Inversion by conjugation

[Christopher Mowla]

This appears not to be true:

gap> g := (2,3)(4,8)(5,7)(6,9)(10,11);
(2,3)(4,8)(5,7)(6,9)(10,11)
gap> (g^-1)*(1,2,3)(4,5,6)(7,8,9)(10,11,12)*g;
(1,3,2)(4,6,5)(7,9,8)(10,12,11)


gap> g := (1,4)(2,3)(6,10)(7,9)(8,11);
(1,4)(2,3)(6,10)(7,9)(8,11)
gap> (g^-1)*(1,2,3,4,5)(6,7,8)(9,10,11)*g;
(1,5,4,3,2)(6,8,7)(9,11,10)

Ahh, I didn't think to cross same length cycles to increment/decrement the number of 2-cycles in the permutation part of B! I of all people should know this.

Since all corner cycle types can all be odd permutations and since you have shown that the only two exceptions (without crossing same length cycles) can be handled with odd permutations, then there is no conflict at all.

For the nxnxn supercube, the following 5 cycle types remain which must be an even permutation.
{{11,7,5},{11,9,3},{13,7,3},{15,5,3},{9,7,5,3}}

and the permutation portion of B for all of the following 6 cycle types must be an odd permutation:
{{23},{13,11},{15,9},{17,7},{19,5},{21,3}}.

Speaking of crossing same length cycles,

I still think you CAN have twisted cycles - you just have to have the same number of clockwise and counterclockwise cycles of each length.

Here's a position with T+(2) = T-(2) = 1:
A = R2 F2 R' B' R F2 R2 F R B R' F'.
Let B = R U2 R2 F D' F D F2 R2 U' B U' B' R'.
Then B A B' A = solved.

Again, of all people, I should have taken this into account. I should have done a disjoint cycle example, as my method does handle this case, but I had to cross same length cycles to do it.

What beautiful theory! Thanks a lot you guys for reminding me of same length cycle crossing! So the only restriction that remains (for the non-supercube) is what qqwref said.