Group Action Induces Equivalence Relation

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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.

$\blacksquare$


Also see


Sources