The symmetries of the cube and the octahedron

 In the previous post, I stated that the cube and the octahedron are duals of one another. A cube has six faces and eight vertices, while an octahedron has eight faces and six vertices. Since six is less than eight, my natural laziness says it should be represented as a subset of S6 instead of S8, which means for a cube, we are permuting the faces, while for an octahedron, we are permuting the vertices. Same diff, the group will still have 24 elements.

 

Six-sided dice have a convention for the placement of the numbers represented, opposite sides have a sum of 7, meaning 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4. 

 

The 12 even permutations first.

 

The identity is always even.

(1)

 

Pick two opposite faces to be fixed, and turn the cube 180° on the axis that runs through the centers of the fixed faces. There are three pairs of opposite faces, so three elements here.

(25)(34), (16)(34), (25)(16)

 

 

Pick two opposite vertices to be the axis of rotation, and turn the cube 120° and 240°. There are four pairs of opposite vertices, so this accounts for eight elements.

 

Spin 1-2-3 and 4-5-6

(123)(654), (132)(645)

 

Spin 1-3-5 and 2-4-6

(135)(642), (153)(624)

 

Spin 1-5-4 and 3-2-6

(154)(623), (145)(632)

 

Spin 1-4-2 and 5-3-6

(142)(635), (142)(653) 

 

Composing two even permutations creates an even permutation since even+even = even. This means these 12 permutations are a subgroup because of closure under the group operation and the existence of the identity in the set. Because 12 = 24/2, This must be a normal subgroup of the the larger group.

The 12 odd permutations.

Pick two opposite faces to be fixed, and turn the cube 90° or 270° on the axis that runs through the centers of the two fixed faces. There are three pairs of opposite faces, so this accounts for six permutations.

(2354), (2453), (1364), (1463), (2156), (2651)

 

Swap two opposite faces, and turn the cube 90° or 270° on the axis that runs through the centers of the swapped faces. There are three pairs of opposite faces, so this accounts for six permutations.

(16)(24)(35)

(16)(23)(45)

(25)(13)(46)

(25)(14)(36)

(34)(12)(56)

(34)(15)(26)

 

If we add diagonals to all six faces with the proviso that every vertex is either the endpoint of three diagonals or the endpoint of none, we get a regular tetrahedron embedded in our cube. Even permutations will permute the pink tetrahedron, while odd permutations will send the pink tetrahedron's vertices to the vertices that have zero diagonals in the original position.

 

This means the symmetries of the tetrahedron is a normal subgroup of the symmetries of the cube, and by duality, a normal subgroup of the symmetries of the octahedron.

 

Symmetries within symmetries. The beauty of group theory.

 

We learned about the group representation trick that tells us if we have k conjugacy classes, we can find k perfect squares that add up to the order of the group, with the proviso that at least one of the squares is 1², which is matched with the identity, always standing alone as a singleton conjugacy class. We have at least five conjugacy classes, based on the five different permutation shapes listed above.

Question 1.

Can you find five perfect squares such that 1²+a²+b²+c²+d²=24?

Question 2.

If yes to Question 1, does this mean there are exactly five conjugacy classes?

 

Question 3.

Find the conjugacy classes of the symmetries of the cube, and find the k squares that sum to 24.

 

Answers on Friday, when I will post more information about the structure of this group.