   Date: Sun, 13 Feb 94 16:59:22 -0500 (EST)   From: Jerry Bryan <BRYAN%WVNVM.BITNET@mitvma.mit.edu >
~~~  Subject: Re: Some Proposed Terminology

On 01/21/94 at 18:32:15 hoey@AIC.NRL.Navy.Mil said:
>I welcome Jerry Bryan's <BRYAN%WVNVM.BITNET@mitvma.mit.edu> efforts to
>improve the terminology of the groups associated with Rubik's cube.
>But there is some additional clarification I think is necessary.

Let G\C be the corners with edges without centers group. I intend
for the notation to indicate G reduced by C, where C is the rotation
group for the cube....

Let GC\C be the corners without edges without centers group....
Let GE\C be the edges without corners without centers group....

First, these are not, strictly speaking, groups. Well, you can make
them groups, by defining what the group operation is. But I don't
know any way of doing that without losing the symmetrical nature of
the problem.

Second, I would suggest that G/C, GC/C, and GE/C are more standard
names for these objects. The elements are nominally 24-element sets,
each of which is an equivalence class when two positions are
considered equivalent when they differ by their position with respect
to the corners. The classes are called the cosets of C in G, GC, and
GE, respectively.

Dan Hoey's criticism's are quite valid. I will attempt to repair
the damage as follows: 1) accept the Gx/C notation in lieu
of Gx\C, 2) define an operation within Gx/C such that
Gx/C is a group, and 3) use Gx/C as a model for cubes without
centers in such a way that the symmetrical nature of the problem
is retained.

Let C be the set of twenty-four whole cube rotations of the cube, and
let G be the standard 3x3x3 cube group. We observe that if X is a cube
in G, then c'Xc is also a cube in G for every c in C. We could call
this operation C-conjugancy. However, there is seldom (if ever) any
reason to speak of C-conjugancy. That is, C is just a subset of M,
the set of forty-eight whole cube rotations and reflections. Indeed,
C is half of M, and the other half of M is the reflection of C. Hence,
M-conjugancy of the form m'Xm is more powerful than C-conjugancy, and
there is normally no reason to speak of C-conjugancy. I only bring it
up to emphasize that if X is in G, then c'Xc is in G.

On the other hand, if I understand correctly the model most people use
for G, elements of the form Xc or cX are not in G except for the
trivial case where c=I. The problem is that C is considered to move
the centers, but G is generated by Q, the set of quarter-turns of the
faces, and Q does not move the centers. For example, there is a
c in C such that F=c'Rc, but there is not a c in C such that
F=Rc or F=cR. And indeed, neither Rc nor cR are in G at all unless
c=I.

As we said, G is generated as G=<Q>, where Q is the set of
quarter-turns Q={F,B,U,D,L,R,F',B',U',D',L',R'}. Elements of Q
move the corners and edges, but Q is the identity on the centers.
C, on the other hand, is generally considered to move the centers.
Hence, the group generated as <Q,C> is a supergroup of G, and there
are elements of the supergroup which are not in G. (This
supergroup, by the way, is not The Supergroup. The Supergroup is
generated by Q alone, but with orientations of the (otherwise fixed)
centers considered.) Therefore, our first order of business is to
make C into a sub-group of G.

We observe that since the elements of Q are the identity on the
centers, the primary function of the centers is to provide a
frame of reference. But we can provide a frame of reference
without the centers actually being there.

For example, consider the group GC consisting of cube centers
and corners. You can model this group by removing the edge
labels from a physical cube. Establish the cube at Start and
perform RL'. The corners will be rotated forward, and will be
positioned properly with respect to each other, but the cube is
clearly not solved. You can tell that the cube is not at Start
because the corners are not aligned properly with the centers.

Now, do the same thing except remove both the edge and center
labels. If you perform RL' at Start, the cube "looks"
solved but rotated forward. However, we can adopt the convention
that the cube is solved only if the Up color is Up, the Front
color is Front, etc. With this convention in place, RL' is
clearly seen not to be solved; it is two moves from Start.
The convention provides the fixed frame of reference.
Furthermore, RL' (which is in GC) is equal to an element of C, and
indeed all elements of C are in GC, as are all elements of the
form Xc or cX for c in C and X in GC. Hence, we have <Q>=<Q,C>.

Similar comments apply to GE, the group of edges and centers, except
that processes composed from elements of Q to accomplish rotations
in C are not quite so short in GE as they are in GC.

G, the full 3x3x3 cube group consisting of corners, edges, and centers
is a bit more difficult. The problem is that if X is in G, then
objects of the form Xc or cX are in G only if c is even. Twelve
elements of C are even and twelve are odd. Indeed, C[even] is
a sub-group of C, but C[odd] is not.

We will deal with this situation (as circumstances require) in two
different ways. One is simply to restrict ourselves to C[even]
when dealing with G. The other is to define a new group we will
call GS. In our model for G in which the centers are implied by
a frame of reference convention rather than by actual physical
centers, we can easily add slice moves to the standard face moves.
If the centers were physically present, then the slice moves would
move the centers, but without the physical centers there is
no problem.

If S is the set of slice moves, then GS is generated as <Q,S>.
GS is essentially G with parity restrictions removed. Hence we
observe that |G|=|GC|*|GE|/2, |GS|=|GC|*|GE|, and |GS|=|G|*2.
Also, if X is in G or in GS, then elements of the form cX or
Xc are in GS for all c in C. In those occasions where we are
willing to think of GS rather than G, we can use C rather than
C[even].

At this point, we can say that GS/C, G/C[even], GC/C, and GE/C are
cosets of C in GS, C[even] in G, C in GC, and C in GE, respectively.
To be a little more conformant with standard coset notation, we will
write cube elements as lower case letters for the remainder of this
note, and hence for a particular cube x a coset of C is denoted as
Cx={y: y=cx} or xC={y: y=xc}.

Now, we propose a group operator for the cosets: Cx Cy = C(xy) and
xC yC = (xy)C. Showing that we have a group is easy. I originally
included a proof in this note, but there is a proof in Chapter 8 of
Frey and Singmaster's _Handbook of Cubik Math_. Hence, I will defer
to their proof instead.

According to Frey and Singmaster, G/C is called the factor group of
C in G, or the quotient group of G by C. Of most significance to
us right now is the fact that the identity of the factor group is
Ci or iC, where i is the identity of G. But Ci or iC is just C.
Hence, the identity of the factor group is C. This justifies our
identification of G/C with a centerless cube. In English, it means
that we can rotate a centerless cube in space without changing
anything. I think this would comply with most people's intuitive
sense of what it means for a cube to be centerless.

Finally, as to whether this model retains the "symmetrical nature of
the problem", I will have to leave that as an open question,
depending on precisely what we mean by "symmetrical". It seems to
me that this model does a better job of being "symmetrical" than
a model which includes only seven corner cubies or only eleven
edge cubies, but maybe not. What does "symmetrical" mean when
it comes to centerless cubes?

``` = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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?     