On 13 Feb 1994, I proposed a model for centerless cubes which
I claimed met two criteria: 1) it was a group, and 2) it
maintained the symmetrical nature of the problem. On
23 May 1994, I retracted the claim that the proposed model
was a group.

I am now of the opinion that it is impossible to satisfy
both criteria simultaneously. I can make a very small modification
to the proposed model to make it a group, but the small
modification costs the model its cubic symmetry.

G is the full cube group, GC is the corners only cube group,
and GE is the edges only cube group. The proposed model
for centerless cubes consisted of partitioning any of G,
GC, or GE into sets of the form {Xc} for all c in C, where
C is the set of twenty-four rotations of the cube and X is
a cube. The sets are the elements of the proposed group.
The sets are called cosets and can also be denoted as
xC. The partitions are denoted as G/C, GC/C, and GE/C,
respectively.

Originally, the proposed group operator was {Xc} * {Yc} = {XYc}.
This operator fails to maintain closure, and hence fails to
define a group.

In order to illustrate the slight modification which will define
a group, we will start by restricting ourselves to GC. An
operator which works to define GC/C as a group is
{Xc} * {Yc} = {VWc}, where V is the unique element of {Xc} such
that the urf cubie is properly positioned in the urf cubicle,
and W is the unique element of {Yc} such that the urf cubie is
properly positioned in the urf cubicle.

Any other corner could have been used instead of urf, but once
you choose a corner the problem loses its symmetric nature.
Well, I guess it still has symmetry, but it is not the uniform
symmetry of the cube any more, because there is a preferred
orientation.

I have found only limited discussion in the archives, but
previous investigators have modeled a corners only, centerless
cube by leaving one corner fixed. Such a model is clearly
a group. For example, if we leave the urf corner fixed,
we can generate the group JC as JC=<L,L',D,D',B,B'>, where
we omit all twists of the U, R, and F faces from the set
of generators.

It is easy to find an isomorphism between GC/C and JC. I would
express it as something like {Xc} = {Wc} <--> W, where W is
defined as before. W is an element of JC, and as well is an
element of {Xc} = {Wc}. {Xc} = {Wc} is an element of GC/C.
But W is a particular element of {Xc} = {Wc}, whereas X is
an arbitrary element. Also, X is in GC, but X is not in JC
unless X = W. The mapping {Wc} <--> W is clearly one-to-one and
onto in both directions.

For the edges GE, we need to keep one edge cubie fixed, so the
centerless cube could be generated by something like
JE=<D,D',L,L',R,R',B,B'>, where we keep the uf cubie fixed by
omitting all twists of the U and F faces from the set of
generators. The isomorphism between GE/C and JE is expressed
as {Xc} = {Wc} <--> W, where X is an arbitrary element of
GE, and W is the unique element of {Xc} such that the uf
cube is properly placed in the uf cubicle. As before, any
edge cube would do as well, but once chosen, it is no longer
arbitrary.

For the whole cube G, at first blush it appears we could model
centerless cubes either by keeping a corner cubie fixed, or by
keeping an edge cubie fixed. But if we keep a corner cubie
fixed, the three immediately adjacent edge cubies are never moved
by any Q-turns. We could solve the difficulty by admitting
slice turns. But slice quarter-turns are odd on edges and
even on corners, so we have to restrict ourselves to slice
half-turns. I find this ugly, plus I would prefer to generate
G with Q-turns only. Hence, I would prefer to model a
centerless full cube as J=<D,D',L,L',R,R',B,B'>, where it
is an edge cubie which is held fixed rather than a corner cubie.

I said at the beginning that I thought it was impossible for a
model of centerless cubes both to be a group and also to
maintain cubic symmetry. The reason is as follows: it seems
to me that for any model which is a group, it should be
possible to find an isomorphism between the model and
J (or JC or JE, as appropriate). But J and JC and JE
do not have cubic symmetry because there is a preferred
orientation.

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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?