# Decomposition formulas for rotational symmetries of a cube

#### kalish

##### Member
I have a problem that I would like to check my work on. I am also stuck on the verifications for $E$ and $F$. Any help would be greatly appreciated. Thanks in advance.

**Problem statement:** Let $G$ be the group of rotational symmetries of a cube, let $G_v, G_e, G_f$ be the stabilizers of a vertex $v$, an edge $e$, and a face $f$ of the cube, and let $V, E, F$ be the sets of vertices, edges, and faces, respectively. Determine the formulas that represent the decomposition of each of the three sets into orbits for each of the subgroups.

**Proposed solution:**

...so if $G$ acts transitively on the vertices, the orbit of one (and thus any) vertex $v$ has order $8$, which means that the index of the stabilizer $[G:G_v]$ = $8$, so there are $8$ cosets of $G_v$ in G.

If one knows that the rotational symmetry group of the cube is $S_4$, this tells you that $G_v$ has order $3$.

$G$ also acts transitively on the faces, so $G_f$ (for any face $f$) has order 4.

Finally, $G$ also acts transitively on the edges, so $G_e$ (for any edge $e$) has order $2$.

The class equation for these subsets of {faces, vertices, edges} is particularly simple.

It occurs to me, that I'm being asked to compute the orbits of $V,F,E$ under the respective actions induced by each group:

$G_v,G_f,G_e$ for some particular stabilizer in each set. This is somewhat of a different matter.

For example, say $V = {v_1,v_2,v_3,v_4,v_5,v_6,v_7,v_8}$. $G_{v_1}$ fixes $v_1$, so it's orbit is: ${v_1}$. If the opposite vertex is $v_7$, $G_{v_1}$ also fixes $v_7$, so its orbit is: ${v_7}$. The other two orbits have to have order $3$, since there are no other points fixed by $G_{v_1}$, except $v_1$ and $v_7$ and the size of the orbits has to divide $|G_{v_1}|$ (it helps to think about WHICH rotations $G_{v_1}$ must be: rotations about the axis between $v_1$ and $v_7$).

What about for E and F?