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.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 7.4$. Kernel and image: Example $142$
• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 81$
• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): alternating group
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): alternating group
• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): alternating group