Symmetric Group is Group
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 1
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.
$\blacksquare$
Proof 2
A direct application of Set of Invertible Mappings forms Symmetric Group.
$\blacksquare$
Also see
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \alpha$
- 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
- 1992: William A. Adkins and Steven H. Weintraub: Algebra: An Approach via Module Theory ... (previous) ... (next): $\S 1.1$ Example $6$