Show that the center of \(S_n\) for \(n\geq 4\) is trivial.
Solution
1. Let \(\sigma \in S_n\) with \(\sigma \neq \id\), and fix \(i\) with \(\sigma(i) = j \neq i\).
2. There is a point \(k \notin \ts{i, j}\).
\(\ts{1, \dots, n}\) has \(n \geq 4\) elements and \(\ts{i,j}\) has two.
3. \(\sigma\) does not commute with the transposition \(\tau = (j\, k)\).
3.1. \((\sigma\tau)(i) = \sigma(\tau(i)) = \sigma(i) = j\), since \(i \notin \ts{j,k}\) and so \(\tau\) fixes \(i\).
3.2. \((\tau\sigma)(i) = \tau(\sigma(i)) = \tau(j) = k\).
3.3. \(j \neq k\) by the choice of \(k\), so \(\sigma\tau \neq \tau\sigma\).