# Fundamental theorem of group actions

## Name

This result is also sometimes termed Burnside's theorem.

## Statement

### 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.

## Proof

### 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.