Which elements of $S_8$ are in the subgroup of rigid motions of a cube?

Question

Let the set $S\colon= \{ 1, 2, 3, 4, 5, 6, 7, 8 \}$. Then which permutations of $S$ will appear in the group of rigid motions of a cube, which is a subgroup of $S_8$, the symmetric group on 8 letters? Here the elements of $S$ are to represent the eight different vertices of the cube when in its original position.

In the original position of the cube, we assume that, going counter-clockwise, the vertices $1$, $2$, $3$, and $4$ form the bottom face; the vertices $5$, $6$, $7$, and $8$ the top face; and the vertices $1$, $2$, $7$, and $6$ the front face. This way, $1$, $8$; $2$, $5$; $3$, $6$; and $4$, $7$ are the four pairs of opposite vertices.

Is it a correct way of beginning? And if so, then what are the elements alongwith their respective geometric interpretations?

Answer 1

Yes, you've got a correct beginning. The group of permutations on $8$ letters is indeed $S_8$. The elements of $S_8$ are the $8!$ possible permutations of these letters. And you've got a good start on identifying these $8$ elements with the eight vertices of a cube. Now, only some of the permutations of $S_8$ belong to the group of rigid motions of a cube: these can be thought of the possible ways in which a cube can be rotated so that it starts snug in a box, is rotated about any number of axes of rotation, and placed back in the snug box, with vertices permuted (or not, if it's placed back in the box with vertices in the exactly same position as they were to begin with.).

Hints:

• The group of rigid motions of the cube is isomorphic to $S_4$. And $|S_4| = 4! = 24$, hence the subgroup of elements of $S_8$ that is isomorphic to $S_4$ (the group of rigid motions of a cube) has 24 elements.

• Consider the permutations of your named vertices when the cube is rotated about each of three axes which pierce opposite faces. There are 9 such permutations.

• Consider the permutations of your named vertices when the cube is rotated about each of six axes which penetrate the midpoints of opposite edges. There are 6 of such permutations.

• Consider the permutations of your named vertices when the cube is rotated about each of four long diagonals connecting opposite vertices of the cube. There are 8 such permutations.

• Don't leave out the identity: the "do nothing" action in which the cubes vertices remain, or return to, their original locations.

Answer 2

One way to prove the number of rotational symmetries of the cube is 24 is as follows. We first move vertex 1 to an arbitrary vertex, and then we rotate about this new vertex - all rotational symmetries can be achieved using these two steps. The number of rotational symmetries that map the vertex 1 to itself is 3 since when we fix 1 we can rotate the cube 120, 240 or 360 degrees. Now, the vertex 1 can be mapped to any of the 8 vertices. Thus, the total number of rotational symmetries of the cube is $3 \times 8=24$. This is essentially the Orbit-Stabilizer Lemma, which says that to find the order of a permutation group, find the size of the stabilizer of vertex 1 (this value is 3), and the size of the orbit of vertex 1 (this value is 8), and multiply these two values.

Now there are many groups of order 24. A proof that the rotation group $H \le S_8$ is isomorphic to $S_4$ is given at the link Automorphism group and congruences of the cube. This proof considers the action of $H$ on the set of four diagonals (opposite pairs of vertices) $\{1, 8\},\{2,5\},\{3,6\},\{4,7\}$. Every rotational symmetry of the cube effects a permutation of the four diagonals, and it is shown that this homomorphism from $H$ onto the symmetric group of these 4 diagonals is faithful and onto, hence gives an isomorphism to $S_4$.