Icosa is a simple group

Let G = Icosa, the group of rotation symmetries of the icosahedron. Just as we did for the group of rotation symmetries of the cube, we can find all the normal subgroups of G by looking at the actions of G on the sets of vertices, edges, and faces of the icosahedron. The icosahedron has 12 vertices, 30 edges, and 20 faces.

There are three types of rotation symmetries of the icosahedron: rotations with axis through a pair of opposite vertices, rotations with axis through the midpoints of a pair of opposite edges, and rotations with axis through the centers of a pair of opposite faces. We count all the rotation symmetries by using this classification.

Symmetry types

(I) Identity: total 1 rotation

(V) Rotation with vertex axis: 6 axes (6 pairs of opposite vertices), angles 72, 144, 216, 288 degrees, total 6x4 = 24 rotations

(E) Rotation with edge axis: 15 axes (15 pairs of opposite vertices), angle 180 degrees, total 15x1 = 15 rotations

(F) Rotation with face axis: 10 axes (10 pairs of opposite faces), angles 120, 240 degrees, total 10x2 = 20 rotations

TOTAL: 1 + 24 + 15+ 20 = 60 rotations, so |G| = 60.

Conjugacy classes

(I) Identity: 1

(V1) Vertex rotations, angles 72, 288 degrees: 12

(V2) Vertex rotations, angles 144, 216 degrees: 12

(E) Edge rotations: 15

(F) Face rotations: 20

We must prove that type (V) splits into two conjugacy classes.

Lemma: Let G be the group of rotation symmetries of the icosahedron. Suppose that the rotation g in G has the following property:

(*) If g fixes the vertex V and A is a face incident to V, then g takes A to a face adjacent to A. (In other words, if g.V = V and A > V, then g.A ∩ A ≠ {V}.)

If g' is conjugate to g, then g' also has property (*).

Now rotations of type (V1) have property (*), and rotations of type (V2) do not have property (*), so rotations of type (V1) are not conjugate to rotations of type (V2).

Now we prove that the only normal subgroups of G are {e} and G. The order of a subgroup H of G must divide |G| = 60, so

|H| = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

If H is normal, then H is a union of conjugacy classes, including (I), so

|H| = 1 + (a sum of the integers 12, 12, 15, 20).

The only numbers |H| satisfying these two conditions are |H| = 1 and |H| = 60.