Symmetric Group is Group/Proof 1

Proof
Taking the group axioms in turn:

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

Thus $\struct {\map \Gamma S, \circ}$ is closed.

From Set of all Self-Maps under Composition forms Monoid, we have that $\struct {\map \Gamma S, \circ}$ is associative.

From Set of all Self-Maps under Composition forms Monoid, we have that $\struct {\map \Gamma S, \circ}$ has an identity, that is, the identity mapping.

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.