Fundamental theorem of group actions

Name
This result is also sometimes termed Burnside's theorem.

Statement for transitive group actions
Suppose a group $$G$$ acts transitively on a nonempty set $$S$$. Suppose $$s \in S$$ is a point, and $$G_s$$ denotes the isotropy subgroup of $$s$$ in $$G$$, i.e.:

$$G_s = \{ g \in G \mid g.s = s \}$$.

Then, there exists a unique bijective map between the left coset space of $$G_s$$ in $$G$$ and the set $$S$$:

$$\varphi:G/G_s \to S$$

satisfying the property that it is $$G$$-equivariant with respect to the natural action on the left coset space; in other words, for any $$g \in G$$ and any left coset $$hG_s$$, we have:

$$\varphi(ghG_s) = g.\varphi(hG_s)$$.

Note that this yields:

$$|G/G_s| = |S|$$.

Combining this with Lagrange's theorem, we obtain that:

$$|G| = |G_s||S|$$.

Statement for more general group actions
Suppose $$G$$ is a group acting on a set $$S$$. Let $$x \in S$$, and $$K$$ be the orbit of $$x$$ under the action of $$G$$. Then, if $$G_x$$ denotes the stabilizer of $$x$$ in $$G$$, we have a bijection:

$$G/G_x \to K$$.

Thus:

$$|G/G_x| = |K|$$

and

$$|G| = |G_x||K|$$

Note that this follows directly from the statement about transitive group actions, because the action of $$G$$ restricted to the orbit of $$x$$ is transitive.

Related facts about group actions

 * Group acts on left coset space of subgroup by left multiplication
 * Orbit-counting theorem
 * Class equation of a group
 * Class equation of a group action

Applications

 * Size of conjugacy class equals index of centralizer, size of conjugacy class divides order of group

Related facts about group homomorphisms

 * Normal subgroup equals kernel of homomorphism
 * Fundamental theorem of group homomorphisms

Construction of the map
We first describe the map $$\varphi$$.

For a left coset $$hG_s$$, define:

$$\varphi(hG_s) = h.s$$.

We need to prove that this is well-defined, and independent of the choice of coset representative. Thus, suppose that $$h_1, h_2$$ are in the same left coset of $$G_s$$. Then, there exists $$g \in G_s$$ such that $$h_2 = h_1g$$. Thus:

$$h_2.s = (h_1g).s = h_1.(g.s) = h_1.s$$

proving that the map is well-defined and independent of the choice of coset representative.

Proof that the map is injective
Suppose $$h_1,h_2 \in G$$ are such that $$h_1.s = h_2.s$$. Then, applying $$h_1^{-1}$$ to both sides yields:

$$h_1^{-1}.(h_1.s) = h_1^{-1}.h_2.s \implies s = (h_1^{-1}h_2).s$$

Thus, $$h_1^{-1}h_2 \in G_s$$, forcing $$h_1,h_2$$ to be in the same left coset of $$G_s$$. Thus, two elements from different left cosets cannot send $$s$$ to the same element.

Proof that the map is surjective
Since the action of $$G$$ on $$S$$ is transitive, every element of $$S$$ is expressible as $$h.s$$ for some $$h \in G$$, and hence as $$\varphi(hG_s)$$.

Proof that the map is equivariant
We need to prove that:

$$\varphi(ghG_s) = g.\varphi(hG_s)$$.

The left side is $$gh.s$$ while the right side is $$g.(h.s)$$. The two are clearly equal, by the definition of a group action.