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However, the product of two or more commutators need not be a commutator. A generic example is [a,b][c,d] in the free group on a,b,c,d. It is known that the least order of a finite group for which there exists two commutators whose product is not a commutator is 96; in fact there are two nonisomorphic groups of order 96 with this property.

True, I believe that was already mentioned. However, since we have in this case that the product is always a commutator, does that make it possible to construct the product? I suspect not, but it would be interesting to know for sure.

If [a,b] and [c,d] are able to be merged together with my method, then for permutations at least, [a,b][c,d] = [c,bd] != [c,db].

Note that I would have to solve all middle edge and corner cases in an alternate manner (a manner in which I thought was impractical when I started, as it is even more involved than my current one) in order to prove that it holds true for orientations of corners and middle edges mixed with their permutations.

By intensive research (brain storming, as I didn't use any existing sources) for 1.5 years. Well, my method itself took about half that time, but it did take that long to both create/fully develop the method and effectively solve every possible case that can arise on the nxnxn cube and to be certain that my proof is correct.

I actually thought that myself, especially when I was solving all corner and middle edge orientation cases by hand (and verifying that they are all correct with software that I wrote in Mathematica which is over 200 pages of code and documentation if printed. I needed to write software because there are more than 10 things that can be wrong with a solution)! I never thought I would finish...it's hard to believe it's only been a year and a half since I seriously started figuring all of this out from scratch.

If I try to picture all of the details of my proof at once, I feel very intimidated and overwhelmed, because this project is bigger than I am, and its applications to cubing (as well as possible derivative works from it) are huge! My proof is only an "introduction" to commutator theory, because it introduces an entirely new universe, and we just travel from the earth to the sun. The way I chose to solve the corner and edge orientation cases, for example, was the easiest (and therefore I could solve the cases as fast as possible) because I had to solve every case there is. When I release my method and theory, we can all work together to find a variety of solutions for any given case, form new hypotheses, and find ways to prove/disprove them.

What i will write now might be obvious to You cmowla, but...

I was thinking about Your theorem when i was asleep and i think that it is possible to prove it using lemma:

Every even permutation P can be written as P=(A*B), that B has the same length as A.

So, on cube with applied algorithm P, when i do algorithm A, then on the cube i would have remaining situation B-1 which could be solved with C A-1 C, where alg C is setting up the remaining elements of the cube.

Every even permutation P can be written as P=(A*B), that B has the same length as A.

So, on cube with applied algorithm P, when i do algorithm A, then on the cube i would have remaining situation B-1 which could be solved with C A-1 C, where alg C is setting up the remaining elements of the cube.

This general idea is correct for permutations (and is where my proof begins), and I believe at the beginning of this thread, Ravi implied something like this, but it's much more involved than this for permutations of all orbits on the nxnxn cube, and even more so for orientations of corners and middle edges.

In fact, I prove this lemma in my proof independently of any old proofs of it (I couldn't find any on the web, to be honest).

Scramble: U2 F' B' D2 L F2 L2 F U' B2 R2 L2 U2 F' U2 F' U2 B' L2 D2
Solution: [F2 L2 B2 D2 R' L' D' L' D F' R' U2 D2 F2 R2 D2 F2, F2 L2 U D R D B2 D' L U' B' D R'] (view at alg.cubing.net)

Whoa, it actually works!
It was my first attempt, and also pretty successful. Well, I think it was not too hard. Below is how it works:

Spoiler

Scramble: U2 F' B' D2 L F2 L2 F U' B2 R2 L2 U2 F' U2 F' U2 B' L2 D2

1st cycle: UFR > DBR > FDR > LBD > UFR // UF > DF > RB > RU > UB > FU + DR > DL > RD
solve alg: F2 L2 B2 D2 R' L' D' L' D F' R' U2 D2 F2 R2 D2 F2 (17f*)

2nd cycle: UFR > BLU > FUL > FLD > UFR // UB > DB > LF > UL > LB > BU + UF > FR > FU
setup alg: R B U' D R2 B' U' D2 R D2 F U' (12f*)

Final Output: [F2 L2 B2 D2 R' L' D' L' D F' R' U2 D2 F2 R2 D2 F2, F2 L2 U D R D B2 D' L U' B' D R']

First, I searched for the whole cycle on the scramble, each corners and edges. It's similar with the memorization on 3x3 BLD. (not 3OP, but BH)

Second, I cut the cycle into half. But, the two cycles(which was cut) should be same one, because the first part of the commutator will excute the same cycle. (the normal one, and the inverse one.)

Third, I chose any between the two cycles, then I solved the cycle. And this is the first part of the commutator. - That's what the '1st cycle' does.
(I didn't want to do FMC with this, so I just ran 'Cube Explorer' here. Of coursely, doing without computer program is posslble.)

Forth, I put setup moves at the second part of the commutator. Each pieces of '2nd cycle' should correspond to other pieces of '1st cycle', but in inverse way. Then it will solve the '2nd cycle' - That's what the '2st cycle' does.
For example: