   Date: Wed, 21 Sep 94 11:31:42 -0400 (EDT)   From: Jerry Bryan <BRYAN@wvnvm.wvnet.edu >
~~~ ~~~ 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     