# Group Action Induces Equivalence Relation

## Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $X$ be a set.

Let $*: G \times S \to S$ be a group action.

Let $\RR_G$ be the relation induced by $G$, that is:

$\forall x, y \in X: x \mathrel {\RR_G} y \iff y \in \Orb x$

where:

$\Orb x$ denotes the orbit of $x \in X$.

Then:

$\RR_G$ is an equivalence relation

and:

the equivalence class of an element of $X$ is its orbit.

## Proof

Let $x \mathrel {\RR_G} y \iff y \in \Orb x$.

Checking in turn each of the criteria for equivalence:

### Reflexivity

 $\ds \exists e \in G: \,$ $\ds x$ $=$ $\ds e * x$ Group Action Axiom $\text {GA} 2$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds \Orb x$ Definition of Orbit (Group Theory)

Thus $\RR_G$ is reflexive.

$\Box$

### Symmetry

 $\ds y$ $\in$ $\ds \Orb x$ $\ds \leadsto \ \$ $\ds \exists g \in G: \,$ $\ds y$ $=$ $\ds g * x$ Definition of Orbit (Group Theory) $\ds \leadsto \ \$ $\ds g^{-1} * \paren {g * x}$ $=$ $\ds g^{-1} * y$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds \paren {g^{-1} \circ g} * x$ $=$ $\ds g^{-1} * y$ Group Action Axiom $\text {GA} 1$ $\ds \leadsto \ \$ $\ds e * x$ $=$ $\ds g^{-1} * y$ Group Axiom $\text G 2$: Existence of Identity Element $\ds \leadsto \ \$ $\ds \exists g^{-1} \in G: \,$ $\ds x$ $=$ $\ds g^{-1} * y$ Group Action Axiom $\text {GA} 2$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds \Orb y$ Definition of Orbit (Group Theory)

Thus $\RR_G$ is symmetric.

$\Box$

### Transitivity

 $\ds y$ $\in$ $\ds \Orb x$ $\, \ds \land \,$ $\ds z$ $\in$ $\ds \Orb y$ $\ds \leadsto \ \$ $\ds \exists g_1, g_2 \in G: \,$ $\ds y$ $=$ $\ds g_1 * x$ Definition of Orbit (Group Theory) $\, \ds \land \,$ $\ds z$ $=$ $\ds g_2 * y$ Definition of Orbit (Group Theory) $\ds \leadsto \ \$ $\ds z$ $=$ $\ds g_2 * \paren {g_1 * x}$ $\ds \leadsto \ \$ $\ds \exists g_2 \circ g_1 \in G: \,$ $\ds z$ $=$ $\ds \paren {g_2 \circ g_1} * x$ Group Action Axiom $\text {GA} 1$ $\ds \leadsto \ \$ $\ds z$ $\in$ $\ds \Orb x$ Definition of Orbit (Group Theory)

Thus $\RR_G$ is transitive.

$\Box$

So $\RR_G$ has been shown to be an equivalence relation.

Hence the result, by definition of an equivalence class.

$\blacksquare$