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    The center of $S_n$ is trivial for $n \geq 3$

    P-JPTMN

    Area
    Algebra
    Topics
    Centralizers and Normalizers, Permutations
    Status
    draft

    Prove that the center of the symmetric group \(S_n\) is trivial for all \(n \ge 3\): \[Z(S_n) = \{e\} \quad \text{for all } n \ge 3.\]

    Solution

    Let \(\sigma\in Z(S_n)\) with \(n\ge3\). Suppose \(\sigma\ne1\). Then for some \(i\), \[ \sigma(i)=j\ne i. \] Choose \(k\) distinct from both \(i\) and \(j\), and let \(\tau=(jk)\). Since \(\tau\) fixes \(i\), \[ (\sigma\tau)(i)=\sigma(i)=j, \] whereas \[ (\tau\sigma)(i)=\tau(j)=k. \] Thus \(\sigma\tau\ne\tau\sigma\), contradicting centrality. Therefore \(\sigma=1\), so \[ Z(S_n)=\{1\} \] for every \(n\ge3\).

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