Center of Symmetric Group is Trivial/Proof 1

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Let $n \in \N$ be a natural number.

Let $S_n$ denote the symmetric group of order $n$.

Let $n \ge 3$.

Then the center $\map Z {S_n}$ of $S_n$ is trivial.


From its definition, the identity (here denoted by $e$) of a group $G$ commutes with all elements of $G$.

So by definition of center:

$e \in \map Z {S_n}$

By definition of center:

$\map Z {S_n} = \set {\tau \in S_n: \forall \sigma \in S_n: \tau \sigma = \sigma \tau}$

Let $\pi, \rho \in S_n$ be permutations of $\N_n$.

Let us choose an arbitrary $\pi \in S_n: \pi \ne e, \map \pi i = j, i \ne j$.

Since $n \ge 3$, we can find $\rho \in S_n$ which interchanges $j$ and $k$ (where $k \ne i, j$) and fixes everything else.

It follows that $\rho^{-1}$ does the same thing, and in particular both $\rho$ and $\rho^{-1}$ fix $i$.


\(\ds \map {\rho \pi \rho^{-1} } i\) \(=\) \(\ds \map {\rho \pi} i\)
\(\ds \) \(=\) \(\ds \map \rho j\)
\(\ds \) \(=\) \(\ds k\)


$\map {\rho \pi \rho^{-1} } i = k \ne j = \map \pi i$

From Conjugate of Commuting Elements, 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 $S_n$.


$\map Z {S_n} = \set e$