Proposition 1
Not all elements of S8 * ℤ38 * S12 * ℤ212 are attainable configurations of the Rubik’s cube. In fact, to be an element of the group R, X = (α, a, β, b) must satisfy the following three conditions:
1.1 The summation of the terms in the orientation vector for the vertices must be 0 mod 3.
1.1 The summation of the terms in the orientation vector for the vertices must be 0 mod 3.
1.2 The summation of the terms in the orientation vector for the edges must be 0 mod 2.
1.3 The permutations for the edges and vertices must either both be even or both be odd. (See definition below)
Let α = (i1, i2,… im), β = (j1, j2,… jn). Then
sgn(α) = sgn(β) where sgn(α) = {+1 if α is even; -1 if α is odd}
Note: To determine whether a permutation α is even or odd, it first must be broken down into 2-cycles, or transpositions. A cycle, α = (i1, i2, … im), can be expressed as a product of transpositions in the following way
(i1, i2, … im) = (i1, im)(i1, i(m-1)),…(i1, i2)
Permutations can be written as a product of cycles, therefore all permutations can be written as a product of transpositions. A permutation that is the product of an even number of transpositions is called an even permutation and a permutation that is the product of an odd number of transpositions is called an odd permutation. An m-cycle
α = (i1, i2, … im) is the product of (m-1) transpositions so the sgn(α) also can be found using (-1)(m-1). Even though the product of transpositions for a distinct permutation is not unique, the number of transpositions used is either always even or always odd.
The proof of proposition 1 requires two steps. First, it must be shown that the generators of the group satisfy these three properties. Next, that these properties are preserved by the group operation, or that the group is closed. In other words, given any two elements of the group, X = (α, a, β, b) and Y = (δ, d, ε, e) that satisfy these conditions, the combination of these moves X⋅Y = (αδ, a+αd, βε, b+βe) will also satisfy these conditions and be an element of the group.
Let α = (i1, i2,… im), β = (j1, j2,… jn). Then
Note: To determine whether a permutation α is even or odd, it first must be broken down into 2-cycles, or transpositions. A cycle, α = (i1, i2, … im), can be expressed as a product of transpositions in the following way
Permutations can be written as a product of cycles, therefore all permutations can be written as a product of transpositions. A permutation that is the product of an even number of transpositions is called an even permutation and a permutation that is the product of an odd number of transpositions is called an odd permutation. An m-cycle
α = (i1, i2, … im) is the product of (m-1) transpositions so the sgn(α) also can be found using (-1)(m-1). Even though the product of transpositions for a distinct permutation is not unique, the number of transpositions used is either always even or always odd.
The proof of proposition 1 requires two steps. First, it must be shown that the generators of the group satisfy these three properties. Next, that these properties are preserved by the group operation, or that the group is closed. In other words, given any two elements of the group, X = (α, a, β, b) and Y = (δ, d, ε, e) that satisfy these conditions, the combination of these moves X⋅Y = (αδ, a+αd, βε, b+βe) will also satisfy these conditions and be an element of the group.
Proof
The generators satisfy these conditions:
The properties are preserved under the group operation:
1.1 The summation of the terms in the orientation vector a + αd must be 0 mod 3.
Since summation rules allow separation of addition, the above equation is equivalent to
Rearranging the elements of d does not change their sum so the alpha inverse may be dropped.
Each of which is 0 mod 3 by hypothesis. The result is that the original summation is congruent to 0 mod 3.
1.2 The summation of the terms in the orientation vector b + βe must be 0 mod 2.
Since summation rules allow separation of addition, the above equation is equivalent to
Rearranging the elements of e does not change their sum so the beta inverse may be dropped.
Each of which is 0 mod 2 by hypothesis. The result is that the original summation is congruent to 0 mod 2.
1.3 The permutations for the edges, αδ, and the vertices, βε, have the same sign.
sgn(αδ) = sgn(βε)
If α is the product of m transpositions and δ is the product of n transpositions, then αδ is a product of (m+n) transpositions.
sgn(αδ) = (-1)(m+n)=(-1)m (-1)n = sgn(α)sgn(δ)
Similarly, if β is the product of o transpositions and ε is the product of p transpositions, then βε is a product of (o+p) transpositions.
sgn(βε) = (-1)(o+p)=(-1)o (-1)p = sgn(β)sgn(ε)
Therefore, the original equation sgn(αδ) = sgn(βε) is equivalent to
sgn(α)sgn(δ) = sgn(β)sgn(ε)
If α is the product of m transpositions and δ is the product of n transpositions, then αδ is a product of (m+n) transpositions.
Similarly, if β is the product of o transpositions and ε is the product of p transpositions, then βε is a product of (o+p) transpositions.
Therefore, the original equation sgn(αδ) = sgn(βε) is equivalent to