Jerry Bryan tried to communicate about this to cube-lovers, but
there's apparently a technical difficulty.
On 20 May, Jerry said:
I doubt that Mark's theory about GAP using C-conjugacy for slice
instead of M-conjugacy is correct. I already have 50 positions
to 23 for GAP, and using C-conjugacy would just make my results
larger. For example, RL' and R'L are M-conjugate positions,
but not C-conjugate positions.
I emailed him to note to the contrary that RL' and R'L are indeed
C-conjugates, for example under 180 degree rotation around the F-B
axis. I did wonder, though, whether that meant that there could be
C-conjugate slice positions that were not M-conjugate.
He emailed me:
We can observe that R and R' are not C-conjugates, nor are L' and L,
which suckered me into stating that RL' and R'L are not. But
rewrite R'L as LR' since opposite face moves commute. Now, RL'
and LR' are clearly C-conjugate.
In fact, I have now verified with a quick search program that
all M-conjugates in the slice group are also C-conjugates. Hence,
there are 50 C-conjugate classes in slice, just as there are
50 M-conjugate classes.
In retrospect, I don't think the search program was necessary....
and continues with an argument that did not convince me, but the
First, consider the central inversion v, which maps each point of the
cube to its diametric opposite. Conjugation by v maps each face-turn
(e.g. F) with its diametric opposite in opposite sense (B'). Since
these are the pairs that constitute a slice move, and they commute, we
v' FB' v = (v' F v) (v' B' v) = B' F = F B',
and similarly for the other slice moves, showing that each slice move
is its own v-conjugate. This extends to a proof that each position in
the slice group is its own v-conjugate:
v' s1 s2 ... sn v = (v' s1 v) (v' s2 v) ... (v' sn v) = s1 s2 ... sn.
Suppose that we have two M-conjugate positions X, Y in the slice
group. So X = m' Y m for some m in M. If m is in C, then X and Y are
C-conjugate and we are done. Otherwise take the central inversion v;
we know that mv is in C. We also know that
X = v' X v = v' m' Y m v = (mv)' Y (mv).
So X and Y are C-conjugate in this case as well. QED.
Note: "Being its own v-conjugate" might as well be called "being