Let C be a cube and let G be its rotational symmetry group. show that G is isomorphic to $S_4$

Question

Im trying to answer this question using the answer given in

Let C be a cube and let G be its rotational symmetry group. Outline a proof that G is isomorphic to Sym(4)

What I did is - I numbered the vertexes and I have the group of long diagonals

$S = {(1,7), (3,5), (4,6) ,(2,8)} $

where 1,2,34 is the button of the cube and 5,6,7,8 is the top (5 above 1)

now i know that all elements of $S_4$ can be made from 2-cycles.

so I need to find a $ g \in G$ that is correspond to every possible 2-cycle in

$S_4$ . I can't find this g. is it one g for every 2-cycle ? or one for all?

please help me solve that.

also I would like to know how to solve it (also with G operating on the diagonals) with the Orbit-stabilizer theore.

thanks a lot

Answer 1

The rotational group of the cube certainly turns a (here always: main) diagonal into a diagonal, hence acts on the set of the four diagonals. This action gives us a homomorphism $\phi\colon G\to \operatorname{Sym}(4)$. On the other hand, $G$ certainly has order $24$: We can pick one of six faces as "ground" face and rotate it in steps of $90^\circ$, thus giving us at least $6\times 4=24$ group elements. Once the ground face is fixed, the whole cube is fixed, thus indeed $|G|=24$. To establish that $\phi$ is an isomorphism it suffices to show that it is one-to-one, or alternatively to show that it is onto.

One quickly verifies that $G$ acts transitively on the diagonals (just rotate in steps of $90^\circ$ around the "vertical" axis). In fact the stabilizer of one diagonal still acts transitively on the remaining three diagonals (per rotation by $120^\circ$ around the fixed diagonal, and finally the stabilizer of two diagonals still acts transitively on the remaining two diagonals (rotation by $180^\circ$ around the axis perpendicular to the plane spanned by the two fixed diagonals). We conclude that $\phi[G]$ is all of $\operatorname{Sym}(4)$, as desired.

Answer 2

Using the orbit-stabilizer theorem: first of all, we can see that we have "some" homomorphism:

$G \to S_4$ by considering the fact that any such rotation of $G$ permutes the main diagonals.

Now let's consider the action of $G$ on the set of faces of the cube, we have $6$ of these. $G$ clearly always takes a face to a face. Since it is possible, by just using rotations about the center of two opposite faces (which is certainly a subset of $G$) to put any face in any desired position (a six-sided die can be rolled so that "any number is up"), we conclude that $G$ is transitive on this set of $6$ faces, that is, the orbit of any face under $G$ is the entire set of faces.

Hence if $x$ is any particular face, its orbit $G.x$ has cardinality $6$.

Now the orbit-stabilizer theorem says:

$|G| = |G.x|\ast|\text{Stab}(x)|$.

If a rotation of the cube fixes a face, it must be a rotation about the center of that face (its face is a square, and this is the full rotational symmetry group of a square-reflections of the square don't work, because changing the orientation of any (single) plane in $\Bbb R^3$ changes the orientation of $\Bbb R^3$ itself, and rotations in $\Bbb R^3$ preserve orientation).

So the stabilizer of the face $x$ has order $4$, and thus $|G| = 24$.

Similar arguments can be made by using the set of edges, or the set of vertices, but the elements of the stabilizers are harder to visualize. Basically, the $120^{\circ}$ rotations stabilize a vertex, and the $180^{\circ}$ rotations stabilize an edge, leading to:

$|G| = 8\cdot 3$ or $|G| = 12 \cdot 2$.