Group Action defines Permutation Representation

Theorem
Let $\map \Gamma X$ 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:
 * $\map {\phi_g} x = \map \phi {g, x}$

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

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

Proof
Note that, by Group Action Determines Bijection, $\phi_g \in \map \Gamma X$ for $g \in G$.

Let $g, h \in G$.

From the definition of group action:
 * $\forall \tuple {g, x} \in G \times X: \map \phi {g, x} \in X = g \wedge x \in X$

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

Thus:

Also, we have:
 * $e \wedge x = x \implies \map {\phi_e} x = x$

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

Therefore, we have shown that $\tilde \phi: G \to \map \Gamma X: g \mapsto \phi_g$ is a group homomorphism.