# Order of Symmetric Group

## Theorem

Let $S$ be a finite set of cardinality $n$.

Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$.

Then $\struct {\map \Gamma S, \circ}$ has $n!$ elements (see factorial).

## Proof

A direct application of Cardinality of Set of Bijections.

$\blacksquare$

## Examples

### $3$rd Symmetric Group

Let $S = \set {1, 2, 3}$.

There are $3 \times 2 \times 1 = 6$ permutations on $S$:

$\begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{bmatrix} \qquad \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{bmatrix} \qquad \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{bmatrix}$

$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \qquad \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{bmatrix} \qquad \begin{bmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{bmatrix}$

and so $\struct {\map \Gamma S, \circ}$ has $6$ elements.