Center of Symmetric Group is Trivial

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 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
In the below, let $a, b, c \in \N_n$ and $g, h \in S_n$, which is the Group of Permutations on $\N_n$.

Clearly the identity lies in the center.

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

As the order is greater than $2$, we may choose $c \notin \left\{{b, g \left({b}\right)}\right\}$ and set $h = \left({b \ c}\right)$ (see cycle notation).

Then, $h g \left({b}\right) = g \left({b}\right) \ne g\left({c}\right) = g h \left({b}\right)$.

This proves that $g$ cannot be in the center.

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

Then $\pi \left({j}\right) \ne j$ since permutations are injective.

Since $n \ge 3$, we can find $ k \ne j, k \ne \pi \left({j}\right)$ and $\rho \in S_n$ which interchanges $j$ and $k$ and leaves everything else fixed.

Let $\pi \left({j}\right) = m$. Then $m \ne j, m \ne k$ so $\rho$ leaves $m$ fixed.

Then $ \rho \pi \left({j}\right) = \rho \left({m}\right) = m = \pi \left({j}\right)$ since $\rho$ leaves $m$ fixed.

Now $k = \rho \left({j}\right)$ by definition of $\rho$.

So $\pi \left({k}\right) = \pi \rho \left({j}\right)$.

But $\pi \left({j}\right) \ne \pi \left({k}\right)$ since permutations are injective.

Thus $\rho \pi \left({j}\right) \ne \pi \rho \left({j}\right)$.

So arbitrary $\pi \ne e$ is not in the center since there exists a $\rho$ which it does not commute with.

Thus only $e$ is in the center, which, by definition, is trivial.