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    $[S_n,S_n]=A_n$

    E-AMD-Z7OGMPKS

    Area
    Algebra
    Topics
    Commutators, Permutations
    Status
    draft

    Show that \[ [S_n,S_n]=A_n \] for all \(n\ge1\).

    Solution

    Since \[ S_n/A_n\cong C_2 \] is abelian, the commutator subgroup satisfies \[ [S_n,S_n]\le A_n. \]

    For \(n=1,2\), both sides are trivial. Assume \(n\ge3\). The group \(A_n\) is generated by \(3\)-cycles. For distinct \(a,b,c\), \[ [(bc),(ab)]=(abc) \] with the convention \([x,y]=xyx^{-1}y^{-1}\). Hence every \(3\)-cycle lies in \([S_n,S_n]\), so \[ A_n\le[S_n,S_n]. \] Therefore \([S_n,S_n]=A_n\).

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