From:

~~~ Subject:

Date: 9 December 1980 1638-EST (Tuesday) From: Dan Hoey at CMU-10AThere is a twelve-element subgroup T of M which will suffice

instead of M for this argument. Representing elements as

permutations of faces, T is generated by the permutations

(represented as cycles):(F L U)(R D B) -- Rotating the cube about the FLU-RBD axis (F B)(U R)(L D) -- Rotation exchanging corners FLU and RBD (L U)(R B) -- Reflection in the LU-RB plane

AH! Excellent! (I believe you mean that last permutation to be

(L U)(R D).) It took me a while to realize that this is the subgroup

of M that leaves the FLU-RBD "diagonal" fixed.

Question: Does there exist a position other than the solved

position and the Pons Asinorum which is T-symmetric or

R-symmetric?

Hmm. I hadn't realized that we don't really know that many symmetric

positions. I have another favorite pattern that happens to be fully

M-symmetric. It is the pattern obtained by "flipping" all of the edge

cubies:

U B U L U R U F U

L U L F U F R U R B U B B L F L F R F R B R B L L D L F D F R D R B D B

D F D L D R D B D

This pattern has another interesting property, it is the only other

permutation besides the identity that commutes with every other

element of the cube group! I have often thought that this position is

a good candidate for maximality. Dave Plummer has shown that this

position can be also be reached in 28 moves...