The Skewb Group

SM485c project by J. Montague and G. Gomes

(both from the class of 1999)

Before discussing the group structure of the skewb, an orientation must be constructed. First, orient the cube and label the sides by R,L,U,D,F,B. A 120 degree clockwise rotation of a corner is denoted by the letters of the three faces which meet at that corner. For example, UFR is the 120 degree clockwise rotation of the up-front-right corner. Such moves are the basic moves of the skewb and are called "generators" since the eight moves of this type generate the skewb group. Any generator of the skewb will permute three corners while leaving the opposite side of the cube alone. It should be noted that twisting about a corner and twisting about the antipodal opposite corner are equivalent moves up to a rotation of the entire cube.

Let

G=< UFR,UFL,BRU,BLU,DFR,DFL,DBR,DBL > denote the group of all legal skewb moves. For the purposes of the following proof, it is helpful to have a standard reference marking established for the skewb. First, label the center facets with numbers 1 through 4 starting with the front face and moving counterclockwise to the right, then back, then left face. The up face is labeled 5 and the bottom 6. Now label the up faces of the top four corners 1 through 4, starting with the UFR corner and moving counterclockwise. Similarly, label the down faces of the bottom four corners 5 through 8 starting with the DBR corner and moving counterclockwise.

Let C denote the set of square center facets of the skewb and S_C denote the symmetric group on C. Let V denote the set of vertices of the skewb and S_V denote the symmetric group on V. Finally, let y(g) be an element of C₃⁸ denoting the twist on all eight corners.

Claim: There are homomorphisms

p: G -> S_C, q: G -> S_V, for each move g (an element of G), such that p(g)=permutation of the center facets associated with g, and q(g)=permutation of the vertices associated with g. Let H= S_C x S_V x C₃⁸ and define *: HxH -> H by (r,s,y)*(r',s',y')= (rr',ss',y+r^-1(y'))

Claim: G is a subgroup of H.

Let G_C be the group that acts only on the center facets of the skewb, and G_V the group that acts only on the vertices. Now,

G= G_C x G_V. Every generator of G is a 3-cycle on the center facets. This means that r is an element of S_C (even). It is a fact that the elements (i,j,k), of S_n, generate A_n (for i,j,k elements of {1,2,...,n}). Therefore, G_C = A₆.

The group that acts on the vertices of the skewb is slightly more complicated. Unlike the Rubik's cube, there is no condition like conservation of twists which applies to the entire vertex set. Instead, we must split the vertices of the skewb into two 4-corner orbits. This idea is borrowed from Bandelow's notes on Mickey's Challenge (a puzzle similar to the Masterball). An orbit is constructed by starting with one corner and including the opposite corner of each face that meets at the first corner. Referring back to our original labeling of the skewb, the orbits are the odd corners {1,3,5,7}, and the even corners {2,4,6,8}. Let the orbit of odd corners be denoted by V(odd), and let V(even) denote the orbit of even corners. We now partition G_V so that

G_V = G_V(odd) x G_V(even). We know that each orbit maps to a permutation on 4 vertices and an orientation on 4 vertices. So G_V is a subgroup of the semi-direct product (S₄ x C₃⁸ )x (S₄ x C₃⁸ ).

Let h=(s,y(h),t,z(h)) be an element of G_V.

First we will examine the permutations of the vertices of both orbits. Each generator produces a 3-cycle on the vertices, whether they are in the odd or even orbit. Therefore, we can say that the permutations of each orbit generate A₄ by the same argument used for the center permutations. Now,

G_V is a subgroup of the semi-direct product (A₄ x C₃⁴)x (A₄ x C₃⁴).

Claim 1: There exist h, such that s is an element of A₄₄, y(h)= (0,0,0,0), t is an element of A₄, and z(h)= (0,0,0,0). We know this is true because there are clean skewb moves which only permute 3 vertices.

Claim 2: There exist h, such that s=1, y(h)= any even permutation of (1,2,0,0), t=1, z(h)= any even permutation of (1,2,0,0). This is true because there are clean skewb moves which only twist vertices.

If we combine the moves of Claims 1 and 2, we should generate all of the possible moves of G_V. The condition on each of the vertex 4-tuples will drop them in dimension to elements of C₃³. So we can conclude that G_V is a subgroup of index 9 of the semi-direct product

(A₄ x C₃⁴)x (A₄ x C₃⁴). Note: This claim is verified by GAP.

Since G_C= A₆, and G_V= (A₄ x C₃³)x (A₄ x C₃³), we can conclude that

G = A₆ x (A₄ x C₃³)x (A₄ x C₃³) and |G|= (6!/2)*(4!/2)^2*(3^6) = 37,791,360.

In conclusion, it is interesting to note that if we let G' denote the illegal skewb group - where reassembly is permitted - then

G'= S₆ x S₈ x C₃⁸ and |G|/|G'| = .0001984127... This means that if you could take apart the skewb and reassemble it however you wanted, only .02% of all possible reassemblies would be solvable.

Permutation and Orientation Tables

Move      Center Permutation       Vertex Permutation
UFR            (1 5 2)                  (2 6 4)
UFL            (1 4 5)                  (1 7 3)
DFR            (1 2 6)                  (1 5 7)
DFL            (1 6 4)                  (4 6 8)
BRU            (2 5 3)                  (1 3 5)
BLU            (3 5 4)                  (2 4 8)
DBR            (2 3 6)                  (2 8 6)
DBL            (3 4 6)                  (3 7 5)

Move           Vertex Orientation
UFR            (1 2 0 2 0 2 0 0)
UFL            (2 0 2 1 0 0 2 0)
DFR            (2 0 0 0 2 1 2 0)
DFL            (0 0 0 2 0 2 1 2)
BRU            (2 1 2 0 2 0 0 0)
BLU            (0 2 1 2 0 0 0 2)
DBR            (0 2 0 0 1 2 0 2)
DBL            (0 0 2 0 2 0 2 1)
UFR*UFL        (0 2 2 2 0 0 2 0)
DFR*DFL        (2 0 0 2 0 0 2 2)

Note: The orientations for the generator moves contain two repeated orbits - permutations of (1 0 0 0) and permutations of (2 2 2 0). Also, the combination moves are equivalent to 0 mod 3.

typed into html by wdj on 4-28-97. Last updated 6-7-2004.