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

• Symmetric Group is Subgroup of Monoid of Self-Maps

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 $6$