# Fundamental theorem of group actions

## Contents

## Name

This result is also sometimes termed Burnside's theorem.

## Statement

### Statement for transitive group actions

Suppose a group acts transitively on a nonempty set . Suppose is a point, and denotes the isotropy subgroup of in , i.e.:

.

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

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

.

Note that this yields:

.

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

.

### Statement for more general group actions

Suppose is a group acting on a set . Let , and be the orbit of under the action of . Then, if denotes the stabilizer of in , we have a bijection:

.

Thus:

and

Note that this follows directly from the statement about transitive group actions, because the action of *restricted* to the orbit of 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

### Related facts about group homomorphisms

## Proof

### Construction of the map

We first describe the map .

For a left coset , define:

.

We need to prove that this is well-defined, and independent of the choice of coset representative. Thus, suppose that are in the same left coset of . Then, there exists such that . Thus:

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

### Proof that the map is injective

Suppose are such that . Then, applying to both sides yields:

Thus, , forcing to be in the same left coset of . Thus, two elements from different left cosets cannot send to the same element.

### Proof that the map is surjective

Since the action of on is transitive, every element of is expressible as for some , and hence as .

### Proof that the map is equivariant

We need to prove that:

.

The left side is while the right side is . The two are clearly equal, by the definition of a group action.