Search the corpus

    $S_4$ is solvable and nonabelian

    P-6URED

    Area
    Algebra
    Topics
    Permutations, Solvable Groups, Subgroup Series
    Status
    draft

    Prove that the symmetric group \(S_4\) is a non-abelian and solvable group.

    Solution

    The group \(S_4\) is nonabelian because, for example, \[ (12)(23)=(123)\ne(132)=(23)(12). \]

    For solvability, consider \[ 1\triangleleft V_4\triangleleft A_4\triangleleft S_4, \] where \[ V_4=\{e,(12)(34),(13)(24),(14)(23)\}. \] The subgroup \(A_4\) is normal of index \(2\) in \(S_4\). The subgroup \(V_4\) is normal in \(S_4\) because conjugation preserves cycle type, hence is normal in \(A_4\) as well. The factors are \[ V_4,\qquad A_4/V_4\cong C_3,\qquad S_4/A_4\cong C_2, \] all abelian. Therefore \(S_4\) is solvable.

    Thus \(S_4\) is solvable but nonabelian.

    Source Collections