Center of Symmetric Group is Trivial/Proof 2

Proof
Let $\pi \in S_n$ such that $\pi \ne e$ be arbitrary.

Let $i, j \in \set {1, 2, \ldots, n}$ such that $\map \pi i = j \ne i$.

Let $m = \map \pi j$.

Then $m \ne j$.

Since $n \ge 3$, there exists $k \in \N$ such that $k \ne j, k \ne m$.

Let $\rho \in S_n$ interchange $j, k$ and fix everything else.

Then $\rho$ fixes $m$.

Then:

By definition of $\rho$:
 * $k = \map \rho j$

So:
 * $\map \pi k = \map {\pi \rho} j$

But because permutations are injective:
 * $\map \pi j \ne \map \pi k$

Thus:
 * $\map {\rho \pi} j \ne \map {\pi \rho} j$

That is, there exists a $\rho$ with which $\pi$ does not commute.

So $\pi \notin \map Z {S_n}$.

As $\pi \ne e$ is arbitrary, it follows that only $e$ is in $\map Z {S_n}$.

That is:
 * $\map Z {S_n} = \set e$

Hence, by definition, $\map Z {S_n}$ is trivial.