a cube (orientation-preserving symmetries). You are asked to analyze the action of G on the set of faces of the cube. A student claims that since there are 6 faces and the group has order 24, the stabilizer of a face must have order 4 by the orbit-stabilizer theorem. You recall that the orbit of a face under G is all 6 faces, so the orbit size is 6. You also know that the stabilizer of a face consists of rotations that map that face to itself. You need to verify the student's claim and then extend the analysis to the action on the set of pairs (face, edge) where the edge is an edge of that face. You must determine the number of orbits and the size of stabilizers, and check consistency with Burnside's lemma.
1. Verify the student's claim about the stabilizer of a face. What is the actual order of the stabilizer of a face, and what is its structure?
Read the worked answer
The group G of rotational symmetries of a cube has order 24. The action on the 6 faces is transitive, so the orbit of any face is all 6 faces. By the orbit-stabilizer theorem, |G| = |Orbit| · |Stabilizer|, so 24 = 6 · |Stab|, giving |Stab| = 4. The student's numerical claim is correct. The stabilizer of a face consists of rotations that map the face to itself. These are rotations about the axis perpendicular to the face through its center: by 0°, 90°, 180°, 270°. This is a cyclic group of order 4, isomorphic to C4. So the stabilizer has order 4 and is cyclic. The student's claim is correct in order but should specify the structure.
2. Consider the action of G on the set S of ordered pairs (f, e) where f is a face and e is an edge of f. Determine the number of orbits of this action and the size of the stabilizer of a typical pair. Use Burnside's lemma or orbit-stabilizer to justify.
Read the worked answer
The set S has 6 faces × 4 edges per face = 24 elements. The group G acts on S by g·(f,e) = (g(f), g(e)). This action is transitive: given any two pairs (f,e) and (f',e'), there is a rotation mapping f to f' and e to e' because the cube's symmetry group acts transitively on flags (face, edge, vertex). Since |G| = 24 and |S| = 24, transitivity implies the stabilizer of any pair has order 1 by orbit-stabilizer: |G| = |Orbit|·|Stab| = 24·1. Indeed, the only rotation fixing a face and one of its edges pointwise is the identity. Thus there is 1 orbit, and the stabilizer of a pair is trivial. Alternatively, Burnside's lemma: the number of orbits is (1/|G|) Σ_{g∈G} |Fix(g)|. The identity fixes all 24 pairs. A 90° rotation about a face center fixes no pair because it moves edges. A 180° rotation about a face center fixes the face but swaps opposite edges, so fixes no pair. A 120° rotation about a body diagonal fixes no face, so fixes no pair. A 180° rotation about an edge-axis? Actually rotations about axes through midpoints of opposite edges: such a rotation swaps two faces and may fix an edge? But it does not fix a face, so no pair. Thus only identity contributes 24, giving 1 orbit. Consistent.