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On 13 Feb 1994, I proposed a way to model centerless cubes which would

(in Dan Hoey's words) retain the symmetrical nature of the problem.

I need to post a partial correction/retraction.

The conventional model for centerless cubes loses the symmetrical

nature of the problem. For example, for a corners-only cube, seven

cubies are modeled rather than eight, and for an edges-only cube,

eleven cubies are modeled rather than twelve.

My proposal in February was to use cosets of the form xC to model

centerless cubes, where x is a cube and where C is the set of

twenty-four whole cube rotations. This proposal in turn requires

an interpretation of C such that C is a subset of G, the entire cube

group.

C is a group, but normally it is not considered to be a subset

of G, hence it is not normally considered to be a subgroup of G.

That is, C moves the centers of the faces, but G does not. The

required interpretation is obtained by removing the centers of each

face, and defining rotational orientation by convention so that the

cube is solved only when the Up color is Up, the Front Color is

Front, and so forth.

Under this interpretation, C is indeed a subset (and hence a subgroup)

of G. More correctly, C[even] is a subset of G, C is a subset of

GC (the corners only cube), C is a subset of GE (the edges only cube),

and C is a subset of GS, where GS=<Q,S> (Q is the set of quarter turns

and S is the set of slice moves). That is, when you start

talking about C as a subset of G, you have to worry about odd and

even permutations. Hence, you have to say C is a subset of GS or

C[even] is a subset of G in order not to violate parity rules.

All of the above was posted in February, and I am still comfortable

with it. However, I went on to say that GS/C, G/C[even], GC/C, and

GE/C were all groups under the operation xC * yC = (xy)C. I find

that I must retract this claim.

In my note in February, I did not give a proof, but rather appealed

to a proof in Frey and Singmaster's _Handbook of Cubik Math_. I now

find that I mis-applied their proof. In order to show the nature of

the problem, I find it useful to go through an attempted proof and

determine the point at which it fails.

Note that the proposed group elements are not cubes, they are cosets.

We proceed as follows:

1. Associativity: (xC * yC) * zC = (xy)C * zC = ((xy)z)C = (x(yz))C = xC * (yz)C xC * (yC * zC)

Note that the associativity of the proposed group G/C derives

directly from the associativity of G.2. Identity: we propose that the identity is iC

iC * xC = (ix)C = xC xC * iC = (xi)C = xC

Note that the identity of the proposed group G/C derives

directly from the identity i of G. Further note that the

identity iC of the proposed group G/C is C, which is

precisely what you would want for the identity of a centerless

cube.3. Inverse: we propose that (xC)'=x'C

xC * x'C = (xx')C = iC x'C * xC = (x'x)C = iC

Note that the inverse of xC in the proposed group G/C derives

from the inverse of x in G.4. Closure: This is where we have our problem. We require that

if xC * yC = (xy)C, then (xy)C must be a coset of C. But the

representation of xC and yC is not unique. That is, xC=(xd)C,

where d is in C, and yC=(ye)C where e is in C. It is the

case that (x(ye))C = (xy)C, but in general it is not the case

that ((xd)y)C = (xy)C. Hence, we can have xC=(xd)C, but have

it be the case that xC * yC is not equal (xd)C * yC. Hence,

we do not have closure.

Strictly speaking, this same problem afflicts our "proof" for

the inverse, but I deferred discussing the problem until I got

to closure. If the problem is repaired for closure, it is also

repaired for inverses (see the next paragraph for a discussion

of normal subgroups).

Cosets of a subgroup H are said to be normal if xH = Hx for all x.

I was implicitly and incorrectly assuming that C is a normal

subgroup of G, but it is not. For normal subgroups, closure of

coset multiplication is readily proven. Frey and Singmaster's proof

is for normal subgroups only, and I was applying it to C, which is

not normal.

It is instructive to consider briefly what xC vs. Cx means for cubes.

We can interpret the left coset xC as simply holding a cube in your

hands and rotating it any way you wish in space without performing

any twists.

The right coset Cx is a little trickier. The cube x

must be pre-multiplied by each c in C to form Cx. If you have cube

x in your hands, there is no obvious thing you can do to form Cx.

The thing that is most intuitive to me personally is to think in

terms of "rotation by color", which is the way I described

pre-multiplication when I first posted some of the results of my

work back in December. That is, think of holding the cube still, but

recoloring it by permuting the colors. The elements of the coset Cx

look the "same" but with the colors permuted. It is not possible to

perform this operation on a real cube (short of pulling off the stickers

and putting them back on), but the operation can be readily performed

on a computer model.

Having said all this, I keep thinking that there must be a way to define

an operation on the cosets xC so that they form a group. However, I

have been unsuccessful in doing so. I would welcome any advice from

the net.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 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?