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.

From Alternating Group is Normal Subgroup of Symmetric Group:
 * $\index {S_n} {A_n} = 2$

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

From Order of Symmetric Group:
 * $\order {S_n} = n!$

The result follows from Lagrange's Theorem.