Group Action defines Permutation Representation

Theorem
Let $\Gamma \left({X}\right)$ be the set of permutations on a set $X$.

Let $G$ be a group.

Let $\phi : G\times X \to X$ be a group action.

For $g\in G$, let $\phi_g: X \to X$ be the mapping defined as:
 * $\phi_g \left({x}\right) = \phi \left({g, x}\right)$

Let $\tilde \phi: G \to \Gamma \left({X}\right)$ be the mapping associated to $\phi$, defined by:
 * $\tilde \phi \left({g}\right) := \phi_g$

Then $\tilde \phi$ is a group homomorphism.

Proof
Note that, by Group Action Determines Bijection, $\phi_g \in \Gamma \left({X}\right)$ for $g\in G$.

Let $g, h \in G$.

From the definition of group action:
 * $\forall \left({g, x}\right) \in G \times X: \phi \left({\left({g, x}\right)}\right) \in X = g \wedge x \in X$

First we show that for all $x \in X$:
 * $\phi_g \circ \phi_h \left({x}\right) = \phi_{g h} \left({x}\right)$

Also, we have:
 * $e \wedge x = x \implies \phi_e \left({x}\right) = x$

where $e$ is the identity of $G$.

Therefore, we have shown that $\tilde \phi: G \to \Gamma \left({X}\right), g \mapsto \phi_g$ is a group homomorphism.