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Date: Mon, 10 Jan 1994 23:06:00 EST
From: "Jerry Bryan"
To: "Cube Lovers List"
Subject: |G\M| - Some Trivial Partial Results
It occurs to me that a small part of my incorrect attempt during
December to calculate |G\M| can be salvaged. In particular,
for those cases where B-conjugate classes are of order 1152,
the calculations are trivial. About 99.9923% of the edge conjugate
classes and about 96.924% of the corner conjugate classes are of
order 1152, so we can calculate the correct number of M-conjugates
of G for a very large percentage of the cases.
Consider some fixed X in GC\B and some fixed Y in GE\B where
|BClass(X)|=1152 and |BClass(Y)|=1152. Form
BClass(X) * BClass(Y). Now, |BClass(X) * BClass(Y)| =
|BClass(X)| * |BClass(Y)| / 2 = 1152 * 1152 / 2.
(The division by 2 takes care of parity). Finally, form
(BClass(X) * BClass(Y))\M, and we have |(BClass(X) * BClass(Y))\M| =
|BClass(X) * BClass(Y)| / 48 = (1152 * 1152) / 2 / 48 = 13,824.
We know the number of BClasses of GC of order 1152 from computer
search (namely 75,392), and we know the number of BCLasses of GE of
order 1152 from computer search (namely 851,493,140). Hence, for
the special case of both BClasses being of order 1152, we have the
total number of elements of G\M being 851,493,140 * 75,392 * 13,824 =
887,442,335,689,605,120.
We can derive similar results if only one of BCLass(X) and BClass(Y)
are of order 1152. For example, there are 4 elements of GE\B
for which |BClass(Y)|=24. Choose such a Y, and choose X in GC\B
such that |BClass(X)|=1152. Form BClass(X) * BClass(Y). It will
be the case that |BClass(X) * BClass(Y)| = 1152 * 24 / 2 = 13,824.
Hence, |(BClass(X) * BClass(Y))\M| = 13,824/48 = 288. There are 75,392
values of X for which |BClass(X)|=1152, 4 values of Y for which
|BCLass(Y)|=24, and hence there are 75,392 * 4 * 288 = 86,851,584
elements of G\M of the form BClass(X) * BClass(Y) for which
|BCLass(X)| = 1152 and |BClass(Y)| = 24.
There are nineteen cases in all for which at least one of BClass(X)
and BCLass(Y) are of order 1152, and this note calculates only
two of the nineteen. Completing the other
seventeen would be trivial but tedious. However, a total solution
to the problem will require coming up with some way to deal with
the cases where neither |BClass(X)|=1152 nor |BClass(Y)|=1152.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) (304) 293-5192
Associate Director, WVNET (304) 293-5540 fax
837 Chestnut Ridge Road BRYAN@WVNVM
Morgantown, WV 26505 BRYAN@WVNVM.WVNET.EDU
If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?