We return to group theory with the language of group actions from Chapter 3 in hand. Understanding which sets a group acts on, and how, reveals the internal structure of itself. In this chapter we count orbits with Burnside’s lemma; let act on itself by conjugation, which leads to the class equation and to the basic structure of -groups; classify the finite groups of rotations of ; study conjugacy in the symmetric and alternating groups, proving that is simple for ; and prove the Sylow theorems. Together with semidirect products, these results classify the groups of several small orders.
11.1.1 Group Actions
We recall the language of Section 3.2. An action of a group on a set (Definition 3.2.1) is a map , , with and for all and ; equivalently, it is a homomorphism , . The action is faithful if is injective; in general it comes from a faithful action of on . The orbit of is . The orbits are the equivalence classes of the relation for some , so they partition . The action is transitive if there is only one orbit; restricting the action to each orbit, we see that every action is a disjoint union of transitive actions. The stabilizer is a subgroup of .
The basic example is the action of on the set of left cosets of a subgroup by left multiplication, . It is transitive ( maps to ), the stabilizer of the coset is itself, and the stabilizer of is . The Orbit–Stabilizer Theorem (Theorem 3.2.9) says that every transitive action looks like this one: if and , then is a bijection. It is moreover equivariant, , so it identifies the action of on the orbit of with the action of on . When is finite this gives so the size of every orbit divides ; and when is finite, is the sum of the sizes of the orbits.
Proposition 11.1.1 (Conjugate stabilizers). Let act on . If , then . In particular, points in the same orbit have conjugate stabilizers, which have the same order.
Proof. If , then , so . Applying this to gives , that is, . ∎
Example 11.1.2 (Rotations of the tetrahedron). Let be the group of rotations of that map a regular tetrahedron centered at the origin to itself, acting on the set of its four faces. Any face can be moved to any other, so the action is transitive and . The stabilizer of a face consists of the rotations by , and about the axis through the center of that face and the opposite vertex, so . Hence : the identity, the eight rotations by about the four axes through a vertex and the center of the opposite face, and the three rotations by about the three axes through the midpoints of opposite edges. Numbering the vertices , the rotations of order 3 permute the vertices by 3-cycles, and the three half-turns by , and . We show in Section 11.3 that .
11.1.2 Burnside’s Lemma
For , the set of fixed points of is . Burnside’s lemma counts the orbits of an action in terms of fixed points.
Theorem 11.1.3 (Burnside’s Lemma (Cauchy–Frobenius)). Let a finite group act on a finite set . The number of orbits is In words: the number of orbits is the average number of fixed points of an element of .
Proof. Count the set in two ways. Summing over gives . Summing over gives . Group this sum by orbits: for in an orbit , by the Orbit–Stabilizer Theorem, so Equating the two counts gives the formula. ∎
Example 11.1.4 (Colorings of a tetrahedron). In how many ways can we color the faces of a regular tetrahedron with 3 colors, up to rotation? Here is the set of colorings of the 4 faces, , and is the rotation group of Example 11.1.2, with 12 elements.
- The identity fixes all 81 colorings.
- Each of the 8 rotations of order 3 fixes one face and permutes the other three cyclically, so a fixed coloring gives those three faces one color: .
- Each of the 3 half-turns swaps the faces in two pairs, so a fixed coloring is constant on each pair: .
So the number of orbits is .
Note. The same method handles colorings that are harder to count by hand. To color the six edges of a tetrahedron with 3 colors: the identity fixes colorings; a rotation of order 3 permutes the edges in two 3-cycles (the three edges at its vertex and the three edges of the opposite face), so it fixes ; a half-turn maps the two edges its axis passes through to themselves and swaps the other four in two pairs, so it fixes . The number of colorings up to rotation is .