Symmetric Group is not Abelian/Proof 1

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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 $Z \paren {S_n}$ of $S_n$.

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

The result follows by Group equals Center iff Abelian.

$\blacksquare$