Center of Symmetric Group is Trivial/Proof 2

Proof
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 fixes everything else.

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

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

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 $\rho$ with which $\pi$ does not commute.

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