Symmetric Group is not Abelian/Proof 1

Theorem

Let $S_n$ be the symmetric group of order $n$ where $n \ge 3$.

Then $S_n$ is not abelian.

Proof

Let $\alpha \in S_n$ such that $\alpha$ is not the identity mapping.

From Center of Symmetric Group is Trivial, $\alpha$ is not in the center $\map Z {S_n}$ of $S_n$.

Thus $S_n \ne \map Z {S_n}$.

The result follows by definition of abelian group.

$\blacksquare$