Fundamental theorem of group actions

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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.


|G/G_x| = |K|


|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

Related facts about group actions


Related facts about 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.