Show that all permutation group $S_{n}$ when $n\geq 3$ is not abelian group

Question

Please check my proof

From Cayley's theorem every group isomorphism with permutation group,then every permutation

group $S_{n}$ when $n\geq 3$ Isomorphism with $S_{3}$

consider P= $\begin{pmatrix} 1&2 &3 \\ 3&2 &1 \end{pmatrix}$

and

$Q=\begin{pmatrix} 1 & 2 & 3\\ 1 & 3 & 2 \end{pmatrix}$

the product PQ = $\begin{pmatrix} 1 &2 &3 \\ 1& 3 &2 \end{pmatrix}$

and QP = $\begin{pmatrix} 1 &2 &3 \\ 2 &1 &3 \end{pmatrix}$

That shows it is not commutative

Since $S_{n}$ When $n> 3$ isomorphism with $S_{3}$

Then $S_{n}$ when $n\geq 3$ are not abelian group

Answer 1

If you insist to use Cayley's theorem, you know that every $S_n , n\geq 3$ contains a copy of $S_3$ in it (in some isomorphic form). Proving $S_3$ is not abelian sufficiently proves your claim and it is easy to show that $S_3$ is evidently not abelian.

Alternatively, the easiest is to note that if two cycles are disjoint, then they commute. One can easily find an example of two non-commuting cycles in $S_3$, and this immediately implies that $Z(S_3) \neq S$, and since $S_3 \subset S_4 \subset ... S_n$,

Then do the case work for $S_1, S_2$ and show they are abelian. They have relatively few elements so it should be easy to analyze.

Answer 2

You can implement the "$S_3$ argument" directly in $S_n$.

As soon as $n\ge 3$, you'll be able to build up $\sigma:=(1i)\in S_n$ and $\tau:=(1ij)\in S_n $. But then: $$\sigma\tau=(ij)$$ whereas: $$\tau\sigma=(1j)$$ and hence $\sigma\tau\ne\tau\sigma$.