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This is a followup on several recent messages concerning the

question of K-symmetric cubes, where K is one of the

98 subgroups of M.

We recall that a permutation is a special kind of function,

namely a one-to-one and onto function on a set. A very

common technique used with functions is to restrict the

domain to a (usually proper) subset of the original domain.

In the paradigm of a function as a general rule, the general rule

is applied to the subset of the domain to obtain the restriction

of the function. In the paradigm of a function as a set of ordered

pairs, the restriction is simply a (usually proper) subset of the

set of ordered pairs. A restriction of a permutation is usually

not a permutation (certainly not a permutation on the original

domain), but it is still a function.

We will treat a permutation on the cube as a set of restrictions

(functions) of the several cubicles. We take as our first example

the function UFL->UFL. Unlike cycle notation, it is not

assumed that the other cubicles are fixed; rather, the other

function values are undefined (e.g., URF->?).

Let X be any function (not necessarily a permutation) whose domain

is some subset of the cubicles. We define Symm(X) in the standard

fashion -- Symm(X) is the set of all m in M such that

m'Xm=X. If X is the function UFL->UFL, we have Symm(X)=AT4, not

Symm(X)=M as you might expect. AT4 is a subgroup in Dan's taxonomy

containing six elements, and which has an axis of symmetry along

the UFL-DBR axis.

This definition of Symm(X) perhaps requires a minor bit of

justification. In a function composition such as FG (left-to-

right notation) or G(F(x)) (right-to-left "calculus" notation),

it is sometimes taken as a convention that the range of F must

match the domain of G. But we can also take the restriction of

G to the intersection of the range of F with the domain of G,

and we do so. Having done so, Symm(X) is well defined.

We wish to build an M-symmetric permutation containing the

function UFL->UFL, but Symm(UFL->UFL)=AT4 is not a very

auspicious start. Rather, we define the conditions under which

a function is K-symmetric is follows. A function X is K-symmetric

if the union of the K-conjugates k'Xk is a function.

Given this definition, the only M-symmetric function on UFL is

in fact UFL->UFL, so we really have made a good start. Furthermore,

any M-symmetric permutation that contains UFL->UFL must also

contain all the M-conjugates of UFL->UFL, and the union of the

M-conjugates is simply the identity permutation on the corners.

The fact that Symm(UFL->UFL)=AT4 can be of some benefit in our

investigations. In particular, the fact that |AT4|=6 means that

there are 8 M-conjugates, so taking all the M-conjugates of

UFL->UFL means that all 8 corners are specified.

Our next example will be UFL->LUF (a twist of the corner). In

this case, we have Symm(X)=ET4. ET4 is the subgroup of M in

Dan's taxonomy which contains 3 elements including the identity

plus the 1/3 and 2/3 rotations around the UFL-DBR axis.

Of more import, UFL->LUF is not M-symmetric. However, it is both

C-symmetric (C is the set of 24 rotations) and H-symmetric

(H is the set of 12 even rotations and 12 odd reflections). As

an aside, we note that it is not A-symmetric, where A is the set

of 24 even rotations and reflections. But since it is both

C-symmetric and H-symmetric, there is not a unique largest

subgroup K for which we can say it is K-symmetric.

It is easy to see that UFL->LUF is not M-symmetric. The set of

M-conjugates contains both UFL->LUF and UFL->FLU, so the union

of the M-conjugates is not a function. We can see the same thing

from the fact that |ET4|=3. Since |ET4|=3, there are 16 M-conjugates,

but there are only 8 corners to represent the 16 M-conjugates.

Since UFL->LUF is C-symmetric, let's see if we can build a

C-symmetric permutation. There are 8 C-conjugates (a good

start!), and the 8 C-conjugates twist each of the 8 corners

by 1/3 in the same direction. Hence, this is a C-symmetric

but not M-symmetric permutation. Of course, it is an "illegal"

position in the sense that it is not in the same orbit

as Start.

In the same manner, we can build a K-symmetric permutation for

any K. We start with a K-symmetric function on a single

cubicle. (A function which is K-symmetric is L-symmetric for

any L which is a subgroup of K). We include all

K-conjugates. If the cube is completely specified, we stop.

Otherwise, we choose another K-symmetric function for any

previously unspecified cubicle, add in the new K-conjugates,

and so forth, repeating until the entire permutation is

specified.

Needless to say, this construction process suffers from not

preserving orbit. Additional steps must be taken to assure that

the constructed position is in the desired orbit (usually, the

Start orbit). And some orbits do not have representatives from

some subgroups, for example it is well known that there are

no C-symmetric but not M-symmetric permutations in the Start

orbit.

To use Dan's adjectives, this process very quickly can become

long, mechanical, and boring. But I now see how to build a

K-symmetric permutation for any K.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 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