Analyzing Rubik's Cube with GAP

This is an updated GAP 4 version of a GAP 3example by Martin Schönert, 1993.An almost classical permutation group of small degree is examined withsome elementary GAP commands.

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The output given here has been produced by GAP 4,the input is available in form of a plain GAP 4 input file.

Ideal Toy Company stated on the package of
the original Rubik cube that there were more than
three billion possible states the cube could attain.
It's analogous to Mac Donald's proudly announcing
that they've sold more than 120 hamburgers.
(J. A. Paulos, Innumeracy)

We consider the group of transformations of Rubik's magic cube. If we numberthe faces of this cube as follows

then the group is generated by the following generators, corresponding tothe six faces of the cube.

First we want to know the size of this group.

Since this is a little bit unhandy, let us factorize this number.

(The result tells us that the size is 2^27 3^14 5^3 7^2 11.)
Next let us investigate the operation of the group on the 48 points(we reduce the line length to get a more appropriate output format).

The first orbit contains the points at the corners, the second those at theedges; clearly the group cannot move a point at a corner onto a point at anedge.
So to investigate the cube group we first investigate the operation onthe corner points. Note that the constructed group that describes thisoperation will operate on the set [1..24], not on the original set[1,3,17,14,8,38,9,41,19,48,22,6,30,33,43,11,46,40,24,27,25,35,16,32].

Now this group obviously operates transitively, but let us test whetherit is also primitive.

Those eight blocks correspond to the eight corners of the cube; on theone hand the group permutes those and on the other hand it permutes thethree points at each corner cyclically.
So the obvious thing to do is to investigate the operation of the groupon the eight corners. The action gives a homomorphism to a permutation groupon the corners:

Now a permutation group of degree 8 that has order 40320 must be the fullsymmetric group S(8) on eight points.
The next thing then is to investigate the kernel of this operation on blocks,i.e., the subgroup of cube1 of those elements that fix theblocks setwise.

We can show that the product of this elementary abelian group 3^7 with theS(8) is semidirect by finding a complement, i.e., a subgroup that has trivialintersection with the kernel and that generates cube1 togetherwith the kernel.

We verify the complement properties:

There is even a more elegant way to show that cmpl1 is acomplement.

Of course, theoretically it is clear that cmpl1 must indeed be acomplement.
In fact we know that cube1 is a subgroup of index 3 in thewreath product of a cyclic 3 with S(8). This missing index 3 tells us thatwe do not have total freedom in turning the corners. The following testsshow that whenever we turn one corner clockwise we must turn another cornercounterclockwise.

More or less the same things happen when we consider the operation of thecube group on the edges.

Inmr 6 2 2 Rubix Games

So there are even 4 classes of complements here.This time we get a semidirect product of a 2^11 with an S(12), namely asubgroup of index 2 of the wreath product of a cyclic 2 with S(12). Herethe missing index 2 tells us again that we do not have total freedom inturning the edges. The following tests show that whenever we flip oneedge we must also flip another edge.

Since cube1 and cube2 are the groups describingthe actions on the two orbits of cube, it is clear thatcube is a subdirect product of those groups, i.e., a subgroupof the direct product. Comparing the sizes of cube1,cube2, and cube we see that cubemust be a subgroup of index 2 in the direct product of those two groups.

This final missing index 2 tells us that we cannot operate on corners andedges totally independently. The following tests show that whenever weexchange a pair of corners we must also exchange a pair of edges (andvice versa).

As a last part of the structure analysis of the cube group let us computethe centre of the cube group, i.e., the subgroup of those operations thatcan be performed either before or after any other operation with the sameresult.

We see that the centre contains one nontrivial element, namely theoperation that flips all 12 edges simultaneously.
Finally we turn to the original idea connected with the cube, namely tofind a sequence of turns of the faces that will transform the cube backinto its original state. This amounts to a decomposition of a givenelement of the cube group into a product of the generators. For thispurpose we introduce a free group and a homomorphism of it onto the cubegroup.

Using this homomorphism, we can now decompose elements into generators. Themethod used utilizes a stabilizer chain and does not enumerate all groupelements, therefore the words obtained are not the shortest possible,though they are short enough for hand solutions.
First we decompose the centre element:

Next we decompose some element arbitrarily chosen by us:

Inmr 6 2 2 Rubix Game

Last we let GAP choose a random element ...

Inmr 6 2 2 Rubix 2x2

... and we verify that the decomposition is correct:

Inmr 6 2 2 Rubix Cube

This concludes our example. Of course, GAP can do much more, but demonstrating them all would take too much room.