From:

~~~ ~~~ Subject:

Dan Hoey writes in his e-mail message of 1994/11/08

... [conjugation by] M fixes the set of the generators of G and

their inverses. M is fact the largest subgroup of the outer

autmorphism group with this property, which makes it rather

important.In a 1983 Cubic Circular article (of which I know only Stan Isaacs's

summary) David Singmaster observed that the group is larger for larger

cubes, provided we work what I call the ``theoretical invisible

group''. That is, we solve not only the surface of the cube, but the

hypothetical interior (n-2)^3 cube, and all the smaller (n-2k)^3 cubes

as well. I blithered at length about this in my article of 1 June

1983 archived (I think I've got it right this time) at

<ftp://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-5.gz>.

Try

http://www.math.rwth-aachen.de:8000/~mschoene/Cube-Lovers/

Dan_Hoey__Eccentric_Slabism,_Qubic,_and_S&LM.html

instead (must be on a single line).

Dan continues

The idea is that a mapping called evisceration allows us to permute

the layers of the cube. On the 4x4x4 cube, this for instance allows

us to exchange each inner slab with its adjacent outer slab. It also

allows us to conjugate each inner slab move by central inversion,

while leaving the outer slab moves alone. In general, evisceration of

a d-dimensional cube by f maps each feature (cubie, colortab, or

face-center arrow) at coordinates (x[1],x[2],...,x[d]) to

(f(x[1]),f(x[2]),...,f(x[d])), where f is a permutation of the

intervals between the cleavage coordinates of the cube. I believe

that if f commutes with the central inversion, then conjugation by

evisceration is an outer automorphism of the Rubik's cube group. (I

think I have proved this for d=3, and I think the proof in higher

dimensions should not be difficult given the right notation.)The group of all eviscerations includes the central inversion; we can

of course augment it by the rotation group in d-space. Is this the

maximum outer automorphism group that respects generators of the

Rubik's cube? For this we take the generators to be turns of slabs

between adjacent cleavage planes. (Turns are direct d-1-dimensional

isometries.)

Allow me to reformulate your description again slightly.

Let P be a d-dimensional n*n*...*n cube.

Let m be (n-1)/2 if n is odd and n/2 if n is even.

The position of each cubie is described by its position vector

(p_1,p_2,...,p_d). If n is odd, then each p_i comes from [-m..m]. If n

is even, then each p_i comes from [-m..-1,1..m] (no middle slice in this

case). The orientation of a cubie is described by its orientation vector

(o_1,o_2,...,o_d), where each o_i comes from the set [-1,1].

I consider the puzzle solved, if each cubie is in its original position

and in its original orientation (this is stronger than we usually require

for the 3x3x3 Rubik's cube, where we ignore the orientation of the centre

cubies, but remember, we *see* the usually invisible faces).

If F is a bijection on [-m..m] (n odd) or [-m..-1,1..m] (n even), then F induces a permutation of the cubies (ignoring orientation) via F( (p_1,p_2,...,p_d) ) = (F(p_1),F(p_2),...,F(p_d)).

Let I_k be defined by I_k(k) = -k, I_k(-k) = k, and I_k(l) = l for l <> k.

The permutation of the cubies induced by I_k is the inversion of the

k-th slab of the cube.

If A is a permutation on [1..m], we write l^A for the image of l under A,

and we define (-l)^A := -(l^A) (you enforce the condition (-l)^A = -(l^A)

by requiring that the permutation commutes with the central inversion).

The permutation induced by A permutes the slabs of the cube.

The group of all eviscerations is generated by all the I_k and all

permutations A of [1..m]. Each evisceration first inverts certain slabs,

and then permutes the slabs. Put differently, the group of all

eviscerations is the wreath product of {-1,1} and S_m.

Repeating the definition of the wreath product, this means that the

group of all eviscerations is the semidirect product of the normal

subgroup generated by the I_k and the symmetric group generated by the A.

This group, together with the group C of symmetries of the d-dimensional

cube P, is a subgroup of the automorphism group of P, which fixes the set

of generators (with any reasonable definition of what the generators of P

are), and thus respects the distance of elements in the Cayley graph.

Is this the largest such group? In the case that d = 3, this is true.

In the case of larger d, I am quite certain it is true if the generators

are rotations of 2 dimensional subsets of P. If we choose the generators

to be symmetries of d-1 dimensional subsets of P, then I still believe it

is true. But I have been fooled often enough in such situations, to not

trust my intuition without a proof.

If we can agree on a precise definition of what the generators of the d

dimensional cube are, I would be happy to compute the largest subgroup of

the automorphism group that respects the distance of elements in the

Cayley graph.

Have a nice day.

Martin.

-- .- .-. - .. -. .-.. --- ...- . ... .- -. -. .. -.- .- Martin Sch"onert, Martin.Schoenert@Math.RWTH-Aachen.DE, +49 241 804551 Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany