(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?