Group of Permutations is Group

Theorem
Let $S$ be a set.

The set of all permutations on $S$ is denoted $\Gamma \left({S}\right)$.

The structure $\left({\Gamma \left({S}\right), \circ}\right)$, where $\circ$ denotes composition of mappings, forms a group.

It is a subgroup of the set $S^S$ of all mappings on $S$.

This is called the group of permutations on $S$.

Proof 1
Taking the group axioms in turn:

G0: Closure
A Composite of Permutations on $S$ is itself a permutation on $S$, and thus $\left({\Gamma \left({S}\right), \circ}\right)$ is closed.

G1: Associativity
From Set of All Mappings is a Monoid, we already have that $\left({\Gamma \left({S}\right), \circ}\right)$ is associative.

G2: Identity
Also from Set of All Mappings is a Monoid, we already have that $\left({\Gamma \left({S}\right), \circ}\right)$ has an identity, that is, the identity mapping.

G3: Inverses
By Inverse of Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.

Proof 2
The Set of All Mappings is a Monoid.

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

By Bijection iff Inverse is Bijection, it follows that all the invertible elements of $S^S$ are exactly the permutations on $S$.

The result follows from Invertible Cancellable Elements of Monoid form Group.

Notation
An alternative way to denote the set of all permutations which is sometimes seen is $\mathfrak S_S$.

However, the fraktur font is rarely used nowadays as it is cumbersome to reproduce and awkward to read.

Also see

 * If $S$ is finite, and has cardinality $n$, then:
 * From Number of Permutations, we see that the order of $\left({\Gamma \left({S}\right), \circ}\right)$ is $n!$
 * We also have that $\left({\Gamma \left({S}\right), \circ}\right)$ is isomorphic to the symmetric group on $n$ letters.