Try
A group of twelve rotations is not identified merely by its order. Give it something concrete to permute. For the tetrahedron, use its four vertices. For the octahedron, using its six vertices produces permutations of six objects—but another choice gives just four objects. For the icosahedron, we will construct a five-object action from cosets rather than invent five vertices.
This lesson uses the exact models and orders proved in polyhedral coordinate models. Let T, O and I denote their orientation-preserving rotation groups, of orders 12, 24 and 60 respectively. These letters name groups here; I is not an identity matrix. Write e for a group identity.
Predict
An action on a finite set gives a homomorphism to its permutation group. Its kernel consists of the elements fixing every object. An action is faithful when that kernel is just e. Predict which conclusions require faithfulness: “the group acts on four objects,” “the group embeds in S₄,” and “the group equals S₄.” The last statement also needs its image to have order 24.
For an action on lines, fixing a line allows its vectors to change sign. This makes a line action different from a vertex action. We will check this possible kernel explicitly.
Experiment
Label the tetrahedral vectors d₁=(1,1,1), d₂=(1,−1,−1), d₃=(−1,1,−1), d₄=(−1,−1,1). Scaling all four to unit length does not change their permutations. Apply A=diag(1,−1,−1) and C(x,y,z)=(z,x,y). Compute each image before revealing the table entries in your own notes.
| Map | Images of labels 1,2,3,4 | Permutation |
|---|
| A | 2,1,4,3 | (1 2)(3 4) |
| C | 1,3,4,2 | (2 3 4) |
| AC, applying C first | 2,4,3,1 | (1 2 4) |
Both generators are even permutations. This computes parity from an actual action; it does not assume that a determinant-one matrix must induce an even permutation on every set it acts on.
Understand
The tetrahedral group is A₄
The coordinate construction gives the twelve matrices consisting of a cyclic coordinate permutation followed by a diagonal sign change whose three signs multiply to 1. The two independent sign changes act as double transpositions on the dᵢ; C acts as a three-cycle. They generate the twelve matrices, so every induced permutation is even. The vertex action is faithful: a linear map fixing all dᵢ fixes a spanning set of ℝ³ and is the identity. Its image therefore has twelve elements inside A₄, which itself has twelve elements. Thus T is isomorphic to A₄.
The octahedral group is S₄
The determinant-one signed permutation matrices preserve the octahedron with vertices ±e₁, ±e₂, ±e₃. They also permute the four unoriented body-diagonal lines ℝd₁,…,ℝd₄ of the dual cube. This defines a homomorphism O→S₄.
Suppose Q fixes each line. Since it preserves lengths, Qdᵢ=εᵢdᵢ with εᵢ equal to 1 or −1. Any three dᵢ are independent: for example the determinant of columns d₁,d₂,d₃ is 4, and the other triples have determinant ±4. Since d₁+d₂+d₃+d₄=0, this is the only linear relation up to a scalar. Applying Q to that relation shows all εᵢ are equal. Thus Q is either the identity or its negative. In dimension three the negative identity has determinant −1, so it is excluded. The line action is faithful. The image has 24 elements, hence is all S₄.
Recover the icosahedral conjugacy classes geometrically
The group I contains the twelve-element coordinate subgroup K generated by the even sign changes and C. This subgroup preserves no line. Indeed a line invariant under the two diagonal sign changes must be one of their simultaneous one-dimensional eigenspaces, the coordinate axes; C permutes those axes without fixing one. Therefore I preserves no line either.
The proved pole-count theorem now identifies its data. At order 60 a noncyclic group could have data (2,2,30) or (2,3,5). The first case preserves a line by the dihedral reconstruction, so I has data (2,3,5). The corresponding pole orbits have sizes 30, 20 and 12. Antipodal poles have the same stabilizer, and each stabilizer size occurs in just one of these orbits. Thus these give 15 axes of order two, 10 of order three and 6 of order five.
Each nonidentity rotation has exactly one axis. There are therefore 15 half-turns, 20 rotations of order three and 24 rotations of order five. Together with e these exhaust the sixty elements.
A rotation of angle other than 0 or π has centralizer equal to its oriented-axis stabilizer within I. To prove this, a commuting map preserves its fixed axis. If it reverses the axis, conjugation reverses the rotation angle, which would require θ=−θ modulo 2π, excluded here. If it fixes the oriented axis, both maps restrict to plane rotations and commute. The centralizer consequently has order three or five. Conjugacy classes have sizes 60/3=20 and 60/5=12. The twenty order-three elements form one class; the twenty-four order-five elements form two classes. The two angles with cosines cos(2π/5) and cos(4π/5) distinguish those classes by matrix trace.
All half-turns are conjugate because their axes form one orbit. Equivalently, the setwise axis stabilizer has size 60/15=4 and is their centralizer: either orientation of the axis describes the same half-turn. Their class has size 15. We have derived the class sizes 1,15,20,12,12 without presupposing an isomorphism with A₅.
Use simplicity to obtain a faithful five-object action
A normal subgroup contains e and whole conjugacy classes. Its possible orders are 1 plus a subset sum of 15,20,12,12. The distinct resulting numbers are
1, 13, 16, 21, 25, 28, 33, 36, 40, 45, 48, 60.
Only 1 and 60 divide 60. Lagrange’s theorem therefore proves that I is simple: it has no proper nontrivial normal subgroup. This repeats the class-union method used for solvable groups and A₅, but supplies the classes independently for the geometric group.
The concrete subgroup K has order twelve and index five. Let I act by left multiplication on its five left cosets. This action is transitive and not trivial. Its kernel is normal; simplicity forces it to be trivial. Hence I embeds in S₅. Composing with the sign map to the two-element group gives a homomorphism whose kernel is normal. If that kernel were trivial, sixty elements would inject into two, impossible. Thus every image permutation is even. The image has sixty elements inside A₅, also of order sixty, and equals A₅. This proves I≅A₅.
The five objects are cosets of K. We did not need a chosen arrangement of five cubes inside an icosahedron, and an unsupported drawing of such cubes would not replace the kernel argument.
Apply
Change the geometric group from rotations of the octahedron to all orthogonal symmetries, allowing determinant −1. Does its action on the same four body-diagonal lines remain faithful? Determine its kernel, image and order. Then explain why “every rotation gives an even permutation” was not a valid shortcut in Experiment.
Check the changed symmetry group and its kernel
There are 3!·2³=48 signed permutation matrices. The same relation argument makes the line-action kernel exactly the two scalar matrices ±identity. Its image has order 48/2=24 and is S₄. The action is therefore not faithful. Restricting to determinant-one matrices removes the negative identity and gives the faithful rotation action proved above. Moreover O maps onto S₄, so some determinant-one rotations act as odd permutations of these four lines. Permutation parity depends on the chosen action; it is not matrix determinant in disguise.
Continue
Combine the coordinate constructions with polyhedral orbit reconstruction and the cyclic/dihedral argument. Together they identify the finite subgroups of SO(3) up to geometric conjugacy and give the abstract types Cₙ, Dₙ, A₄, S₄ and A₅. Here Dₙ has order 2n; the identity group is C₁. The abstract isomorphism proved in this lesson is a different statement from the geometric conjugacy proved by reconstruction. Full orthogonal groups containing reflections require the additional orientation-reversing analysis; they are not included in this SO(3) classification.
