I've taken a basic stab at this problem. I feel like I am missing something big. Please help. Thanks!
Q: 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.
Three Orbits:
- Edges. $|O_e| = 12, |G_e| = 2, |O_e| |G_e| = 24$
- Vertices. $|O_v| = 8, |G_v| = 3, |O_v| |G_v| = 24$
- Faces. $|O_f| = 6, |G_f| = 4, |O_f| |G_f| = 24$
Is that it? Am I missing anything?
I think that for each group $G_v$, $G_e$ and $G_f$ you must decompose $\mathcal{V}$, $\mathcal{E}$ and $\mathcal{F}$ into orbits, here you just apply some class formula to the whole group $G$. I will do it for $G_f$ only.
You need to make $G_f$ act on $\mathcal{V}$, $\mathcal{E}$ and $\mathcal{F}$. For instance take the following cube f being the face $ABCD$ :
Then $G_f$ is the group generated by an order $4$ rotation around the axis on the picture. Then the orbit decomposition of $\mathcal{V}$ with respect to $G_f$ is :
$$\{A,B,C,D\}\cup\{E,F,G,H\} $$
Then the orbit decomposition of $\mathcal{E}$ with respect to $G_f$ is :
$$\{(A,B),(B,C),(C,D),(D,A)\}\cup\{(E,F),(F,G),(G,H),(H,E)\}\cup\{(A,G),(B,H),(C,E),(D,F)\} $$
Then the orbit decomposition of $\mathcal{F}$ with respect to $G_f$ is :
$$\{(A,B,C,D)\}\cup\{(E,F,G,H)\}\cup\{(A,B,H,G),(B,C,E,H),(C,D,F,E),(D,A,G,F)\}$$
Edit : If you just need the number of orbits you can apply the Burnside's formula for $G$ acting on $X$ you have :
$$|X/G|=\frac{1}{|G|}\sum_{g\in G}|Fix_X(g)| $$
If $g$ is a rotation of order $4$ around the axis, then $G_f=\{Id,g,g^2,g^3\}$
$$G_f\text{ acting on } \mathcal{V}$$
$$Fix_{\mathcal{V}}(Id)=\mathcal{V}\text{, }Fix_{\mathcal{V}}(g^k)=\emptyset\text{ for } 1\leq k\leq 3$$
So :
$$|\mathcal{V}/G_f|=\frac{1}{4}(8+0+0+0)=2 $$
$$G_f\text{ acting on } \mathcal{E}$$
$$Fix_{\mathcal{E}}(Id)=\mathcal{E}\text{, }Fix_{\mathcal{E}}(g^k)=\emptyset\text{ for } 1\leq k\leq 3$$
So :
$$|\mathcal{E}/G_f|=\frac{1}{4}(12+0+0+0)=3 $$
$$G_f\text{ acting on } \mathcal{F}$$
$$Fix_{\mathcal{F}}(Id)=\mathcal{V}\text{, }Fix_{\mathcal{F}}(g^k)=\{(A,B,C,D),(E,F,G,H)\}\text{ for } 1\leq k\leq 3$$
So :
$$|\mathcal{F}/G_f|=\frac{1}{4}(6+2+2+2)=3 $$