Center of Symmetric Group is Trivial

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

Definitions:

The Symmetric Group of Order $$n$$ is denoted $$S_n$$ and is isomorphic to the group of all permutations of the $$n$$ elements of $$\mathbb {N}_n$$, that is, $$\left\{ {0, 1, 2, \ldots, n-1, n}\right\}$$.

The centre of a group $$G$$ (American: "center"), denoted $$Z\left({G}\right)$$ is the set of elements of $$G$$ that commute with all other elements of $$G$$.

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