Center of Symmetric Group is Trivial/Proof 2
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
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:
\(\ds \map {\rho \pi} j\) | \(=\) | \(\ds \map \rho m\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds m\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \pi j\) |
By definition of $\rho$:
- $k = \map \rho j$
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.
$\blacksquare$