   Date: Sat, 10 Dec 94 10:13:33 -0500   From: Jerry Bryan <BRYAN@wvnvm.wvnet.edu >
~~~  Subject: Cubic Symmetry of the 2x2x2 (Again)

The argument has been made that the 2x2x2 cube (or really any 2Nx2Nx2N)
cube cannot have the "symmetry of the cube". In order for a real
2x2x2 cube to have the "symmetry of the cube", you would have to
adopt unreasonable rules, such as no rotations (or if you use
rotations they have a cost of at least 2) and the cube is only solved
when the colors are oriented properly. But a 2x2x2 cube certainly
feels like a real cube when you hold it in your hands. I offer the
following interpretation that "sort of" gives the 2x2x2 cube the
symmetry of the cube. Since I will only be talking about the
2x2x2, I will simplify the notation by talking about C, G, Q, etc.
rather than C[C], G[C], Q[C], etc.

As I have discussed several times before, my favorite model for the
2x2x2 is G/C (or CG/C, if you prefer; G=CG for the 2x2x2).
The group operation is (xC)(yC)=(uv)C, where u and v are the elements of
xC and yC, respectively, which fix a particular corner. (xC)(yC)=(xy)C
doesn't work because C is not normal. There is an obvious isomorphism
between G/C and <q_i, q_j, q_k>, where the three q-turns are those which
fix the same corner as the selection function for u and v.

There are eight corners, and hence there are eight conjugate selection
functions and eight conjugate subgroups G_m of the form <q_i, q_j, q_k>
which fix a particular corner. If you think of mapping G/C
simultaneously and in parallel to the all the elements in the set
{G_1, G_2, G_3, G_4, G_5, G_6, G_7, G_8}, then in a loose sense you
have preserved the cubic nature of the problem. That is, none of
the individual G_m's have the same symmetry as the cube, but in a loose
sense the entire collection {G_m} does.

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Robert G. Bryan (Jerry Bryan)                        (304) 293-5192
Associate Director, WVNET                            (304) 293-5540 fax     