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 $\map Z {S_n}$ of $S_n$.
Thus $S_n \ne \map Z {S_n}$.
The result follows by definition of abelian group.
$\blacksquare$