Using group actions and counting, we can classify all finite subgroups of the rotation group .
Recall from Lecture 9: every element ,, is a rotation about some axis, hence fixes exactly two unit vectors (the “poles” of ). For a finite subgroup , let be the set of all poles. Since , the set is -invariant, so acts on .
For each pole , the stabilizer is a cyclic subgroup of rotations about the axis , of some order . The proof proceeds by counting the set :
Summing over : each has exactly 2 poles, so .
Summing over : for each pole , there are non-identity rotations fixing , so . Grouping by orbits (elements in the same orbit have the same ): Dividing by : Since each , each term is . The left side is , so there are at most 3 orbits (since but ).
Analyzing the cases with and the equation :
2 orbits:. Since and each , one finds . The two poles form one orbit of size 2, with stabilizer of order . So is a cyclic group of rotations about a single axis: .
3 orbits:. With , necessarily (else ). Then , forcing .
:. This gives the dihedral group .
:. Rotational symmetries of the tetrahedron, .
:. Rotational symmetries of the cube (equivalently, octahedron), .
:. Rotational symmetries of the icosahedron (equivalently, dodecahedron), .
Theorem 11.4.1 (Classification of finite subgroups of ). The finite subgroups of are exactly: the cyclic groups , the dihedral groups , the tetrahedral group , the octahedral group , and the icosahedral group .
Note. The cube and octahedron are dual polyhedra (vertices of one correspond to centers of faces of the other), so they share the same symmetry group. Similarly the dodecahedron and icosahedron are dual, sharing . For regular polyhedra, the poles at midpoints of edges have , poles at vertices have faces meeting at a vertex, and poles at centers of faces have edges of the face.