Set of Invertible Mappings forms Symmetric Group

Theorem
Let $S$ be a set.

Let $\left({S^S, \circ}\right)$ be the monoid of all the mappings from $S$ to itself.

Let $\mathcal G$ be the set of all invertible mappings from $S$ to $S$.

Then $\left({\mathcal G, \circ}\right)$ is the group of permutations on $S$.

Proof
We have Set of All Self-Maps is Monoid.

By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.

By Bijection iff Inverse is Bijection, it follows that all the invertible elements of $S^S$ are exactly the permutations on $S$.

The result follows from Invertible Elements of Monoid form Subgroup.