The purpose of our paper is to show that the cross group of the Rubik's Cube is isomorphic to the group PSL_{2}(F_{7}). 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. PSL_{2}(F_{7}) is the projective special linear group of degree 2 over F_{7}. F_{7} is the field of numbers {0,1,. . . ,6} under addition and multiplication mod 7. PSL_{2}(F_{7}) will map each member of the projective line of F_{7} to another line in P^{1}(F_{7}). (There are 8 lines, 7 have slopes equal to a corresponding element of F_{7}, 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

[ a b ] [ c d ]then this matrix corresponds an element of PSL

f(A proof of this may be found in Rotman's book.)_{1}= [ 0 -1 ] f_{2}= [ 2 1 ] f_{3}= [ 2 0 ] [ 1 0 ] [ 0 1 ] [ 0 1 ]

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

inf -------------- 1 |\ B /| | \ / | | \5_____3/ | | | | | | L | U | R | | |_____| | | / 2 4\ | | / F \ | |/___________\| 6 0Under this labeling, we can show that

- the image of the move m
_{1}= (UD^{-1})^{2}will permute the vertices in this way: ((inf) 0)(1 6)(2 3)(4 5). The same permutation is associated to the Mobius transform f_{0,-1,1,0}(x) =(0x - 1)/(x + 0) acting on P^{1}(F_{7}). - m
_{2}= UR^{-1}gives us the permutation: (0 1 3)(2 5 4), which is given by f_{2,1,0,1}(x) = (2x + 1)/(0x + 1) acting of P^{1}(F_{7}). - m
_{3}= (BU^{-1}LB^{-1}) gives us the permutation (1 2 4)(3 6 5), which is given by f_{2,0,0,1}(x) = (2x + 0)/(0 + 1) acting on P^{1}(F_{7}).

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 PSL_{2}(F_{7}). Now we will define a homomorphism q: C --> PSL_{2}(F_{7}), such that q(m_{1})=f_{1}, q(m_{2})=f_{2}, q(m_{3})=f_{3}. 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 PSL_{2}(F_{7}), which can be written as a product of generators {f1,f2,f3} (where q(m_{2})=f_{2}, q(m_{3})=f_{3}). Now take f as some element of PSL_{2}(F_{7}). f can be broken down as a product of its generators, f_{1}, f_{2}, f_{3}, we'll say

To show that q is one to one we need to know that PSL_{2}(F_{7}) 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 c_{1} and c_{2} are elements of C, such that q(c_{1})=q(c_{2}), and c_{1} is not equal to c_{2}. |PSL_{2}(F_{7})| = |q(C)|, by surjectivity. We now subtract c_{2} from C , and |q(C)|=|q(C-c_{2})| because q(c_{1})=q(c_{2}). Now we can say that |q(C-c_{2})|< or = |C-c_{2}| because we know that q is a well-defined function between finite sets. Since we have taken c_{2} out of C, we know |C-c_{2}| < |C|, which by transitivity implies |PSL_{2}(F_{7})| < |C|. This is a contradiction because we know |PSL_{2}(F_{7})| = |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 PSL_{2}(F_{7}). 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 PSL_{2}(F_{7}). In fact, there is a labeling of the vertices of the cube by the elements of P^{1}(F_{7}) such that the isomorphism

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