Group of orientable symmetries of the 3-dimensional unit cube.

Question

Here is the question I am trying to solve:

Let $G$ be the group of orientable symmetries of the $3$-dimensional unit cube. Prove that $G$ is isomorphic to the symmetric group $S_4.$

I also got the following hint:

The unit cube has four diagonals that pass through the opposite corners, and $G$ acts on the set of diagonals.

Still, I do not know how to solve this problem. I have the following questions:

1- What is the meaning of orientable symmetry?

2- What is the relation between the action of $G$ and the isomorphism I should find?

Could someone help me answer these questions and solve the problem, please?

EDIT:

Do we actually need to know all the rotations of the cube and conclude from them the corresponding permutations of the vertices to write the isomorphism in this question?

Answer 1

Your book's hint is to consider the four diagonals of the cube (the long diagonals, lines passing through a pair of antipodal vertices), which are permuted by every symmetry of the cube, and particularly by every rotation. In the hope a sketch helps:

Enumerating all the rotations is not necessary, but it might be a good exercise depending on your background and interests. Rotations of the cube can be described, for example, by

• Geometric action as rotation about some axis;
• Effect as a permutation on the set of four diagonals;
• As signed $3 \times 3$ permutation matrices of positive determinant: $3 \times 3$ real matrices whose entries are all $0$, $1$, or $-1$, that have precisely one non-zero entry in each row and each column, and whose determinant is $1$.

With an eye toward minimizing computation, we can show successively that

1. Every rotation of the cube permutes the diagonals (diagonals map to diagonals, and distinct diagonals map to distinct diagonals).
2. The mapping from rotations to permutations is a homomorphism (performing successive rotations effects performing successive permutations on the diagonals).
3. The mapping from rotations to permutations is injective (no non-identity rotation fixes every diagonal).
4. The number of cube rotations is equal to the number of permutations of the set of diagonals.

To count the number of rotations without enumerating possibilities, it helps to think about:

• If we fix a vertex $0$, how many locations are there for the image of $0$ under a rotation of the cube?
• How many rotations fix the vertex $0$ (or any particular vertex)?

Answer 2

(Feel free to skip the next two paragraphs and go to the construction, if you want.)

I will show you how to concretely construct an isomorphism $S_4 \to G$. I assume that by the word concrete, you mean to have the elements of $S_4$ written as permutations of $\{1, 2, 3, 4\}$ and the elements of $G$ written as rotation matrices. If this is what you mean, you should understand the reason you will not find this information in a book: different labelings of the diagonals with $\{1, 2, 3, 4\}$ and different bases of $\mathbb{R}^3$ will give different concrete maps, so there is no concrete map that can be used as a reference. (There is a "standard basis", but there is no standard labeling of the diagonals. After solving your problem, you might like to think about what happens if you use a different labeling of diagonals, or a different basis.) What can be used as a reference is the idea that the permutations of the diagonals correspond exactly to rotations of a cube, so that's in the book.

Many mathematical constructions can only be given in this way, and you should be able to make them concrete whenever you want to. There is nothing fancy here, it's just brute force. If it's difficult, you should ask if you know how to make all the ingredients in the problem concrete. (For example, do you know how to concretely construct rotations? Rotations that send a given point to another? Permutations of a set?)

For the construction: first, orient the cube along the standard axis and center it at the origin. The diagonals are in the directions $[1, 1, 1], [1, 1, -1], [1, -1, -1], [1, -1, 1]$, and let us call them $1, 2, 3, 4$ in that order. Then, name each face by the outward normal vector and list which diagonals each vertex belongs to, running through the vertices in a clockwise order (only the relative order of the vertices matters here):

• $\hat{x}$: $(1, 2, 3, 4)$
• $\hat{y}$: $(1, 3, 4, 2)$
• $\hat{z}$: $(1, 4, 2, 3)$
• $-\hat{x}$: $(1, 4, 3, 2)$
• $-\hat{y}$: $(1, 2, 4, 3)$
• $-\hat{z}$: $(1, 3, 2, 4)$

Next, consider an arbitrary permutation $\pi$ of $\{1, 2, 3, 4\}$. I'll choose $1 \mapsto 4, 2 \mapsto 1, 3 \mapsto 3, 4 \mapsto 2$, How should the corresponding rotation $R$ transform the faces? We know how the clockwise order of the vertices changes, and that should uniquely tell us which faces are transformed into which other faces. (Note that we can cyclically translate the ordering of the vertices, and it will still correspond to the same face.)

• $R(\hat{x})$: $\pi(1, 2, 3, 4)$ : $(4, 1, 3, 2)$ : $(1, 3, 2, 4)$ : $-\hat{z}$
• $R(\hat{y})$: $\pi(1, 3, 4, 2)$ : $(4, 3, 2, 1)$: $(1, 4, 3, 2)$ : $-\hat{x}$
• $R(\hat{z})$: $\pi(1, 4, 2, 3)$ : $(4, 2, 1, 3)$: $(1, 3, 4, 2)$ : $\hat{y}$.

So we have that the permutation $1 \mapsto 4, 2 \mapsto 1, 3 \mapsto 3, 4 \mapsto 2$ corresponds with the rotation $\begin{pmatrix} 0 & -1 & 0 \\ 0 & 0 & 1 \\ -1 & 0 & 0 \end{pmatrix}$.

If you draw the setup above, you will be able to find a matrix for the remaining $23$ permutations (and you should do this, if you want to see them concretely). If you have doubts on why this is an isomorphism, consider why you get one, and exactly one, matrix for every permutation following the process above.