# Order of Alternating Group

## Theorem

Let $n \in \Z$ be an integer such that $n > 1$.

Let $A_n$ be the alternating group on $n$ letters.

Then:

$\order {A_n} = \dfrac {n!} 2$

where $\order {A_n}$ denotes the order of $A_n$.

## Proof

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

$\index {S_n} {A_n} = 2$

where $\index {S_n} {A_n}$ denotes the index of $A_n$ in $S_n$.

$\order {S_n} = n!$

The result follows from Lagrange's Theorem.

$\blacksquare$