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 $S$:

$\phi :G/G_s\to S$

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 $h{G}_s$, we have:

$\phi (gh{G}_s)=g.\phi (h{G}_s)$.

Note that this yields:

$|G/G_s|=\left|S\right|$.

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

$\left|G\right|=\left|G_s\right|\left|S\right|$.

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:

$|G/G_x|=\left|K\right|$

and

$\left|G\right|=\left|G_x\right|\left|K\right|$

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

• 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

Proof

Construction of the map

We first describe the map $\phi$.

For a left coset $h{G}_s$, define:

$\phi (h{G}_s)=h.s$.

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 $h_2=h_{1}g$. Thus:

$h_2.s=(h_{1}g).s=h_1.(g.s)=h_1.s$

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:

$h_1^{-1}.(h_1.s)=h_1^{-1}.h_2.s⟹s=(h_1^{-1}{h}_2).s$

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 $\phi (h{G}_s)$.

Proof that the map is equivariant

We need to prove that:

$\phi (gh{G}_s)=g.\phi (h{G}_s)$.

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.