Group Action of Symmetric Group

Theorem
Let $S$ be a set.

Let $\left({\Gamma \left({S}\right), \circ}\right)$ be the symmetric group on $S$.

The mapping $*: \Gamma \left({S}\right) \times S \to S$ defined as:
 * $\forall \pi \in \Gamma \left({S}\right), \forall s \in S: \pi * s = \pi \left({s}\right)$

is a group action.

Proof
The group action axioms are investigated in turn.

Let $\pi, \rho \in \Gamma \left({S}\right)$ and $s \in S$.

Thus:

demonstrating that group action axiom $GA1$ holds.

Then:

demonstrating that group action axiom $GA2$ holds.