Center of Symmetric Group is Trivial/Proof 1

Theorem
The center of the Symmetric Group of order $3$ or greater is trivial.

Thus, when $n > 2$, the symmetric group of order $n$ is not abelian.

Proof
From its definition, the identity of a group (here denoted by $e$) commutes with all elements of a group.

So $e \in Z \left({G}\right)$.

By definition:
 * $Z \left({S_n}\right) = \left \{ {\tau \in S_n: \forall \sigma \in S_n: \tau \sigma = \sigma \tau} \right\}$

Let $\pi, \rho \in S_n$ be permutations of $\N_n$.

Let us choose an arbitrary $\pi \in S_n: \pi \ne e, \pi \left({i}\right) = j, i \ne j$.

Since $n \ge 3$, we can find $\rho \in S_n$ which interchanges $j$ and $k$ (where $k \ne i, j$) and fixes everything else.

It follows that $\rho^{-1}$ does the same thing, and in particular both $\rho$ and $\rho^{-1}$ fix $i$.

So:

So $\rho \pi \rho^{-1} \left({i}\right) = k \ne j = \pi\left({i}\right)$.

From Conjugate of Commuting Elements, if $\rho$ and $\pi$ were to commute, $\rho \pi \rho^{-1} = \pi$. But they don't.

Whatever $\pi \in S_n$ is, you can always find a $\rho$ such that $\rho \pi \rho^{-1} \ne \pi$.

So no non-identity elements of $S_n$ commute with all elements of $S_n$.

Hence, $Z \left({S_n}\right) = \left\{ {e}\right\}$.