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.

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

Proof
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 definition, a permutation is a bijection.

All bijections have a unique inverse.

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.