Symmetric Group is Group/Proof 2

Theorem
Let $S$ be a set.

Let $\Gamma \left({S}\right)$ denote the set of all permutations on $S$.

Then the group of permutations on $S$ $\left({\Gamma \left({S}\right), \circ}\right)$ forms a group.

It is a subgroup of $\left({S^S, \circ}\right)$, where $S^S$ is the set of all mappings on $S$.

Proof
A direct application of Invertible Mappings form Group of Permutations.