Symmetric Group is not Abelian

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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 1

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.

$\blacksquare$


Proof 2

Let $a, b, c \in S$.

Let $\alpha$ be the transposition on $S$ which exchanges $a$ and $b$.

Let $\beta$ be the transposition on $S$ which exchanges $b$ and $c$.


Then:

$\alpha \circ \beta$ maps $\tuple {a, b, c}$ to $\tuple {c, a, b}$

while:

$\beta \circ \alpha$ maps $\tuple {a, b, c}$ to $\tuple {b, c, a}$

Thus $\alpha, \beta \in S_n$ such that $\alpha$ does not commute with $\beta$.


Hence the result by definition of abelian group.

$\blacksquare$


Sources

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