# Nonconstructive proof of the size of the cube group?



## meichenl (Mar 10, 2010)

I'm wondering if anyone can give me some insight on why the cube group is as large as it is.

I know, for example, that the parity of the permutation of the corners and edges must be the same because each quarter turn is an odd permutation of both (excluding slice turns, double layer turns, and cube rotations). This excludes half of the conceivable permutations of the cubies.

It remains to be shown that the other half of the permutations can be solved. One way to do this is to note that any odd permutation becomes even with a single quarter turn. Also, all even permutations can be factored into 3-cycles. Next, find one 3-cycle, and then conjugate it to produce all conceivable 3-cycles. This shows that the other half of the permutations of the cubies can be solved.

Similarly for orientations, once the orientation of a cubie not at home is defined, it is easy to see the restrictions on cubie orientation. To prove there are no more restrictions we're missing, though, the only way I know is to give commutators that flip two edges or that flip two corners, and by conjugating those show how to solve any remaining orientation of the cubies.

These aren't viscerally satisfying to me, but I haven't found any better way to approach the problem.


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## Lucas Garron (Mar 10, 2010)

meichenl said:


> These aren't viscerally satisfying to me


Really?
I have no problem thinking of it that way.
It's a little unsatisfying to show that all the permutations are possible (while staying formal), but I find it the more intuitive part.

(Rather than, say, demonstrating L2 B2 D U2 R2 U L2 F2 U B2 U' B U2 R2 U2 B' L2 D, F2 D' R2 D2 U' R2 B2 R2 B' U' F' L2 F U B' R2, and D F2 L2 D2 B2 R2 D' L2 D B2 D F' L2 D2 L2 F D F2)


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## Ton (Mar 10, 2010)

You can solve the cube with the T perm and use setup moves to solve any permutation/orientation on corners or edges . So you do not factor the cube in 3-cycles but in 2-Cycles to prove it


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## meichenl (Mar 11, 2010)

Ton,

That's a good point. I chose 3-cycles because there are 56 corner 3-cycles and 220 edge 3-cycles, making 276 total.

There are 28 corner pairs and 66 edge pairs, making 1848 total edge/corner pairs. (Of course there are far fewer of each accounting for symmetries.) So you have fewer conjugations to look at.

Lucas,

I guess a 3-cycle that's a commutator is fairly intuitive. Doing the proof using T-perms, I'd be asking, "why can you do a T-perm?".


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