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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?

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    $\begingroup$ 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. $\endgroup$ – reuns Jun 16 at 21:18
  • $\begingroup$ So there is only one proper normal subgroup? Can you please give a reference where I can see this detailed? Thank you! $\endgroup$ – DesmondMiles Jun 16 at 21:30
  • $\begingroup$ $H$ is normal iff $L^H/Q$ is Galois. $\endgroup$ – reuns Jun 16 at 21:58
  • $\begingroup$ 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 $H$? $\endgroup$ – DesmondMiles Jun 16 at 22:01

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