Symmetric Group is Group/Proof 1

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.

Proof
Taking the group axioms in turn:

G0: Closure
By Composite of Permutations is Permutation, $S$ is itself a permutation on $S$.

Thus $\left({\Gamma \left({S}\right), \circ}\right)$ is closed.

G1: Associativity
From Set of all Self-Maps is Monoid, we have that $\left({\Gamma \left({S}\right), \circ}\right)$ is associative.

G2: Identity
From Set of all Self-Maps is Monoid, we have that $\left({\Gamma \left({S}\right), \circ}\right)$ has an identity, that is, the identity mapping.

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

Thus all the group axioms have been fulfilled, and the result follows.