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Date: Tue, 06 Aug 1996 08:58:23 -0500 (EST)
From: Jerry Bryan
Subject: Commuting Sets
To: Cube-Lovers
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If X and Y are sets of permutations, we define XY to be the set
{xy | x in X and y in Y}. In my various search programs, I have
encountered a number of cases where we have XY=YX, even though we do not
in general have xy=yx. For example, let Q[n] be the set of all positions
which are n quarter turns from Start. My standard breadth first search
is essentially Q[n+1] = Q[n]Q[1] - Q[n-1]. But we could just as well say
Q[n+1] = Q[1]Q[n] - Q[n-1] because Q[n]Q[1] and Q[1]Q[n] are the same set.
I have been wondering, what are the necessary and sufficient conditions
for XY = YX? Note that X and Y are not necessarily groups.
I really don't know the answer, and I wondered if anybody out there does.
I have some suspicions it has something to do with conjugacy. In all the
cases I have worked with, it it the case that if x in X and y in Y, then
all the K-conjugates of x are also in X and all the K-conjugates of y are
also in Y -- where K is usually M, the set of 48 rotations and reflections
of the cube. For other searches such as __, K is the symmetry group
associated with the group being searched.
It is trivial to make an X and Y that don't "commute" in this matter.
That is, pick x and y that don't commute and have sets X and Y containing
only the single elements x and y, respectively.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7127
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
__