# Group Action Induces Equivalence Relation

## Theorem

Let $G$ be a group whose identity is $e$.

Let $X$ be a set.

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

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

$x \mathrel {\mathcal R_G} y \iff y \in \Orb x$

where:

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

Then:

$\mathcal R_G$ is an equivalence relation.
The equivalence class of an element is its orbit.

## Proof

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

Checking in turn each of the critera for equivalence:

### Reflexivity

$x = e * x \implies x \in \Orb x$ from the definition of group action.

Thus $\mathcal R_G$ is reflexive.

$\Box$

### Symmetry

 $\ds y$ $\in$ $\ds \Orb x$ $\ds \leadsto \ \$ $\ds \exists g \in G: y$ $=$ $\ds g * x$ $\ds \leadsto \ \$ $\ds g^{-1} * \paren {g * x}$ $=$ $\ds g^{-1} * y$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds g^{-1} * y$ $\ds \leadsto \ \$ $\ds \exists g^{-1} \in G: x$ $=$ $\ds g^{-1} * y$ $\ds \leadsto \ \$ $\ds x$ $\in$ $\ds \Orb y$

Thus $\mathcal R_G$ is symmetric.

$\Box$

### Transitivity

 $\ds y$ $\in$ $\ds \Orb x, z \in \Orb y$ $\ds \leadsto \ \$ $\ds \exists g_1 \in G: y$ $=$ $\ds g_1 * x, \exists g_2 \in G: z = g_2 * y$ $\ds \leadsto \ \$ $\ds z$ $=$ $\ds g_2 * \paren {g_1 * x}$ $\ds \leadsto \ \$ $\ds z$ $=$ $\ds \paren {g_2 g_1} * x$ $\ds \leadsto \ \$ $\ds z$ $\in$ $\ds \Orb x$

Thus $\mathcal R_G$ is transitive.

$\Box$

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

Hence the result, by definition of an equivalence class.

$\blacksquare$