In his message of Sat, 8 Jan 1994 08:46:20 EST, Jerry Bryan
<BRYAN%WVNVM.BITNET@mitvma.mit.edu> considers his use of the term "B"
``to indicate various aspects of the conjugacy class generated by
m'Xmc.''

I don't think that's properly called a conjugacy class, but a
different sort of equivalence class. A conjugacy class is a special
kind of equivalence class (just as a coset is a special kind of
equivalence class) but this B is a little bit of both, so I don't
think it is correct to call it either.

Let X be any cube. Then the set of B-conjugacy classes of X is
the set of all m'Xmc for all m in M and all c in C. We denote
this set as BClass(X). B is the function B(X)=min(BClass(X)).

That's a little unfortunate--I'd prefer to use B(X) for the
equivalence class, and min(B(X))--or repr(B(X))--for the canonical
representative. That's because the representative is not the
important thing here, it's just a convenient way to represent (!) the
class in a computer.

Note that we could have defined BClass(X) equivalently as the set of
all mXm'c, or as the set of all cm'Xm, or as the set of all
cmXm'.... This is the justification for the assertion in a previous
note that Gx\B = (Gx\M)\C = (Gx\C)\M.

Not quite. The justification for (Gx\Conj(M))/C = (Gx/C)\Conj(M)
is that instead of m'Xmc we could choose m'Xcm, a possibility you
didn't list.

In his message of "Sat, 8 Jan 1994 10:52:22 EST", Jerry continues with
discussion on combining conjugacy classes. We've exchanged some
private email on the subject material, but in case anyone on the list
is following this stuff....

There are only 10 distinct values for |BClass(X)| and for
|BClass(Y)|, namely 24, 48, 72, 96, 144, 192, 288, 384, 576, and
1152. (By the way, I have never figured out why it is *exactly* the
same set of values for both the corners and for the edges. It is
easy to see why it is approximately the same set of values....

I'm not sure what kind of approximation you mean, but certainly those
ten values are all that are possible:

Proof: For if m1,m2 are in the same coset of M/CSymm(X), then
(m1 m2') is in CSymm(X), so X' (m1 m2')' X (m1 m2') = c0 in C so
m1' X m1 = m2' X c0 m2.  It's then clear that
     { m1' X m1 c : c in C } =  { m2' X m2 c : c in C }      (*)
are equal 24-element sets.  The same manipulation in reverse shows
that if (*) holds for some m1,m2 in M, then m1 and m2 are in the
same coset of M/CSymm(X).  So |BClass(X)|=24 |M/CSymm(X)|.
|M/CSymm(X)| must be a divisor of |M|=48, QED.

It wouldn't have been all that surprising to see one of the possible
sizes of |CSymm(X)| fail to appear as a symmetry group of the corners
or edges, but it's not surprising that they all do, either.

[For the original approach] I needed to be able to prove that for a
fixed m and n, that |(BClass(X)[m] * BClass(Y)[n]| had the same
value for all X in GC[m]\B and all GE[n]\B.

That is to say, that the sizes |CSymm(X)| and |CSymm(Y)| might
determine {|Symm(X*Y)|} for X in GC\B, Y in GE\B, and so (X*Y) in
G\Conj(M). It doesn't, but the situation is even worse. Jerry goes
on to suppose that perhaps CSymm(X) and CSymm(Y) themselves might
determine {|Symm(X*Y)|}, and even that isn't true. I've discovered
this by a computer search of GC\B. (A search of GE\B is in progress,
but for the current result we can take Y=I in GE\B). I have found
that AllSymms(X) is not determined, even up to subgroup sizes, by
CSymm(X).

According to the search, the following are the only positions of GC\B
for which |CSymm(X)|=16.

      X1                       X2                       X3
    +---+                    +---+                    +---+
    |F F|                    |B F|                    |F B|
    |B B|                    |F B|                    |F B|
+---+---+---+            +---+---+---+            +---+---+---+
|R R|D D|L L|            |L L|D T|R R|            |L L|D T|R R|
|R R|T T|L L|            |L L|T D|R R|            |L L|D T|R R|
+---+---+---+            +---+---+---+            +---+---+---+
    |B B|                    |F B|                    |B F|
    |F F|                    |B F|                    |B F|
    +---+                    +---+                    +---+
    |T T|                    |T D|                    |T D|
    |D D|                    |D T|                    |T D|
    +---+                    +---+                    +---+

Coincidentally, I have been (privately) calling the CSymm(Xi)
subgroups the "X subgroups" of M, an X subgroup being the subgroup
that maps an orthogonal axis of the cube (in the above examples, the
L-R axis) to itself. X1 is a notable position, in that each corner
has been swapped with its opposite corner. Symm(X1) is an X subgroup
as well, and there is another X subgroup in AllSymms(X1). There is,
however, no 16-element subgroup in AllSymms(X2) or AllSymms(X3). (We
have seen X2 before: it is the corners of the Laughter (or 4/)
position). In fact, my program says that

AllSymms(X1) contains two occurrences of 16-element X subgroups,
                      two occurrences of the 8-element HX subgroup,
                      two occurrences of 8-element R subgroups,
                      two occurrences of 8-element S subgroups,
                      eight occurrences of 4-element HS subgroups, and
                      eight occurrences of the 2-element HV subgroup.

AllSymms(X2) and AllSymms(X3) each contain
                      two occurrences of 8-element CX subgroups,
                      two occurrences of 8-element AX subgroups,
                      two occurrences of 8-element P subgroups,
                      two occurrences of 8-element Q subgroups,
                      eight occurrences of 2-element ES subgroups, and
                      eight occurrences of 2-element HW subgroups.

The names of these groups are part of a taxonomy of the subgroups of M
I've developed, which I won't go into just now. But the point that I
find surprising here is that AllSymms(X1) and AllSymms(X2) are
completely disjoint. While that can't happen all the time (smaller
CSymm() groups have many occurrences of the one-element "I" subgroup)
I think the tendency to disjointness is too pronounced to be simple
anti-coincidence.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil