Center of Symmetric Group is Trivial

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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.

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 $i$.


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


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

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 $S_n$.


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


Proof 2

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 $m$.


\(\ds \map {\rho \pi} j\) \(=\) \(\ds \map \rho m\)
\(\ds \) \(=\) \(\ds m\)
\(\ds \) \(=\) \(\ds \map \pi j\)

By definition of $\rho$:

$k = \map \rho j$


$\map \pi k = \map {\pi \rho} j$

But because permutations are injective:

$\map \pi j \ne \map \pi k$


$\map {\rho \pi} j \ne \map {\pi \rho} j$

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.


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