# Commutator elements of the cube

I was wondering if the number of commutator elements of the cube is known?

Commutator elements are of the form ABA'B' where A and B are some sequences.

It seems that the subgroup generated by the commutator elements is half the size of the cube, but is it the case that every element of the commutator subgroup is a commutator element? if not how many are they and how far they are from solved?

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### I do not know, but you can compute

I do not know, but you can compute it, I guess.

Since g(aba'b')g' = (gag')(gbg')(gag')'(gbg')', you only need to find commutators for single representants of conjugacy classes. What the conjugacy classes are can be found in this topic, as well as how to split them in conjugacy classes for the corner cube and the edge cube.

In order to find commutators for even corner cube or edge cube positions, the idea is to try all or at least many pairs (a,b), but not unnecessary pairs. At first, we loop through the conjugacy classes and take for a the standard representant of the conjugacy class at hand. Next, we do not need to try all b either, since after trying b, we do not need to try other elements of bC(a), where C(a) is the centralizer of a. But since it is hard to loop though cosets of C(a), let us choose b randomly a couple of times as a first attempt.

That is it more or less. For conjugacy classes that are not parity sensitive, where parity sensitive is as defined here, one can invert the parities of a and b, so one only needs to store pairs (a,b) for each of the two possible differences between the parities of a and b.

Positions for which only the indices are different here can by obtained form each other by conjugating with the twist/flip of a single cubie. Since such a conjugation is possible in this context, we may ignore positions with positive indices. For edge positions with even cycles that are parity-sensitive, the parity sensitivity is in fact a lifted orientation sensitivity. Since this orientation sensitivity vanishes after conjugating with a single edge flip, we may ignore the parity sensitivity of these positions.

EDIT: one can loop though all conjugacy classes for a and all cosets of C(a) for b as follows. Loop through all elements g of the group at hand. Write g = bab', where a is the standard representant of the conjugacy class of g.

To see that this works, we must show the following:

• Every g in a fixed conjugacy class gives another coset bC(a).
This is clear: if bC(a) = cC(a), then bab' = cac'.
• For each a, b loops through all cosets of C(a).
This follows form the fact that the size of a conjugacy class of an element a is equal to the number of cosets of C(a), see the fourth property here.