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:
Group Axiom $\text G 0$: Closure
By Composite of Permutations is Permutation, $S$ is itself a permutation on $S$.
Thus $\struct {\map \Gamma S, \circ}$ is closed.
$\Box$
Group Axiom $\text G 1$: Associativity
From Set of all Self-Maps under Composition forms Monoid, we have that $\struct {\map \Gamma S, \circ}$ is associative.
$\Box$
Group Axiom $\text G 2$: Existence of Identity Element
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.
$\Box$
Group Axiom $\text G 3$: Existence of Inverse Element
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.
$\blacksquare$
Sources
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