# Symmetric Group is Group

## Contents

## 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:

### 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.

$\blacksquare$

## 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 $6$