Alternating Group on More than 3 Letters is not Abelian

Theorem

Let $n$ be an integer such that $n > 3$.

Then the $n$th alternating group $A_n$ is not abelian.

Proof

Let $\tuple {1, 2, 3}$ and $\tuple {2, 3, 4}$ be elements of $A_n$.

Then we have:

\(\ds \tuple {1, 2, 3} \tuple {2, 3, 4}\)

\(=\)

\(\ds \tuple {1 2} \tuple {3 4}\)

\(\ds \tuple {2, 3, 4} \tuple {1, 2, 3}\)

\(=\)

\(\ds \tuple {1 3} \tuple {2 4}\)

So:

$\tuple {1, 2, 3} \tuple {2, 3, 4} \ne \tuple {2, 3, 4} \tuple {1, 2, 3}$

and so $A_n$ is not abelian.

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 84$