What is the center \(Z(S_n)\) of the symmetric group \(S_n\)? Prove it.
Solution
For \(n=1\), \(S_1\) is trivial, so \(Z(S_1)=S_1\). For \(n=2\), \(S_2\) is abelian, so \(Z(S_2)=S_2\).
Now let \(n\ge3\) and suppose \(\sigma\in Z(S_n)\) is nontrivial. Choose \(a\) with \(\sigma(a)=b\ne a\), and choose \(c\) distinct from \(a,b\). Let \(\tau=(bc)\). Then \(\tau(a)=a\), so \[ (\sigma\tau)(a)=\sigma(a)=b, \] whereas \[ (\tau\sigma)(a)=\tau(b)=c. \] Thus \(\sigma\tau\ne\tau\sigma\), contradicting \(\sigma\in Z(S_n)\).
Therefore \[ Z(S_n)= \begin{cases} S_n,&n=1,2,\\ \{1\},&n\ge3. \end{cases} \]