S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 7: Finite Groups

23 Finite Subgroups of SO(3)

· Denis Auroux

Original Professor Auroux’s handwritten notes (PDF), behind a password for now:

Recall:

  • An action of on a set is a map , , such that and ; or equivalently a homomorphism .
  • Given , the orbit and the stabilizer subgroup are related by ( iff ).

Today: use these ideas to classify the finite subgroups of (and hence classify regular polyhedra, as well).

Rotations of

Recall: an inner product space the orthogonal group Elements of have , and (transformations that preserve orthonormality and orientation).

We’ve seen: a decomposition , , , (this uses: there exists an invariant subspace, and if is invariant then so is ). If , ; if , = a rotation.

  • In dimension 3, either The condition narrows it down to every element of is a rotation; if , it has an axis (the -eigenspace = a line) and rotates by some angle in the plane the axis.
  • Given a subset , we can look at the symmetry group . It could be infinite (e.g. if is a circle in a plane, all rotations with axis the plane will be symmetries), or it could be finite.

Example 23.1. = a regular -gon in a plane (centered at the origin):

  • rotations (axis plane, angle );
  • flips = rotations by with axis plane.

this is isomorphic to . (Special cases , : see the right of the figure.)

A regular pentagon lying in a plane, shown twice: once with a rotation axis perpendicular to the plane through its center (n rotations by 2 pi k over n), once with an axis lying in the plane through a vertex and the center (n flips by pi); to the right, the degenerate polygons for n = 1 (a loop with one marked vertex) and n = 2 (a lens with two marked vertices).

Example 23.2. To only keep in the above example, consider a cone on a regular -gon in a plane:

A pyramid (cone) over a regular pentagon, hidden edges dashed, with an arrow below indicating rotation by 2 pi k over n about its vertical axis.

Example 23.3. Symmetries of regular polyhedra:

The five regular polyhedra side by side with hidden edges dashed: tetrahedron, cube, octahedron, dodecahedron, icosahedron.
  • tetrahedron;
  • cube and octahedron: these have the same symmetries, by duality: vertices centers of faces (or: );
  • dodecahedron (12 pentagonal faces) and icosahedron (20 triangular faces): duals, have the same symmetries.

These give respectively:

  • (seen last time): the action on the vertices or faces of the tetrahedron;
  • (HW 2!): the action on the 4 diagonals of the cube;
  • : the action on what?

Theorem 23.4. This is the complete list of finite subgroups of :

Poles

The key observation is that every , , is a rotation about some axis, hence fixes exactly two unit vectors , called the poles of .

A sphere with a rotation axis T passing through its center; the axis meets the sphere in two antipodal points, the poles v and minus v, and a great circle perpendicular to the axis is drawn on the sphere.

For finite, let = the set of all poles of elements of : Now, if is a pole of , and given any , is a pole of (since ). So acts on ! This is the key to understanding the group .

Example 23.5. In the case of symmetry groups of regular polyhedra: (normalized to unit vectors). These form 3 different orbits for the action of on (namely, acts separately on vertices, on faces, and on edges; for a regular polyhedron each of these actions is transitive).

The next observation is that for , consists of the rotations with axis ! These form an abelian, in fact cyclic, subgroup of . So: (since is a pole of some element of , must be nontrivial). (I.e. the angles of rotations through form a finite subgroup of , so they must be all multiples of .)

With this understood, the proof of the theorem is a counting argument.

Proof. Let be a nontrivial finite subgroup, the set of poles as above. Let For each element of , there are exactly 2 poles. So . For each element , there are rotations in fixing . So: Now, the elements of an orbit of have conjugate stabilizers (), hence the same ’s: . Thus: Now remember orbit/stabilizer: , so The rhs gets quickly if there are too many orbits!! Each term is since , hence .

We now analyze each case based on the number of orbits.

  • 1 orbit: impossible: lhs (since ) vs. rhs .

  • 2 orbits: , i.e. . Since each divides , we must have . Hence , i.e. there are 2 poles , each fixed under all of , and , the cyclic subgroup of rotations with axis .

  • 3 orbits: . Assume . Observe: (a) If : then . Thus : two poles form an orbit. These poles are necessarily , and

    • half of = the rotations by about (the stabilizer of );
    • the other half of = rotations by (since these poles have ) swapping (since preserves the orbit ).

    = a dihedral group.

    (b) If : . These 3 cases give the tetrahedron, cube, and icosahedron.

∎

Note. For regular polyhedra:

  • poles at midpoints of edges have ;
  • poles at vertices: = the number of faces meeting there;
  • poles at centers of faces: = the number of edges of the face.
tetrahedron
cube, octahedron
dodecahedron, icosahedron

The icosahedral group is

Question 23.6. What is the 5-element set that the symmetries of the dodecahedron act on?

Answer: the 20 vertices of a dodecahedron can be partitioned into 5 sets of 4 forming regular tetrahedra (in 2 different ways, which are mirror images but not related by a rotation). A rotation of the dodecahedron then permutes the 5 tetrahedra.

A dodecahedron drawn in gray with its 20 vertices colored in five colors, four of each color; the four blue vertices are connected, showing that they span a regular tetrahedron, and the other color classes are the other four tetrahedra.

(The notes show a picture of the five tetrahedra inside the dodecahedron, image by Greg Egan.)

  • Rotations about centers of faces 5-cycles (24 of them).
  • Rotations about vertices 3-cycles etc. (20 of them).
  • Half-rotations about midpoints of edges etc. (15 of them).