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 $\left({a, b, c}\right)$ to $\left({c, a, b}\right)$

while:
 * $\beta \circ \alpha$ maps $\left({a, b, c}\right)$ to $\left({b, c, a}\right)$

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

Hence the result by definition of abelian group.