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Why is it true that any solvable quintic polynomial in has a Galois group that is a subgroup on the Frobenius group of order 20?

Thanks in advance.

Justin Zhou
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    The Galois group of a quintic is a subgroup of $S_5$. If the quintic is irreducible then you must have a solvable group of $S_5$ of order a multiple of $5$ (since the action must be transitive on the roots). All such subgroups are isomorphic to subgroups of that Frobenius group. Of course, the statement is false as given, since the Galois group of $(x^3-2)(x^2-1)$ has $6$, and so cannot be a subgroup of a group of order $20$. – Arturo Magidin Dec 06 '11 at 20:30
  • Your title and your question are not equivalent. The title asks about realizing the Frobenius group of order 20 as a Galois group. The body asks why every Galois group of a certain type is a subgroup of a Frobenius group of order 20. Please try to make titles correspond to the question. – Arturo Magidin Dec 06 '11 at 20:31
  • Thanks so much for your response, @ArturoMagidin, and sorry for the misleading title. I'll edit it asap.

    The problem I am working on is to show that when p, q are such that $t^5 + 5p t^3 + 5p^2 t + q$ is irreducible, then the Galois group is $\text{Fr}_{20}$. But I don't see how this is true; I'm looking at the fields

    $\mathbb{Q} \subset \mathbb{Q}(e^{2 \pi i/5}) \subset \mathbb{Q} \left(e^{2 \pi i/5}, \sqrt[5]{\dfrac{-q + \sqrt{q^2 + 4p^5}}{2}} \right)$

    and it seems like the degree of the first extension is $4$ while the degree of the second is $10$, implying $|G| = 40$. Thanks in advance!

     – Justin Zhou Dec 06 '11 at 20:36
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    This is true from a group theoretic perspective: when the polynomial is irreducible, your Galois group is transitive: transitive groups of prime degree are either doubly transitive, or subgroups of the affine group $AGL(1,5)$ [this is your Frobenius group.] A doubly transitive group has order divisible by 20, and hence in $S_5$ can only be $AGL(1,5)$, $A_5$, or $S_5$. –  Dec 06 '11 at 21:15
  • Historical trivia: it seems to me the desired result is equivalent to a special case of the main result of Galois' original essay, "Memoir sur la resolution des equations algebriques." What Galois actually stated and proved was that an irreducible polynomial of prime degree is solvable in radicals if and only if all of its roots are rationally expressible in terms of any two of its roots. – Ben Blum-Smith Jan 15 '12 at 18:54
  • Incidentally, Galois' proof was to show that the Galois group of a solvable irreducible polynomial of prime degree is affine, and that an affine group is solvable. – Ben Blum-Smith Jan 15 '12 at 18:56

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