Galois group of $X^5-5$

Question

I am trying to calculate this galois group of $x^5-5$ over $\mathbb Q$. I know that there is a tower of extensions with groups $C_4$ and $C_5$ respectively, so the group is order 20. I am guessing it isn't $C_{20}$ though. Could I have hint?

Answer 1

Note right away that your Galois group $G$ cannot be $C_{20}$ because $G$ is not abelian (the extension $Q(\sqrt [5] 5)/Q$ is not normal). To determine $G$, allow me a development which could be used in a more general situation.

Here, because 4 and 5 are coprime, an extension $G$ of $C_4$ by $C_5$ is necessarily split, i.e. $G$ is a semi-direct product (this is a classical theorem in group theory). So $G$ admits a subgroup $H\cong C_4$ (here the subgroup fixing $Q(\sqrt [5] 5)$), and you can describe $G$ explicitly by following the action on the roots of $X^5-5$ of $H$ and of the subgroup fixing $Q(\zeta_5)$.

If you are interested only in the isomorphism type of $G$, without any calculation, you need first to determine the action of $C_4$ on $C_5$. Such an action is determined by some $\theta \in Hom (C_4, Aut(C_5))\cong Hom (C_4, C_4)$ (because $Aut(C_5)$ is cyclic of order 4). Fixing a generator $C$ of $C_4$ and writing $\theta (c)=c^k,k=1,2,3$ (exclude $0$ because $G$ is not abelian), you get 3 possible isomorphic types for $G$, which are the dihedral $D_{20}$, the dicyclic $Dic_{20}$ and the general affine $GA(1,5)$, see https://groupprops.subwiki.org/wiki/Groups_of_order_20. Here the Galois action easily determines $k$ and shows that $G\cong GA(1,5)$. See also the transitivity argument given by @user10354138 .