How many rotations does a cube have? Counting them one by one is error-prone. The orbit–stabilizer theorem gets the answer from two easy counts, by the same partition idea that proved Lagrange's theorem.
Why it works
Orbit–stabilizer theorem. If a finite group $G$ acts on $X$ and $x \in X$, then $$|G \cdot x| = [G : G_{x}], \qquad \text{so} \qquad |G| = |G \cdot x| \cdot |G_{x}|.$$ Proof. Match each left coset of the stabilizer with a point of the orbit by $gG_{x} \mapsto g \cdot x$.
Well defined and one-to-one: $g \cdot x = h \cdot x \iff (h^{-1}g) \cdot x = x \iff h^{-1}g \in G_{x} \iff gG_{x} = hG_{x}$ (reading the chain left to right gives one-to-one, right to left gives well defined).
Onto: every point of the orbit is some $g \cdot x$.
So the number of cosets, $[G : G_{x}] = |G|/|G_{x}|$, equals the size of the orbit. $\square$
Well defined and one-to-one: $g \cdot x = h \cdot x \iff (h^{-1}g) \cdot x = x \iff h^{-1}g \in G_{x} \iff gG_{x} = hG_{x}$ (reading the chain left to right gives one-to-one, right to left gives well defined).
Onto: every point of the orbit is some $g \cdot x$.
So the number of cosets, $[G : G_{x}] = |G|/|G_{x}|$, equals the size of the orbit. $\square$
So orbit sizes always divide $|G|$. If any two of $|G|$, the orbit size and the stabilizer size are known, so is the third.
Worked example: The rotations of a cube
How many rotations does a cube have?
- Act on the $6$ faces: any face can be rotated to any other, so the orbit of the top face has $6$ elements.
- Rotations fixing the top face spin about the vertical axis: $0^\circ$, $90^\circ$, $180^\circ$, $270^\circ$. The stabilizer has $4$ elements.
- $|G| = 6 \cdot 4 = 24$. Check with vertices: $8$ vertices, $3$ rotations fix a given vertex, $8 \cdot 3 = 24$.
Answer: $24$.
Worked example: Subsets
$S_{6}$ acts on the 2-element subsets of $\{1, \dots, 6\}$. Use orbit–stabilizer to count them.
- Every 2-subset can be moved to every other, so there is one orbit: all of them.
- The stabilizer of $\{1, 2\}$ permutes $\{1, 2\}$ and $\{3, 4, 5, 6\}$ separately: $2! \cdot 4! = 48$ elements.
- Orbit size $= 720/48 = 15 = \binom{6}{2}$.
Answer: $15$.
Worked example: Conjugacy classes in $S_{4}$
$(1\;2)$ commutes with exactly $4$ permutations of $S_{4}$. How many permutations are conjugate to it?
- Let $S_{4}$ act on itself by conjugation; the stabilizer of $(1\;2)$ is the set of permutations commuting with it, of size $4$.
- Orbit size $= 24/4 = 6$: the six transpositions.
Answer: $6$.
Common mistake
Multiplying when you should divide, or using the size of the whole set in place of the orbit. For the cube acting on its $8$ vertices, the stabilizer of a vertex has $24/8 = 3$ elements only because the vertices form a single orbit; when there are several orbits, use the size of the orbit of that point.
Key idea
For a finite group acting on a set: $|G| = |\text{orbit of } x| \cdot |\text{stabilizer of } x|$. The points of the orbit match the cosets of the stabilizer.