Symmetric Group is not Abelian/Proof 2

Theorem

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

Then $S_n$ is not abelian.

Proof

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

• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Examples of groups $\text{(iii)}$