Center of Symmetric Group is Trivial

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

Thus, when $n > 2$, 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 leaves everything else where it is.

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

So:

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

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

Alternative Proof
Somewhat more tersely, but less clearly:

Clearly the identity lies in the center.

Assume $g$ is not the identity, so that $ g(a) = b \ne a $ for some $a$.

As the order is greater than $2$, we may choose $c \ne b, g(b)$ and set $ h = (b \ c) $.

Then, $hg(b) = g(b) \ne g(c) = gh(b)$ (recall that permutations are one-to-one).

This proves that $g$ cannot be in the center and we're done.