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    An irreducible cubic with Galois group not in $A_3$ has Galois group $S_3$

    P-OIG6K

    Area
    Algebra
    Topics
    Classification, Galois Theory, Permutations
    Status
    draft

    Let \(f\in F[x]\) be an irreducible separable cubic. If its Galois group is not contained in \(A_3\), prove that its Galois group is \(S_3\).

    Solution

    Let \(L\) be the splitting field and \[ G=\operatorname{Gal}(L/F)\le S_3 \] via the action on the three roots.

    Because \(f\) is irreducible, \(G\) acts transitively on its roots. Hence \(3\) divides \(|G|\). The transitive subgroups of \(S_3\) are therefore \(A_3\) and \(S_3\): indeed, a subgroup with order divisible by \(3\) contains the unique subgroup \(A_3\) of order \(3\), and its order is either \(3\) or \(6\).

    By hypothesis \(G\not\subseteq A_3\), so \(G\ne A_3\). Consequently \[ G=S_3. \]

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