From BRYAN@wvnvm.wvnet.edu Wed Sep 21 11:32:38 1994 Return-Path: Received: from WVNVM.WVNET.EDU by life.ai.mit.edu (4.1/AI-4.10) for /com/archive/cube-lovers id AA17783; Wed, 21 Sep 94 11:32:38 EDT Message-Id: <9409211532.AA17783@life.ai.mit.edu> Received: from WVNVM.WVNET.EDU by WVNVM.WVNET.EDU (IBM VM SMTP V2R2) with BSMTP id 7073; Wed, 21 Sep 94 11:31:44 EDT Received: from WVNVM.WVNET.EDU (NJE origin BRYAN@WVNVM) by WVNVM.WVNET.EDU (LMail V1.2a/1.8a) with BSMTP id 6772; Wed, 21 Sep 1994 11:31:44 -0400 X-Acknowledge-To: Date: Wed, 21 Sep 1994 11:31:42 EDT From: "Jerry Bryan" To: "Cube Lovers List" Subject: M-Conjugate Classes as a Group C is the set of twenty-four rotations of the cube. After much bungling (see my notes of 13 Feb 1994, 23 May 1994, and 19 July 1994), I showed that the left cosets of C, denoted by xC or {Xc}, form a group, and that the group is isomorphic to a subgroup of G. I consider this to be important because I use left cosets of C to model centerless cubes. M is the set of forty-eight rotations and reflections of the cube. I often model the cube with M-conjugate classes of the form {m'Xm}. Therefore, it seems that I should try to define an operation such that the M-conjugate classes form a group, and such that the group is isomorphic to a subgroup of G. I would like to start by reviewing briefly the results for left cosets. Two operations were defined: 1.a. {Xc} * {Yc} = {(VW)c} 1.b. V ** W = (VW) where V and W are representative elements of {Xc} and {Yc}, respectively. Further, the mapping V <--> {Vc} defines an isomorphism between the set of left cosets and the operation * on the one hand, and the set of representative elements and the operation ** on the other hand. Since the ** operation is simply normal cube multiplication and since the set of representative elements are a group under **, the set of representative elements form a subgroup of G. I tried to define groups without using representative elements and failed. Not only that, the representative elements had to be selected in a special way rather than arbitrarily. For example, we could choose as the representative element of {Xc} the unique element V such that the ur cubie is positioned properly. Positioning the ur cubie properly is not the only selection function for a representative element which will work, but any selection function must satisfy two criteria in order to work: A. It must select a representative element based on a property which is possessed by exactly one element of each coset. B. There must be closure in the sense that if V is the representative element of {Xc} and W is the representative element of {Yc}, then (VW) must be the representative element of {(VW)c}. Criterion #B merits some additional discussion. First, it is the criterion that really proves you have a group. Associativity for a subset of a group generally follows from the the associativity of the group. For a finite group, closure for a subset implies the identity and the complement for the subset, so closure is the key factor in demonstrating that a set of cubes is a group. Second, criterion #B will bear directly on our attempt to define a group operation for the M-conjugate classes. Suppose we choose not to require criterion #B. We still need to have closure in order to have a group. We could obtain closure by brute force as follows: 2.a. {Xc} * {Yc} = {(Repr{(VW)c})c} 2.b. V ** W = Repr{(VW)c} It is probably a little easier to see what is going on in equation 2.b. than in 2.a., but it is the identical mechanism in both cases. Suppose we don't have closure. That is, suppose the selection function operates in such a way that if V is the representative element of {Xc} and W is the representative element of {Yc} that (VW) is not necessarily the representative element of {(VW)c}. We can still find the representative element of {(VW)c} by simply applying the selection function, which we have done. Equations 2.a and 2.b define groups, where the left cosets are a group under * and the representative elements are a group under **. Furthermore, the mapping V <--> {Vc} defines an isomorphism between the two groups. But even though the set of representative elements is a subset of G, and even though they form a group under **, they are not a subgroup of G. The problem is that the operation ** as defined by equation 2.b. is not the same operation as standard cube multiplication as it was in equation 1.b. Now, let's look at M-conjugate classes. By analogy with the left coset case, there are two possibilities to define a group: 3.a. {m'Xm} * {m'Ym} = {m'(VW)m} 3.b. V ** W = (VW) 4.a. {m'Xm} * {m'Ym} = {m'(Repr{m'(VW)m})m} 4.b. V ** W = Repr{m'(VW)m} As before, X and Y are any cubes in G, and V and W are the representative elements of {m'Xm} and {m'Ym}, respectively. In order to make 3.a. and 3.b. work, we need some characteristic which can be used by the selection function which possesses the properties of uniqueness and closure as defined by #A and #B above. But I can't think of any such property, and I don't think such a property exists (see below). 4.a and 4.b certainly work. That is, they define operations * and ** under which the set of M-conjugate classes and the set of representative elements, respectively, form groups, and the groups are isomorphic under the mapping V <--> {m'Vm}. However, the groups fail to be subgroups of G for the same reason elements of left cosets fail to be subgroups of G under equation 2.b. Namely, the ** operation is not really the same operation as normal cube multiplication. As to the question of whether 3.a. and 3.b. can be made to work, I think we can prove that they cannot. Suppose the contrary. That is, suppose that there is some property such that it is possessed by exactly one element of each M conjugancy class and such that the normal cube product of two such elements also possesses the property. Then, it would be the case that the set of representative elements would be a subgroup of G. But the number of representative elements is the same as the number of M conjugate classes, and the number of M conjugate classes is known not to divide the number of cubes in G evenly. Hence, the set of representative elements of M-conjugate classes is not a subgroup of G. Working backwards contrapositively, the desired property cannot exist. So, the final result is that the set of M conjugate classes can be made into a group, and the set of representative elements of the M conjugate classes can be made into a group. But neither group is a subgroup of G, nor is either group isomorphic to any subgroup of G. = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 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?