Proposition 7
The groups of the Rubik’s cube and the octahedron are related to the group of the cuboctahedron.
7.1 The generators of the Rubik’s cube group can be mapped to elements in the cuboctahedron group.
7.2 The generators of the octahedron group can be mapped to elements of the cuboctahedron group.
7.1 The generators of the Rubik’s cube group can be mapped to elements in the cuboctahedron group.
7.2 The generators of the octahedron group can be mapped to elements of the cuboctahedron group.
Proof
7.1 To show that the generators of the Rubik’s cube can be mapped to elements in the cuboctahedron group, it is useful to understand the geometrical relationship between the cube and the cubocahedron. From embedding a picture of the cuboctahedron into a picture of the cube, we conclude that the edges of the cube correspond to vertices of the cuboctahedron and the vertices of the cube correspond to the three edges of the surrounding triangular face on the cuboctahedron.
This means that moving one edge on the cube would correspond to moving that vertex on the cuboctahedron and moving one vertex on the cube would correspond to moving the three corresponding edges on the cuboctahedron.
For example, rotating the front face of the cube 90 degrees clockwise would correspond to the following edge and vertex permutations on the cuboctahedron in the graph above. (Note that the original labeling of the cuboctahedron has been changed for this example)
Cube:
Vertices α = (1 2 3 4)
Edges β = (1 2 3 4)
Cuboctahedron:
Edges β = (1 4 7 10) (2 5 8 11) (3 6 9 12)
Vertices α = (1 2 3 4)
The move that results in the edge permutation (1 2 3 4) and vertex permutation (1 2 3 4) on the cube is simply F. To show that this generator can be mapped to an element in the cuboctahderon group, the sequence of moves that will result in the edge permutation (1 4 7 10) (2 5 8 11) (3 6 9 12) and vertex permutation (1 2 3 4) must be found. To do this, 3-cycles will be used since in Proposition 6, it was shown that any 3-cycle on edges or vertices on the cubocathedron can be found. First we will look to solve for the edge permutation.
β = (1 4 7 10) (2 5 8 11) (3 6 9 12)
Since this permutation is odd, an extra move must be added on so it can be rewritten in terms of transpositions. The move F will be added, and then removed later.
= (1 4 7 10) (1 4 7 10) (2 5 8 11) (3 6 9 12)
= (1 10) (1 7) (1 4) (1 10) (1 7) (1 4) (2 11) (2 8) (2 5) (3 12) (3 9) (3 6)
The transpositions are then grouped and rewritten as 3-cycles.
= (1 7 10) (1 10 4) (1 4 7) (2 8 11) (2 3 5) (2 3 12) (3 6 9)
Then, each of these 3-cycles can be written as a sequence of moves. Since this is a long process, only one 3-cycle will be shown, (3 6 9). In proposition 6.1, it was shown that the 3-cycle (3 2 13) (which was (19 18 5) based on how the diagram was labeled in that example) on edges can be achieved using C2 where C = [U, UR-1] = UR∙U-1∙UR-1∙U. It was also shown that any 3-cycle on edges of the form (3 z y) could be attained by conjugation. We are looking to attain (3 z y) where y = 9 and z = 6. To do this, first, it must be shown that we can get (3 2 6) and (3 2 9) by moving edges 6 and 9 into position 13 without affecting the other pieces. The following are the sequence of moves used to do just that.
Conjugating C2 by T where T = UB-1∙B∙DB-1∙D gives the result, T-1∙C2∙T = (3 2 6)
Conjugating C2 by S where S = UB-1∙B2∙DL-1∙L gives the result, S-1∙C2∙S = (3 2 9)
Finally, conjugating these 3-cycles in the following way results in (3 6 9).
(S-1∙C2∙S)-1∙T-1∙C2∙T ∙S-1∙C2∙S = (9 2 3) (3 2 6) (3 2 9) = (3 6 9)
The corresponding sequence of moves is below.
L-1∙DL∙B-2∙UB ∙(UR∙U-1∙UR-1∙U)-2∙UB-1∙B2∙DL-1∙L∙D-1∙DB∙B-1∙UB∙ (UR∙U-1∙UR-1∙U)2∙UB-1∙B∙DB-1∙D∙L-1∙DL∙B-2∙UB∙(UR∙U-1∙UR-1∙U)2∙ UB-1∙B2∙DL-1∙L
This process to find the corresponding sequence of moves can be done for each of the 3-cycles in (1 7 10) (1 10 4) (1 4 7) (2 8 11) (2 3 5) (2 3 12) (3 6 9). The resulting sequence of moves combined will be what the move F on the cube can be mapped to in the cuboctahedron group. (Recall an additional move F was added to the initial permutation. When the entire sequence of moves is found, F-1 must be added at the front to negate this.) This can be done for all of the other generators of the group R, (F,R,L,U,D,B) which means that there does exist a mapping from the Rubik’s cube group to the cuboctahedron group.
Cube:
Vertices α = (1 2 3 4)
Edges β = (1 2 3 4)
Cuboctahedron:
Edges β = (1 4 7 10) (2 5 8 11) (3 6 9 12)
Vertices α = (1 2 3 4)
The move that results in the edge permutation (1 2 3 4) and vertex permutation (1 2 3 4) on the cube is simply F. To show that this generator can be mapped to an element in the cuboctahderon group, the sequence of moves that will result in the edge permutation (1 4 7 10) (2 5 8 11) (3 6 9 12) and vertex permutation (1 2 3 4) must be found. To do this, 3-cycles will be used since in Proposition 6, it was shown that any 3-cycle on edges or vertices on the cubocathedron can be found. First we will look to solve for the edge permutation.
β = (1 4 7 10) (2 5 8 11) (3 6 9 12)
Since this permutation is odd, an extra move must be added on so it can be rewritten in terms of transpositions. The move F will be added, and then removed later.
= (1 4 7 10) (1 4 7 10) (2 5 8 11) (3 6 9 12)
= (1 10) (1 7) (1 4) (1 10) (1 7) (1 4) (2 11) (2 8) (2 5) (3 12) (3 9) (3 6)
The transpositions are then grouped and rewritten as 3-cycles.
= (1 7 10) (1 10 4) (1 4 7) (2 8 11) (2 3 5) (2 3 12) (3 6 9)
Then, each of these 3-cycles can be written as a sequence of moves. Since this is a long process, only one 3-cycle will be shown, (3 6 9). In proposition 6.1, it was shown that the 3-cycle (3 2 13) (which was (19 18 5) based on how the diagram was labeled in that example) on edges can be achieved using C2 where C = [U, UR-1] = UR∙U-1∙UR-1∙U. It was also shown that any 3-cycle on edges of the form (3 z y) could be attained by conjugation. We are looking to attain (3 z y) where y = 9 and z = 6. To do this, first, it must be shown that we can get (3 2 6) and (3 2 9) by moving edges 6 and 9 into position 13 without affecting the other pieces. The following are the sequence of moves used to do just that.
Conjugating C2 by T where T = UB-1∙B∙DB-1∙D gives the result, T-1∙C2∙T = (3 2 6)
Conjugating C2 by S where S = UB-1∙B2∙DL-1∙L gives the result, S-1∙C2∙S = (3 2 9)
Finally, conjugating these 3-cycles in the following way results in (3 6 9).
(S-1∙C2∙S)-1∙T-1∙C2∙T ∙S-1∙C2∙S = (9 2 3) (3 2 6) (3 2 9) = (3 6 9)
The corresponding sequence of moves is below.
L-1∙DL∙B-2∙UB ∙(UR∙U-1∙UR-1∙U)-2∙UB-1∙B2∙DL-1∙L∙D-1∙DB∙B-1∙UB∙ (UR∙U-1∙UR-1∙U)2∙UB-1∙B∙DB-1∙D∙L-1∙DL∙B-2∙UB∙(UR∙U-1∙UR-1∙U)2∙ UB-1∙B2∙DL-1∙L
This process to find the corresponding sequence of moves can be done for each of the 3-cycles in (1 7 10) (1 10 4) (1 4 7) (2 8 11) (2 3 5) (2 3 12) (3 6 9). The resulting sequence of moves combined will be what the move F on the cube can be mapped to in the cuboctahedron group. (Recall an additional move F was added to the initial permutation. When the entire sequence of moves is found, F-1 must be added at the front to negate this.) This can be done for all of the other generators of the group R, (F,R,L,U,D,B) which means that there does exist a mapping from the Rubik’s cube group to the cuboctahedron group.
7.2 To show that the generators of the octahedron group can be mapped to elements in the cuboctahedron group, it is useful to understand the geometrical relationship between the octahedron and the cubocahedron. From embedding a picture of the cuboctahedron into a picture of the octahedron, we conclude that the edges of the octahedron correspond to vertices of the cuboctahedron and the vertices of the octahedron correspond to the four edges of the surrounding square face on the cuboctahedron.
This means that moving one edge on the octahedron would correspond to moving that vertex on the cuboctahedron and moving one vertex on the octahedron would correspond to moving the four corresponding edges on the cuboctahedron.
For example, rotating the upper front face of the octahderon 120 degrees clockwise would correspond to the following edge and vertex permutations on the cuboctahedron in the graph above. (Note that the original labeling of the cuboctahedron has been changed for this example)
Octahedron:
Vertices α = (1 3 4)
Edges β = (1 2 3)
Cuboctahedron:
Edges β = (1 5 9) (2 6 10) (3 7 11) (4 8 12)
Vertices α = (1 4 8)
The move that results in the edge permutation (1 2 3) and vertex permutation (1 3 4) on the octahedron is simply UF. To show that this generator can be mapped to the cuboctahderon group, the sequence of moves that will result in the edge permutation (1 5 9) (2 6 10) (3 7 11) (4 8 12) and vertex permutation (1 4 8) must be found. To do this, 3-cycles will be used since in Proposition 6, it was shown that any 3-cycle on edges or vertices on the cubocathedron can be found.
β = (1 5 9) (2 6 10) (3 7 11) (4 8 12)
Each of these 3-cycles can be written as a sequence of moves. Since this is a long process, only one 3-cycle will be shown, (4 8 12). In proposition 6.1, it was shown that the 3-cycle (12 11 21) which was (19 18 5) based on how the diagram was labeled in that example) on edges can be achieved using C2 where C = [U,UR-1] = UR∙U-1∙UR-1∙U. It was also shown that any 3-cycle on edges of the form (12 z y) could be attained by conjugation. We are looking to attain (12 z y) where y = 8 and z = 4. First, it must be shown we can get (12 11 4) and (12 11 8) by moving edge 4 and 8 into position 21 without affecting the other pieces. The following are the sequence of moves used to do just that.
Conjugating C2 by T where T=R∙DR∙F-1 gives the result, T-1∙C2∙T = (12 11 4)
Conjugating C2 by S where S=R∙DR∙F-1∙DF∙L2 gives the result, S-1∙C2∙S = (12 11 8)
Finally, conjugating these 3-cycles in the following way results in (12 4 8). (S-1∙C2∙S)-1∙T-1∙C2∙T ∙S-1∙C2∙S = (8 11 12) (12 11 4) (12 11 8) = (12 4 8)
The corresponding sequence of moves is below.
L-2∙DF-1∙F∙DR-1∙R-1∙(UR∙U-1∙UR-1∙U)^-2∙R∙DR∙F-1∙DF∙L2∙F∙DR-1∙ R-1∙ (UR∙U-1∙UR-1∙U)2∙R∙DR∙F-1∙L-2∙DF-1∙F∙DR-1∙R-1∙(UR∙U-1∙UR-1∙U)2∙R∙DR∙F-1∙DF∙L2
This process to find the corresponding sequence of moves can be done for each of the 3-cycles in (1 5 9) (2 6 10) (3 7 11) (4 8 12). The resulting sequence of moves combined will be what the move U_F on the octahedron can be mapped to in the cuboctahedron group. This can be done for all of the other generators of the group O, (UF, UR, UL, UB, DF, DR, DL, DB) which means that there does exist a mapping from the octahedron group to the cuboctahedron group.
Octahedron:
Vertices α = (1 3 4)
Edges β = (1 2 3)
Cuboctahedron:
Edges β = (1 5 9) (2 6 10) (3 7 11) (4 8 12)
Vertices α = (1 4 8)
The move that results in the edge permutation (1 2 3) and vertex permutation (1 3 4) on the octahedron is simply UF. To show that this generator can be mapped to the cuboctahderon group, the sequence of moves that will result in the edge permutation (1 5 9) (2 6 10) (3 7 11) (4 8 12) and vertex permutation (1 4 8) must be found. To do this, 3-cycles will be used since in Proposition 6, it was shown that any 3-cycle on edges or vertices on the cubocathedron can be found.
β = (1 5 9) (2 6 10) (3 7 11) (4 8 12)
Each of these 3-cycles can be written as a sequence of moves. Since this is a long process, only one 3-cycle will be shown, (4 8 12). In proposition 6.1, it was shown that the 3-cycle (12 11 21) which was (19 18 5) based on how the diagram was labeled in that example) on edges can be achieved using C2 where C = [U,UR-1] = UR∙U-1∙UR-1∙U. It was also shown that any 3-cycle on edges of the form (12 z y) could be attained by conjugation. We are looking to attain (12 z y) where y = 8 and z = 4. First, it must be shown we can get (12 11 4) and (12 11 8) by moving edge 4 and 8 into position 21 without affecting the other pieces. The following are the sequence of moves used to do just that.
Conjugating C2 by T where T=R∙DR∙F-1 gives the result, T-1∙C2∙T = (12 11 4)
Conjugating C2 by S where S=R∙DR∙F-1∙DF∙L2 gives the result, S-1∙C2∙S = (12 11 8)
Finally, conjugating these 3-cycles in the following way results in (12 4 8). (S-1∙C2∙S)-1∙T-1∙C2∙T ∙S-1∙C2∙S = (8 11 12) (12 11 4) (12 11 8) = (12 4 8)
The corresponding sequence of moves is below.
L-2∙DF-1∙F∙DR-1∙R-1∙(UR∙U-1∙UR-1∙U)^-2∙R∙DR∙F-1∙DF∙L2∙F∙DR-1∙ R-1∙ (UR∙U-1∙UR-1∙U)2∙R∙DR∙F-1∙L-2∙DF-1∙F∙DR-1∙R-1∙(UR∙U-1∙UR-1∙U)2∙R∙DR∙F-1∙DF∙L2
This process to find the corresponding sequence of moves can be done for each of the 3-cycles in (1 5 9) (2 6 10) (3 7 11) (4 8 12). The resulting sequence of moves combined will be what the move U_F on the octahedron can be mapped to in the cuboctahedron group. This can be done for all of the other generators of the group O, (UF, UR, UL, UB, DF, DR, DL, DB) which means that there does exist a mapping from the octahedron group to the cuboctahedron group.
Another interesting way to relate the groups of these puzzles would be by mapping the 3-cycles on edges and vertices from one to another. Simply comparing the sequence of moves to get these 3-cycles on each puzzle as shown in the chart below, there seems to be a pattern.
All three puzzles use C2 on two adjacent sides to attain a 3-cycle on edges. The 3-cycle that was used on vertices for the cuboctahedron puzzle is not very similar to the ones used on the cube and octahedron; however, another was found that somewhat relates to the cube and octahedron. All three use C on three adjacent sides – one side commutated with another conjugated by the third. By inspection, there seems to be a pattern between these 3-cycles. To prove that this is in fact true, the same process that was used in 7.1 and 7.2 can be employed. First, find the corresponding edge and vertex permutations for the 3-cycle. Then find the sequence of moves used to attain that permutation using the 3-cycles that are known. Mapping 3-cycles from one puzzle to another is a different method than used above but equally as useful in relating the cube, the octahedron and the cuboctahedron groups. Though mapping 3-cycles hasn’t been executed in this paper, mapping the generators of the puzzles has already indicated that there is a relationship between these three polyhedra.