Show that $S(X)$ is abelian if and only if $|X|\le 2$

Question

I am trying to prove that the symmetric group $S(X)$ is only abelian for sets $X$ with cardinality below $2$. But I only get as far as proving that $S_2$ is abelian, and I can provide examples of $S_3$ and higher of not being abelian, for not sufficient for a proof.

Also, I can prove it by using the principle that the symmetric group is abelian if it consists of disjoint permutations, but the proof is supposed to be done without ant knowledge of permutations.

Can anyone point me in the right direction?

Thanks!

Edit: Obviously it is sufficient to provide an example of $S_3$ not being abelian. Question answered!

Answer 1

Note that $(12)(13)=(132)$ but $(13)(12)=(123)$.

Answer 2

To show a group is non-abelian it is enough to find just one counterexample.

And every set of three or more members has at least one subset with exactly three members.

And within that subset, you have found an example.