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?
Best Answer
The main point : all your subgroups are rather small and generated by a single rotation.
There are at least 24 elements in $G$ :
The identity ${\sf id}$,
Rotations around axes through the middles of one of 3 pairs of opposing faces and with a rotation of $\frac{k\pi}{2}(1\leq k \leq 3)$ :there are $3\times 3=9$ such rotations, and we denote them by $OF(<name\ of\ face>,<angle>)$
Half-turns around axes through the middles of one of 6 pairs of opposing edges : there are $6$ such rotations, and we denote them by $HT(<name\ of\ opposite\ edges>)$.
Rotations with axis $18,26,37$ or $45$, and angle
$\frac{2k\pi}{3}(1\leq k \leq 2)$ :there are $2\times 4=8$ such rotations, and we denote them by $R(<name\ of\ axis>,<angle>)$.
Since there are at most $24$ orientation-preserving of the cube (see here), we see that $G$ consists exactly of the $24$ elements enumerated above.
Let $G_v$ denote the subgroup of $G$ fixing the vertex $1$, $G_e$ the subgroup fixing the edge $12$, and $G_f$ the subgroup fixing the face $1234$.
By the enumeration above, we have :
$$ \begin{array}{lcl} G_v &=& \lbrace {\sf id};R(18,\frac{2k\pi}{3}) (1 \leq k \leq 3) \rbrace \\ G_e &=& \lbrace {\sf id};HT(12,78) \rbrace \\ G_f &=& \lbrace {\sf id};OF(1234,\frac{k\pi}{2}) (1 \leq k \leq 3) \rbrace \end{array} $$
It is easy then to deduce the decompositions into orbits :
$$ \begin{array}{|l|l|} \hline \text{Subgroup} & G_v \\ \hline \text{Orbits in } V & [1] [2,3,5] [4,6,7] [8] \\ \hline \text{Orbits in } E & [12,13,15] [24,37,56] [26,34,57] [48,68,78] \\ \hline \text{Orbits in } F & [1234,1256,1357] [2468,3478,5678] \\ \hline \end{array} $$
$$ \begin{array}{|l|l|} \hline \text{Subgroup} & G_e \\ \hline \text{Orbits in } V & [1,2] [3,6] [4,5] [7,8] \\ \hline \text{Orbits in } E & [12] [13,26] [15,24] [34,56] [37,68] [48,57] [78] \\ \hline \text{Orbits in } F & [1234,1256] [1357,2468] [3478,5678] \\ \hline \end{array} $$
$$ \begin{array}{|l|l|} \hline \text{Subgroup} & G_f \\ \hline \text{Orbits in } V & [1,2,3,4] [5,6,7,8] \\ \hline \text{Orbits in } E & [12, 13, 24, 34] [15, 26, 37, 48] [56, 57, 68, 78] \\ \hline \text{Orbits in } F & [1234] [1256,1357,2468,3478] [5678] \\ \hline \end{array} $$