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    $[A_4,A_4]\cong\ZZ_2^2$

    E-AMD-JFCWSOUA

    Area
    Algebra
    Topics
    Commutators, Permutations
    Status
    draft

    Show that \([A_4,A_4]\cong\mathbb Z_2^2\).

    Solution

    Let \[ V=\{1,(12)(34),(13)(24),(14)(23)\}\cong C_2\times C_2. \]

    1. We have \([A_4,A_4]\le V\).

    The subgroup \(V\) is normal in \(A_4\), and \[ A_4/V\cong C_3 \] is abelian. Therefore the commutator subgroup is contained in \(V\).

    2. We have \(V\le[A_4,A_4]\).

    With \(x=(123)\) and \(y=(124)\), \[ [x,y]=xyx^{-1}y^{-1}=(12)(34). \] The commutator subgroup is normal in \(A_4\), so it contains all \(A_4\)-conjugates of \((12)(34)\), namely the three nonidentity elements of \(V\). Hence it contains \(V\).

    Thus \[ [A_4,A_4]=V\cong\mathbb Z_2^2. \]

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