TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set

Statement

For a left action of $G$ on $X$ , define $x \sim y$ when $y = g \cdot x$ for some $g \in G$ . This is an equivalence relation, its equivalence class at $x$ is $G \cdot x$ , and the distinct orbits partition $X$ .

Facts & Assumptions

Given: A left action of a group $G$ on a set $X$ .

[L1]

The action laws are $e \cdot x = x$ and $(g h) \cdot x = g \cdot (h \cdot x)$ (Left group actions, transitive actions, and faithful actions).

[L2]

The orbit at $x$ is $G \cdot x = {g \cdot x : g \in G}$ (The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action).

[L3]

Equivalence classes of an equivalence relation partition the underlying set (Equivalence relation, equivalence class, and the quotient set $A / \sim$ , The equivalence classes of an equivalence relation are nonempty, cover $A$ , and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

Proof

technique · direct

1.1L1given

The relation is reflexive: $x = e \cdot x$ , so $x \sim x$ .

1.2L1givenalgebra

If $y = g \cdot x$ , then $x = g^{- 1} \cdot y$ , so $x \sim y$ implies $y \sim x$ .

1.3L1givenalgebra

If $y = g \cdot x$ and $z = h \cdot y$ , then $z = (h g) \cdot x$ , so $x \sim y$ and $y \sim z$ imply $x \sim z$ .

2.1step 1.1step 1.2step 1.3L2L3

Steps 1.1–1.3 show that $\sim$ is an equivalence relation. Its class at $x$ is precisely the set of $y = g \cdot x$ , namely $G \cdot x$ .

3.1step 2.1L3∎

Therefore the distinct orbits partition $X$ .

Depends on

Dependency tree · 2 levels

Sources

Brosnan, Orbits and stabilizers