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 $\struct {\Gamma \paren S, \circ}$ denote the symmetric group on $S$.

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

Then $\struct {\Gamma \paren S, \circ}$ is a subgroup of $\struct {S^S, \circ}$.

Proof

By Symmetric Group is Group, $\struct {\Gamma \paren S, \circ}$ is a group.

Let $\phi \in \Gamma \paren S$ be a permutation on $S$.

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

Thus by definition $\Gamma \paren S$ is a subset of $S^S$.

So by definition, $\Gamma \paren S$, is a subgroup of $\struct {S^S, \circ}$.

$\blacksquare$