SKEDSOFT

EQUIVALENCE CLASSES OF FUNCTIONS: Consider two sets D and R, with the number of elements | D | and | R | respectively. Let f be a mapping or function which maps each element d from doamin D to a unique image f(d) in range R. Since each of the | D | elements can be mapped into any of | R | elements, the number of different functions from D to R is | R || D |.
Let there be a permutation group P on the elements of set D. Then define two mappings f1 and f2 as P-equivalent if there is some permutation π in P such that for every d in D we have

f₁(d) = f₂[π(d)] ...(1)

The relationship defined by (1) is an equivalence relation can be shown as follows :

(i) Since P is a permutation group, it contains the identity permutation and thus (1) is reflexive.
(ii) If P contains permutation π, it also contains the inverse permutation π– 1. Therefore, the relation is symmetric also.
(iii) Furthermore, if P contains permutations π1 and π2, it must also contain the permutation π1π2. This makes P-equivalence a transitive relation.

The permutation group P on D is the set of all those permutations that can be produced by rotations of the cube.

These permutations with their cycle structures are :
(i) One identity permutation. Its cycle structure is y₁ ⁸.
(ii) Three 180° rotations around lines connecting the centers of opposite faces. Its cycles structure is y₂⁴.
(iii) Six 90° rotations (clockwise and counter clockwise) around lines connecting the centers of opposite faces. The cycle structure is y₄².
(iv) Six 180° rotations around lines connecting the mid-points of opposite edges. The corresponding cycle structure is y₂⁴.
(v) Eight 120° rotations around lines connecting opposite corners in the cube. The cycle structure of the corresponding permutation is y₁².y₃³.

The cycle index of this group consisting of these 24 permutations is, therefore,