Date:  9 December 1980 1638-EST (Tuesday)
From: Dan Hoey at CMU-10A

There 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...