# Galois group of $x^p - a$, $a$ - squarefree

Let $$p$$ be a prime and $$a > 1$$ be a squarefree positive integer. We wish to understand (at least to some extent) the Galois correspondence in the Galois group $$G$$ of $$x^p - a$$. The splitting field is $$\mathbb{Q}(\omega, \sqrt[p]{a})$$ where $$\omega = e^{\frac{2\pi i}{p}}$$ - one can easily show it is of degree $$p(p-1)$$. I can see that the Galois group is non-Abelian and generated by $$\sigma$$ (fix $$\omega$$, send $$\sqrt[p]{a}$$ to $$\omega \sqrt[p]{a}$$) and $$\theta$$ (fix $$\sqrt[p]{a}$$, send $$\omega$$ to $$\omega^g$$ where $$g$$ is a primitive root mod $$p$$), of orders $$p$$ and $$p-1$$. I guess with some effort one can find suitable relations like $$\theta^{-1}\sigma\theta = ....$$.

What I am particularly interested in - is there a nice way to further the computation and understand which are the normal subgroups of $$G$$ and how many are they in number?

• The Galois group is the group of affine transformations $x \mapsto bx+c$ of $\Bbb{Z/pZ}$, the correspondence is $\omega^x a^{1/p} \mapsto \omega^{bx+c} a^{1/p}$. It is a semidirect product of $\Bbb{Z/pZ}$ with $\Bbb{Z/pZ}^\times$, the translations are a normal subgroup. – reuns Jun 16 at 21:18
• So there is only one proper normal subgroup? Can you please give a reference where I can see this detailed? Thank you! – DesmondMiles Jun 16 at 21:30
• $H$ is normal iff $L^H/Q$ is Galois. – reuns Jun 16 at 21:58
• So you want to describe all normal subgroups by checking the correspondent extensions? If yes, how do we check whether the intermediate extension is Galois in this case in terms of $H$? – DesmondMiles Jun 16 at 22:01