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    The Galois group of an irreducible quintic with two non-real roots is $S_5$

    P-OU7QZ

    Area
    Algebra
    Topics
    Galois Theory, Permutations, Polynomials
    Status
    draft

    Let \(f\in \QQ[x]\) be an irreducible quintic with exactly two non-real roots. Prove that its Galois group over \(\QQ\) is \(S_5\).

    Solution

    Let \(L\) be the splitting field and let \[ G=\operatorname{Gal}(L/\QQ)\le S_5 \] act on the five roots.

    Because \(f\) is irreducible, this action is transitive. Hence \(5\mid |G|\), so by Cauchy’s theorem \(G\) contains an element of order \(5\), necessarily a \(5\)-cycle.

    Since \(f\) has rational coefficients and exactly two non-real roots, complex conjugation fixes the three real roots and swaps the two non-real conjugate roots. Thus complex conjugation acts as a transposition in \(G\).

    A \(5\)-cycle together with any transposition generates \(S_5\): after conjugating the transposition by powers of the \(5\)-cycle one obtains enough adjacent/star transpositions to generate all of \(S_5\). Therefore \[ G=S_5. \]

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