Order of Symmetric Group

Theorem

Let $S_n$ be the symmetric group on $n$ letters.

Then $S_n$ has $n!$ elements (see factorial).

Proof

A direct application of Cardinality of Set of Bijections.

$\blacksquare$

Example

Thus, when $n = 3$, there are $3 \times 2 \times 1 = 6$ permutations:

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

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

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 7$: Example $7.5$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 30 \alpha$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 38$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 78$