The Cross Group of the Rubik's Cube

SM485c project by Mr. Dunivan (class of 1998) and Mr. Conrady (class of 1997)

The purpose of our paper is to show that the cross group of the Rubik's Cube is isomorphic to the group PSL2(F7). An analogous result, for the "Rubik icosahedron", due to Conway, is discussed in Ann Luers' thesis.

The cross group is the subgroup of the symmetric group on the set V of vertices of the cube generated by the moves XY-1, where X and Y are elements of the set of basic moves {R, L, U, D, F, B}. We will call this group C. PSL2(F7) is the projective special linear group of degree 2 over F7. F7 is the field of numbers {0,1,. . . ,6} under addition and multiplication mod 7. PSL2(F7) will map each member of the projective line of F7 to another line in P1(F7). (There are 8 lines, 7 have slopes equal to a corresponding element of F7, and the eighth has an infinite slope.) As a set, we write


Each Mobius transformation from the projective line to itself has the general formula of

fa,b,c,d(x) = (ax+b)/(cx+d),
where x is an element of P1(F7), and ad-bc = 1. For the point at infinity (inf), we define f(inf)= a/c. If we put the constants a,b,c,d into a matrix
                       [ a b ]
                       [ c d ] 
then this matrix corresponds an element of PSL2(F7). It has been shown that PSL2(F7) can be generated by the three matrices:
               f1 = [ 0 -1 ]     f2 = [ 2 1 ]     f3 =  [ 2 0 ]
                    [ 1  0 ]          [ 0 1 ]          [ 0 1 ]
(A proof of this may be found in Rotman's book.)

We will label the vertices of the cube in the following manner:

                 inf  --------------  1
                     |\     B     /|
                     | \         / |
                     |  \5_____3/  |
                     |   |     |   |
                     | L |  U  | R |
                     |   |_____|   |
                     |  / 2    4\  |
                     | /    F    \ |
                   6                 0

Under this labeling, we can show that

You should notice that if the constants in these Mobius transformations ( a, b, c, d ) are written in matrix form, they correspond to the generators of PSL2(F7). Now we will define a homomorphism q: C --> PSL2(F7), such that q(m1)=f1, q(m2)=f2, q(m3)=f3. We want to show that our q is an isomorphism.

To do this we will first show that it is surjective. Let f be a matrix in PSL2(F7), which can be written as a product of generators {f1,f2,f3} (where q(m2)=f2, q(m3)=f3). Now take f as some element of PSL2(F7). f can be broken down as a product of its generators, f1, f2, f3, we'll say

f = (fi1)e1 (fi2)e2 ... (fin)en.
Since we have a homomorphism, we can write it as a product of the images of the generators of C. Again we can rewrite it as
f = q((mi1)e1 (mi2)e2 ... (min)en).
Therefore q is surjective.

To show that q is one to one we need to know that PSL2(F7) has order 168, and that the order of the cross group is also 168. (This fact was proven by computer.) We will prove by contradiction that q is one to one. Now we assume that c1 and c2 are elements of C, such that q(c1)=q(c2), and c1 is not equal to c2. |PSL2(F7)| = |q(C)|, by surjectivity. We now subtract c2 from C , and |q(C)|=|q(C-c2)| because q(c1)=q(c2). Now we can say that |q(C-c2)|< or = |C-c2| because we know that q is a well-defined function between finite sets. Since we have taken c2 out of C, we know |C-c2| < |C|, which by transitivity implies |PSL2(F7)| < |C|. This is a contradiction because we know |PSL2(F7)| = |C|. Therefore q is injective.

Now that we have shown that q is both surjective and injective, it is bijective and an isomorphism. This tells us that each move in the cross group corresponds to a matrix in PSL2(F7). Those matrices define a Mobius transform that will permute one line to another as their related corners are permuted to another.

The labeling used above is the labeling that accomplishes this goal. Thus we have proven the following:

THEOREM: C is isomorphic to PSL2(F7). In fact, there is a labeling of the vertices of the cube by the elements of P1(F7) such that the isomorphism

q: C --> PSL2(F7)
q((UD-1)2 ) = -1/x, q(UR-1) = 2x+1, and q(BU-1LB-1) = 2x

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