Group of Permutations is Group

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 1
Taking the group axioms in turn:

G0: Closure
A Composite of Permutations on $S$ is itself a permutation on $S$, and thus $\left({\Gamma \left({S}\right), \circ}\right)$ is closed.

G1: Associativity
From Set of All Mappings is a Monoid, we already have that $\left({\Gamma \left({S}\right), \circ}\right)$ is associative.

G2: Identity
Also from Set of All Mappings is a Monoid, we already have that $\left({\Gamma \left({S}\right), \circ}\right)$ has an identity, that is, the identity mapping.

G3: Inverses
By Inverse of Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.

As a permutation is a mapping, it follows that $\Gamma \left({S}\right) \subseteq S^S$.

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

Proof 2
The Set of All Mappings is a Monoid.

By Inverse of 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 Group.