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 $$ g(a) = b \neq a $$ for some $$ a $$.

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

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

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