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    $[A_n,A_n]=A_n$ for $n\geq 5$

    P-XEOMT

    Area
    Algebra
    Topics
    Commutators, Permutations, Simple Groups
    Status
    draft
    • Show that \([A_n, A_n] = A_n\) for \(n\geq 5\), so \(A_{n\geq 5}\) is nonabelian.
    Solution

    For \(n\ge5\), the group \(A_n\) is simple and nonabelian. The commutator subgroup \[ [A_n,A_n] \] is characteristic, hence normal, in \(A_n\). It is nontrivial because \(A_n\) is nonabelian: for example, the \(3\)-cycles \[ (123),\qquad(345) \] do not commute.

    By simplicity, any nontrivial normal subgroup of \(A_n\) is all of \(A_n\). Therefore \[ \boxed{[A_n,A_n]=A_n\qquad(n\ge5)}. \] In particular, \(A_n\) is perfect and therefore nonabelian.

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