# Galois group of solvable quintic is subgroup of Fr20

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?

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 6 '11 at 20:30
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 6 '11 at 20:36
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 $S_5$. – user641 Dec 6 '11 at 21:15