Center of Symmetric Group is Trivial

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Theorem

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.


Proof 1

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 $i$.


So:

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


So:

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

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 $S_n$.

Hence:

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

$\blacksquare$


Proof 2

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.

$\blacksquare$


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