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

    E-AMD-74T5EHRR

    Area
    Algebra
    Topics
    Centralizers and Normalizers, Permutations
    Status
    draft

    Show that \(Z(S_n) = 1\) for \(n\geq 3\)

    Solution

    Let \(\sigma\in S_n\) be nonidentity. Choose \(i\) with \(\sigma(i)=j\ne i\). Since \(n\ge3\), choose \(k\notin\{i,j\}\), and let \(\tau=(j\ k)\).

    Then \(\tau(i)=i\), so \[ (\sigma\tau)(i)=\sigma(i)=j, \] while \[ (\tau\sigma)(i)=\tau(j)=k. \] Thus \(\sigma\tau\ne\tau\sigma\), so \(\sigma\notin Z(S_n)\). Therefore the identity is the only central element: \[ \boxed{Z(S_n)=1\qquad(n\ge3).} \]

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