# Classification of the symmetries and antisymmetries of Rubik's cube

M has 98 subgroups, up to conjugacy there are 33 different subgroups of M and hence 33 different types of symmetries ignoring the inversion. Including the inversion we have 420 subgroups, up to conjugacy there are 131 different subgroups which give us 131 essentially different kinds of symmetry/antisymmetry.

The group we use to deal antisymmetries can be thought as the direct product M x C2 of M with the cyclic group C2 ={1,a}. This group has 96 elements. We define a group action of M x C2 on the set of all cubes by

(m,1).y = m*y*m^-1 and (m,a).y = (m*y*m^-1)^-1 for any cube y.

For any cube y, the stabilizer subgroup Stab(y) defines the symmetry/antisymmetry of the cube. Stab(y) contains all elements of M x C2 which leave y fixed.

In case that this subgroup contains an element (m,a) we have (m,a).y = y, which means (m*y*m^-1)^-1 = y or m*y*m^-1 = y^-1. In this case applying a symmetry operation m to y gives the inverse of y and we say (m,a) is a pure antisymmetry.

In case that Stab(y) contains a nontrivial element (m,1) we have (m,1).y=y, which means m*y^m^-1 = y. In this case we say, that (m,1) is a pure symmetry of y.

If it is not known which kind of element we have we just talk of symmetry/antisymmetry.

It is not difficult to characterize the subgroups of M x C2. They are closely related to the subgroups of M. There are three different kind of subgroups:

- If H is any of the 33 essentially different subgroups of M, (H,1) is a subgroup of M x C2 which is isomorphic to H.
- If H is a subgroup of M, (H,1)∪(H,a) also is a subgroup of M x C2. This gives 33 additional cases.
- If H
_{1}and H_{2}are subgroups of M with H_{1}⊂ H_{2}and | H_{2}| = 2 | H_{1}|, then (H_{1},1) ∪( H_{2}\ H_{1},a) is a subgroup of M x C2. H_{2}\ H_{1}denotes the difference set of H_{2}and H_{1.}We have 65 different types of this most interesting case.

The most interesting column is the column "Number of cubes mod M x C2 with exactly this symmetry". To get these numbers in a first step I computed for each of the 4.3*10^19 cubes to which of the 420 possible stabilizer subgroups they belong. Because egdes and corners can be dealt separately and because the stabilizer subgroup of a fully defined cube is the intersection of the stabilizers of the corners and edges, this computation can be done in a couple of hours.

Let's take for example the case 3.39 C3{S6}. There are 4 subgroups of type C3{S6}. The computation showed that each of these subgroups is the stabilizer subgroup of 1456 different cubes. In fact it would mean that something went wrong if the count would be different for the 4 conjugate subgroups. The union U of all these cubes has 4*1456 elements. Since the orbit of any element y in U is completely in U, we can compute the number of orbits in U by dividing 4*1456 by the orbit size.

The orbit-stabilizer theorem tells us that the orbit size is the group size divided by the stabilizer subgroup size, |M x C2|/|Stab(y)| which is 96/6=16.

So we have 4*1456/16=364 different orbits which means we have 364 cubes up to symmetry and inversion which exactly have the symmetry/antisymmetry C3{S6}.

When we sum up the corresponding numbers of all 131 subgroups we exactly get 450541810590509978 in accordance with the result of 2005.

A task not yet solved is to compute the shortest maneuver for a cube which gives an example for each of the 131 possible symmetries/antisymmetries classes. Even more demanding is a distance table for each of these subgroups. Silviu Radu managed to do this in the case of the subgroups of M, I wonder if someone is able to manage this for for the subgroups of M x C2.