Center of Symmetric Group is Trivial

Theorem
The centre of the Symmetric Group of order greater than 3 is trivial.

Proof
We take as an axiom that 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 $$\mathbb {N}_n\,$$.

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:

$$\rho \pi \rho^{-1} \left({i}\right)\,$$

$$= \rho \pi \left({i}\right)\,$$

$$= \rho \left({j}\right)\,$$

$$= k\,$$

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