The alternating group containsThe product of any two different nonidentity elements is the third, and each squares to the identity, so is a Klein four-group. Conjugation by a permutation replaces the labels in each cycle, preserving the property of being a product of two disjoint transpositions. Thus is a normal subgroup of . Its order is , strictly between and . is not simple.
Write for the rotational symmetry group of a regular dodecahedron. Centre the solid at the origin. A rotation in three dimensions is determined by the image of a chosen face and one of its vertices, so there are at most rotations. All these choices occur. The outward unit normal to a face and the direction from its centre to the chosen vertex determine an oriented orthonormal frame. The proper rotation mapping one such frame to another maps the whole regular pentagonal face. At each edge, the common dihedral angle and convexity then determine the adjacent face uniquely; propagating across the connected face-adjacency graph shows that the rotation maps the entire solid to itself. Hence , and the action is transitive on faces, vertices and edges.
Its rotations can be counted geometrically. There are six axes joining opposite face centres, each allowing four nonidentity rotations of order five; ten axes joining opposite vertices, each allowing two of order three; and fifteen axes joining opposite edge midpoints, each allowing a half-turn. The orbit-stabilizer theorem also verifies these possibilities: vertex stabilizers have order and edge stabilizers have order , and their nonidentity elements rotate about the corresponding vertex or edge-midpoint axis. Together with the identity this accounts for all elements. In particular there are no elements of order four.
We now determine the conjugacy classes. The group is transitive on edges, so all fifteen edge half-turns are conjugate. For a rotation whose angle is neither zero nor , a commuting rotation must preserve its oriented axis: conjugation carries the axis and its rotation angle to their images, and reversing the axis would reverse a non-half-turn angle. Thus its centralizer consists of rotations about the same axis. For a face rotation this centralizer has order five, and for a vertex rotation it has order three. The orbit-stabilizer theorem for conjugation therefore gives class sizes and . The twenty vertex rotations form one class, and the twenty-four face rotations form two classes. The class sizes are .
A normal subgroup is a union of these classes and contains the identity. Its possible orders areBy Lagrange's theorem its order must divide . Only and in this list do so. Therefore .
For each involution , its conjugacy class has size fifteen, so its centralizer has order four. A group of order four is a cyclic group or a Klein four-group; the cyclic possibility would contain an element of order four, excluded by the geometric enumeration. Hence each is a Klein four-group containing three involutions.
If is any of the three involutions in , the latter group is an abelian group and hence is contained in . Both centralizers have order four, so they are equal. It follows that the fifteen involutions partition into five triples, with five distinct centralizers. Conjugation by permutes these five subgroups and gives a group homomorphism . This action is nontrivial: otherwise each of these order-four subgroups would be a normal subgroup, contrary to simplicity. The kernel of is normal; since is simple and the action nontrivial, its kernel is trivial. Thus is injective.
The sign of defines a homomorphism . If it were nontrivial, simplicity would make it injective, impossible for a group of order sixty mapping into one of order two. Hence every image permutation is even, and . Both groups have order sixty, so they are equal. This constructs the requested group isomorphism:
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