Symmetric Group is Group/Proof 1

Theorem

Let $S$ be a set.

Let $\map \Gamma S$ denote the set of all permutations on $S$.

Then $\struct {\map \Gamma S, \circ}$, the symmetric group on $S$, 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 $\struct {\map \Gamma S, \circ}$ is closed.

$\Box$

G1: Associativity

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

$\Box$

G2: Identity

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

$\Box$

G3: Inverses

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

$\Box$

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

$\blacksquare$