 Solution

A left group action of $G$ on a set $X$ is a map $(g,x)\mapsto g\cdot x$ satisfying $e\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The group orbit of $x$ is $G\cdot x=\{g\cdot x:g\in G\}$, and its stabilizer subgroup is $G_x=\{g\in G:g\cdot x=x\}$. The stabilizer subgroup contains the identity, is closed under products, and contains $g^{-1}$ whenever it contains $g$, so it is indeed a subgroup.

The orbit-stabilizer theorem for a finite group saysTo prove it, map a left coset $g{G}_x$ to $g\cdot x$. This is well-defined because every element of $G_x$ fixes $x$. Conversely, $g\cdot x=h\cdot x$ precisely when $h^{-1}g\in G_x$, which is precisely equality of the two cosets. Thus this map is a bijection between the cosets and the group orbit. Each coset has $|G_x|$ elements, by the bijection $k\mapsto gk$, and the cosets partition $G$, giving the formula.

Apply this to the rotational symmetry group of a cube acting on its six faces. The action is transitive: rotations through right angles about coordinate axes can send the top face to each of the other faces. A rotation fixing a face must fix its outward normal, so it rotates about the axis through that face's center and the opposite face's center. Preservation of its square boundary permits exactly four angles modulo $2\pi$: $0,\pi /2,\pi ,3\pi /2$. All four preserve the cube, so this face stabilizer subgroup has order four. The group is finite because a rotation permutes the six face normals and is determined by their images. The orbit-stabilizer theorem givesThese are orientation-preserving rotations; reflections are not being counted.