Rotations of a cube isomorphic to $S_4$

Question

I'm trying to prove that the group $G$ of rotations of the cube is isomorphic to $S_4$. I have a very rough sketch of a proof, which has the following three components:

• $G$ has $24$ elements
• There are four main diagonals of the cube, and each rotation permutes these diagonals
• No two rotations correspond to the same rotation of the diagonals, so we get an injective homomorphism to $G$ to $S_4$. As they have the same size, this is also surjective.

The only part I'm having trouble proving is the third piece. An alternative is to prove that the kernel of the homomorphism is trivial, but I don't know how to rigorously do that other than by saying, "the only permutation which does nothing to the diagonals is one that does nothing to the cube at all, which is the identity permutation."

Answer 1

Hint: show that for any pair of diagonals, $G$ contains a rotation that switches those diagonals and fixes the other two. So its image in $S_4$ contains all the transpositions and as the transpositions generate $S_4$ your homomorphism maps onto a generating set and hence is surjective.

Answer 2

You can make the calculation that the kernel is trivial rigorous.

Suppose that the cube has edge length $1$. Label one vertex $A$ and the three adjacent ones $B,C,D$. Label the vertices opposite these $A',B',C',D'$ so that the four diagonals are $\{A,A'\},\{B,B'\},\{C,C'\},\{D,D'\}$.

Suppose now that $\pi$ is a rotation fixing the diagonals. Then two cases arise.

Case (i) $\pi(A)=A$. In this case $\pi(B)$ must be one of $B,B'$ and must be distance $1$ from $A$. Hence $\pi(B)=B$; similarly $\pi(C)=C, \pi(D)=D$ and so $\pi$ fixes the vertices and so is the identity rotation.

Case (ii) $\pi(A)=A'$. In this case $\pi(B)$ must be one of $B,B'$ and must be distance $1$ from $A'$. Hence $\pi(B)=B'$; similarly $\pi(C)=C', \pi(D)=D'$. But no rotation can do this, since $\pi$ acts as the matrix $-I_3$, which has determinant $-1$ and so is not a rotation.

Answer 3

Here's a way to directly prove injecitivity, by directly proving your statement the only permutation which does nothing to the diagonals is one that does nothing to the cube at all, which is the identity permutation.

The cube has 8 vertices which fall into 4 pairs of vertices, each pair being the endpoints of one of the 4 diagonals. For each vertex $V$ let me use $\rho V$ to use the vertex at the other end of the diagonal containing $V$.

Consider now any rotation $R$ of the cube that takes each diagonal to itself. It follows that $R$ preserves the endpoint pair $V$, $\rho V$ of each diagonal, either fixing both of them, or transposing them by the map $R(V)=\rho V$, $R(\rho V)=V$.

Now let's do an argument in two cases.

For the first case, suppose that there exists a vertex $V$ of the cube that is fixed by $R$, so $\rho V$ is also fixed by $R$. Of the remaining six vertices of the cube, there are edges $E_1,E_2,E_3$ of the cube that connect $V$ to three respective vertices denoted $W_1,W_2,W_3$; also, none of the opposite vertices $\rho W_1, \rho W_2, \rho W_3$ are connected to $V$ by a single edge. Since $R$ fixes $V$ and preserves the endpoint pair $W_i,\rho W_i$, it follows that $R$ also fixes $W_i$, and then it follows furthermore that $R$ fixes $\rho W_i$. Having proved that $R$ fixes all six vertices of the cube, it follows that $R$ is the identity.

For the second case, suppose that there exists a vertex $V$ such that $R$ transposes $V$ and $\rho V$, so $R(V)=\rho V$ and $R(\rho V)=V$. By applying the first case, it follows that $R(W)=\rho W$ for every vertex of the cube (otherwise, if $R(W)=W$ then the previous case would tell us that $R(V)=V$, a contradiction). It follows that $R$ is the point reflection map, reflecting across the center of the cube.

However, the point reflection map is not in the group of rotations of the cube. One way to see this is using a tiny bit of linear algebra, representing the cube as $[-1,1]^3$ and using orthonormal matrices to represent each transformation of the cube: rotations have determinant $+1$; but the point reflection which is given by the matrix $\begin{pmatrix} -1 & 0 & 0 \\0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ has determinant $-1$.