# $S_n$ is an abelian group iff $n \le 2$

Assume that $S_n$ consists of all bijective functions like $f:\{1,\dots,n\} \to \{1,\dots,n\}$. Consider the group of $S_n$ which the combination of functions as the operation defined on it.

Question:
Prove that $S_n$ is abelian iff $n\le 2$.

Note 1: I considered two members of $S_n$ like $\sigma$ and $\lambda$. I want to show that if $n \ge 3$, then $S_n$ is not abelian. Assume that it is (the objective is to reach a contradiction.)

If $S_n$ is abelian for some $n \ge 3$, Then for all $i \in \{1,\dots,n\}$, We have:

$\lambda(\sigma(i)) = \sigma(\lambda(i))$

This should result in a contradiction... But i can't reach it... Any help?

Note 2: One of my classmates considered these two members of $S_n$:
$$\sigma = \begin{pmatrix} 1 & 2 & \dots & n\\ n & n-1 & \dots & 1\\ \end{pmatrix}$$ $$\lambda = \begin{pmatrix} 1 & 2 & \dots & n\\ 2 & 1 & \dots & n\\ \end{pmatrix}$$ Then showed that $\sigma \lambda \neq \lambda\sigma$. Because there exist such members in $S_n$ for all $n \ge 3$, He claimed that the problem is solved. But i want a more general answer (If its possible.) By the way, is this solution true?

Saying that $S_n$ is abelian means that $\sigma\lambda=\lambda\sigma$ for all $\lambda$ and $\sigma$.
Of course, you can take $\sigma = (1 2)$ and $\lambda = (2 3)$ with $S_n$, $n \geq 3$.
$S_n$ is not abelian for $n\ge 3$ because $(12)(13)\ne(13)(12)$.