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 $Z \left({S_n}\right)$ of $S_n$.

Thus $S_n \ne Z \left({S_n}\right)$.

The result follows by Group equals Center iff Abelian.