# Math Problem - 12



## Lucas Garron (Aug 15, 2008)

Sune as a single commutator

I think I should post a problem sometime. This, however, will be a cube problem; nevertheless, it's a lot like the others presuming some cube knowledge.

Many cubers know knows that the Sune (RUR'URU2R') can be written as a two commutators ([R, U][U2, R]) or as a conjugated commutator ([R U2; U', R']). But even Johannes didn't know it could be written as a single commutator. In fact, I wasn't even sure until I tried it, and the task turned out to be pretty fun.

So, the problem: Write the Sune as a commutator in the pure form [moves, moves].
(Doesn't have to be pretty unless you want it to be.)

If you've found it or already know it, _please_ use spoiler tags, as seeing a solution even for a glance may be distracting for people who haven't solved it yet.


And if you think that's too easy: It's well-known that commutators generate all the even permutations. Show that any even permutation can be written as a single commutator, or show that it can't be done.


----------



## DavidWoner (Aug 15, 2008)

i found the ones for U-perm, still working on the sune one.

EDIT: im pretty sure these are actually wrong. yes. they are wrong. i will remove them now.


----------



## Tim Reynolds (Aug 15, 2008)

That was fun to work on...I can't really explain my steps very well, but I got


Spoiler



[R U R2, R U2 R2] = R U R2 R U2 R2 R2 U' R' R2 U2 R' = R U R' U2 U' R U2 R'= sune


now working on the general case...


----------



## nitrocan (Aug 15, 2008)

I don't know commutators that well, but this doesn't require so much commutator knowledge I guess. I will give it a try:


Spoiler



R U R' U R U2 R' as A B A' B'
I'm guessing A will start with R U and B will start with R U2 and they should be 3 or 4 moves most. I will call the 3rd (and if it's 4 moves the 3rd and the 4th) move x(A) and y(B)
Our commutator can be written as : R U x R U2 y x' U' R' y' U2 R'
so x R U2 y x' U' R' y' must be R' U R
then our x becomes R2 and then we can see that:
R U R2 R U2 y R2 U' R' y' U2 R' = R U R' U R U2 R'
then U2 y R2 U' R' y' must be U R
then y is R2 so that in cancels the R2 and U2 U' makes U and R' R2 makes R
so R U x is R U R2 and R U2 R2, our commutator becomes:
[R U R2, R U2 R2]
this wasn't so easy so I'm not going to answer the 2nd question


----------



## Stefan (Aug 17, 2008)

Lucas Garron said:


> Show that any even permutation can be written as a single commutator, or show that it can't be done.



Do you know the answer?


----------



## cuBerBruce (Aug 18, 2008)

Lucas Garron said:


> It's well-known that commutators generate all the even permutations. Show that any even permutation can be written as a single commutator, or show that it can't be done.


Actually, every legal position of Rubik's cube is an even permutation, and half of them can't be generated by commutators (of legal Rubik's cube moves). Apparently by "even permutation," you really meant that the edge permutation is even and the corner permutation is even. 



Spoiler



Assuming that's what you really meant, Joyner's book _Adventures in Group Theory_ has the answer, but his proof "cheats" by using another theorem.


----------

