Rotational symmetry

An object has rotational symmetry when a nontrivial rotation matrix leaves it unchanged.

The orientation-preserving rotational symmetry group of a regular icosahedron has order sixty. Its conjugacy classes have sizes $1,15,20,12,12$: identity; half-turns about opposite-edge axes; rotations through $\pm 2\pi /3$ about opposite-face axes; and rotations through $\pm 2\pi /5$ and $\pm 4\pi /5$ about opposite-vertex axes, respectively. Conjugation carries an axis to its rotated axis and preserves the rotation angle. The transitive group action on edges, faces and vertices gives the indicated classes, with reversal of the axis identifying the two signs. A normal subgroup must contain whole conjugacy classes; none of the proper nonidentity sums of these class sizes is a divisor of sixty. Thus this group is a simple group.

The rotations in three dimensions preserving a dodecahedron form a group of order $60$. There are $24$ nonidentity rotations about the six axes through opposite face centres, $20$ about the ten axes through opposite vertices, and $15$ half-turns about the fifteen axes through opposite edge midpoints. Its nonidentity conjugacy classes have sizes $12,12,20,15$: the centralizer of a rotation of order five or three consists of rotations about its oriented axis and has order five or three, respectively, while all edge half-turns are conjugate. A normal subgroup is a union of whole conjugacy classes containing the identity. No proper nontrivial sum of these class sizes plus one divides $60$, proving that it is a simple group. The centralizers of the fifteen involutions are five distinct Klein four-groups, each containing three involutions. Conjugation permutes these five subgroups nontrivially. The kernel is a normal subgroup, so simplicity makes this nontrivial action injective. Its sign map to a group of order two must be trivial, since simplicity would otherwise force an impossible injection of sixty elements into two. Thus its image lies in the alternating group $A_5$, and equality of orders makes the image all of $A_5$.