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    The center of $S_3$ is trivial

    P-UFD6Y

    Area
    Algebra
    Topics
    Centralizers and Normalizers, Permutations
    Status
    draft

    Show that \[ Z(S_3)=\{e\}. \]

    Solution

    Let \(z\in Z(S_3)\). If \(z\) is a transposition, then it does not commute with a different transposition; for example, \[ (12)(23)\ne(23)(12). \] So no transposition lies in the center.

    If \(z\) is a \(3\)-cycle, then it does not commute with a transposition. For instance, \[ (123)(12)\ne(12)(123). \] Thus no \(3\)-cycle lies in the center.

    The elements of \(S_3\) are the identity, three transpositions, and two \(3\)-cycles. The only remaining possibility is the identity. Hence \[ Z(S_3)=\{e\}. \]

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