# Cycle decompositions of the rigid motions of the cube

1. Nov 3, 2009

### gotmilk04

1. The problem statement, all variables and given/known data
Write down cycle decompositions of all 24 of the rigid motions of the cube by regarding each rigid motion as a permutation of the 6 faces.
Then, this is a subgroup of S$$_{6}$$, called S$$_{CUBE}$$.
Prove or disprove that S$$_{CUBE}$$ is a subgroup of A$$_{6}$$

2. Relevant equations

3. The attempt at a solution
I'm not sure how the permutations are represented in regards to the 6 faces of the cube, so I can't do much else until I know that.

2. Nov 4, 2009

### Billy Bob

Would it help to look at some dice?

Place one on a desk in front of you with
6 on top
5 on front
4 on left
3 on right
2 on back
1 on bottom.

Now keep the 6 on top and 1 on bottom, but rotate so that
5 is on left
4 on back
3 on front
2 on right.

You could represent that as (5 4 2 3) meaning that the 5 got sent to where the 4 was, the 4 got sent to where the 2 was, etc.

3. Nov 5, 2009

### Quantumpencil

A cyclic decomposition of a permutation is a way of isolating the things which go to each under repeated application of the Permutation. f Associate to each face of the Cube, and a number {1,...,6}, and then consider an element of the permutation Group on 6 letters. Can you find a symmetry of the cube, which is capable of reproducing this permutation? That is, rotation about some axis (Say, rotation by Pi or Pi/2 about the axis going through the center of the faces (Perhaps you can't; think about this. Why is this impossible?) Now think about the the Group of permutations on 4 letters. Can you identify some part of the cubes geometry which is permuted in a way such that every element of S_4 corresponds to a permutation of these parts of the cube under permutations?

Also try going the other way around. For instance, if your cube is situated at the origin in R^3, what kinds of rotation options do you have? If you start with the top face 1, bottom 2, left 3, right 4, one facing you 5, and one away from you 6, assuming that x runs horizontally, z towards you, and y vertically. Say you rotate the cube by pi/2 about x. What is the image under this permutation of each face? Well, 3 and 4 aren't going to change/ they map to themselves. 1 is going to go to 5, 5 is going to go to 2, 2 is going to go to 6, and 6 is going to go to 1.

This gives you the cyclic decomposition of the permutation

(1526)(3)(4). So this rotation fixes three and four, and repeated applications of it "permutes" the other faces cyclically. Is it obvious to you that this is equally well thought of as the image of the faces under an element of S_6?

Where does that take the faces of your cube? Now think about the cyclic structure of the permutation p associated to this rotation about x (apply the permutation again, see where things go). Think about all the "rotations" and "flips" you can perform on this cube. What cyclic decomposition is associated to each kind (for instance, flips about the diagonals, or flips about the lines dividing the edges)