# 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:

 $\ds \map {\phi_g \circ \phi_h} x$ $=$ $\ds g \wedge \paren {h \wedge x}$ Definition of $\phi_g$, $\phi_h$ $\ds$ $=$ $\ds \paren {g h} \wedge x$ Definition of Group Action $\ds$ $=$ $\ds \map {\phi_{g h} } x$ Definition of $\phi_{g h}$

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.

$\blacksquare$