Symmetric Group is Subgroup of Monoid of Self-Maps

Theorem
Let $S$ be a set.

Let $S^S$ be the set of all mappings from $S$ to itself

Let $\left({\Gamma \left({S}\right), \circ}\right)$ denote the group of permutations on $S$.

Let $\left({S^S, \circ}\right)$ be the monoid of self-maps under composition of mappings.

Then $\left({\Gamma \left({S}\right), \circ}\right)$ is a subgroup of $\left({S^S, \circ}\right)$.

Proof
By Group of Permutations is Group, $\left({\Gamma \left({S}\right), \circ}\right)$ is a group.

Let $\phi \in \Gamma \left({S}\right)$ be a permutation on $S$.

As a permutation is a self-map, it follows that $\phi \in S^S$.

Thus by definition $\Gamma \left({S}\right)$ is a subset of $S^S$.

So by definition, $\Gamma \left({S}\right)$, is a subgroup of $\left({S^S, \circ}\right)$.