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    $S_3$ is not nilpotent

    P-MMYH6

    Area
    Algebra
    Topics
    Counterexamples, Nilpotent Groups, Permutations
    Status
    draft

    Show that \(S_3\) is not nilpotent.

    Solution

    A finite nilpotent group has all Sylow subgroups normal.

    In \(S_3\), the Sylow \(2\)-subgroups are \[ \langle(12)\rangle, \qquad \langle(13)\rangle, \qquad \langle(23)\rangle. \] Thus there are three Sylow \(2\)-subgroups, so none is normal.

    Therefore \(S_3\) is not nilpotent.

    Equivalently, its lower central series does not terminate at the identity: since \[ [S_3,S_3]=A_3 \] and \([A_3,S_3]=A_3\), the lower central series stabilizes at \(A_3\ne1\).

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