True if we have even permutation in every orbital effected by the "total permutation".

If we don't have an even permutation in every orbit, then we aren't in the commutator subgroup. Commutators (and products of commutators) can only do even permutations, even if you restrict to a particular orbit (such as edges or corners on a 3x3). So the commutator subgroup G' is contained in the group H of operations that do even permutations on every orbit. But if we mod out orientations (which, as Lucas pointed out, are easy) then H/{orientations} = H_perm is just a direct product of alternating groups A_n (where n=8, 12, or 24, the sizes of the orbits). Then, as glazik pointed out, every element of A_n is a commutator, so everything in H_perm (and therefore H) should be a single commutator.

There's one catch, though: glazik's article only says that everything in A_n is a commutator of two elements of S_n, not necessarily of A_n. So it could be that some element of H_perm is a commutator of two permutations, but that one or both of these is an odd permutation on some orbit; this may give an impossible position of the NxNxN supercube. The question is: is every element of A_n (at least for n=8, 12, and 24) a commutator of two other elements of A_n?

Perhaps we can use E. Bertram's result from glazik's link (bottom of the first page) to prove this. If we set l = n-1 (which is odd for even n), then we get that every element of A_n can be written as a product of two l-cycles, and therefore as a commutator of one l-cycle (an even permutation) with some tau in S_n. But it may be that tau is an odd permutation--for example, (1234567) and (1234576) are conjugate in S_8, but only by the odd permutation (67). So we must instead use l = n-3. Then we can force tau to be an even permutation, so every element of A_n will be a commutator of two even permutations. The final problem is that in the case n=8, n-3 is less than the bound of 3n/4 required by Bertram. For this case, I went through all the possible cycle structures of permutations in A_8, and with a bit of permutation-matrix-multiplying dirty work from Mathematica, I believe I have found direct commutators for all of them. (Exercise for the reader?) Now let's go back up the chain of reasoning and see what we've proved:

- For n=8 or (from Bertram, using l = n-3) n>=12, every element of A_n is a commutator in A_n.

- Since H_perm is a direct product of groups A_n with n=8, 12, or 24, every element of H_perm is a commutator of elements in H_perm (and therefore in the NxNxN supercube group G).

- Since orientations are easy, every element of H is a commutator in G.

- Since all commutators in G belong to G' by definition of the commutator subgroup, and since G' is contained in H, it follows that {commutators} = G' = H.

Thus the commutator subgroup of the NxNxN supercube group is simply the set of operations that are even permutations on all orbits, and all such operations can be written as single commutators.

It's interesting to note, however, that a similar statement about commutator subgroups of more general groups is not true. Dummit and Foote's

*Abstract Algebra* (page 180 in the third edition) gives a rather convoluted-looking group of order 96 in which one element belongs to the commutator subgroup despite not being a commutator of any two elements in the group.