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Consider $p(x) = x^5-2 \in \mathbb{Q}[x]$.

I have proven that the splitting field of $p(x)$ is $F = \mathbb{Q}(\sqrt[5]{2}, \xi)$, where $\xi = e^{\frac{2\pi i}{5}}$. After doing so, I have proven that $$[F: \mathbb{Q} ]=[F: \mathbb{Q} (\sqrt[5]{2}) ] \cdot [ \mathbb{Q} (\sqrt[5]{2}): \mathbb{Q} ]=4\cdot 5 =20. $$

Now, I would like to prove that if we consider $G = \text{Gal}(F/\mathbb{Q})$.

  1. There exists $\sigma \in G$ such that $\sigma(\sqrt[5]{2}) = \xi\sqrt[5]{2}$ and $\sigma(\xi)=\xi$.

  2. There exists $\tau\in G$ such that $\tau(\sqrt[5]{2})=\sqrt[5]{2}$ and $\tau(\xi)=\xi^2$.

  3. $G \cong \mathbb{Z}/(5) \rtimes \mathbb{Z}/(4)$.

MY THOUGHTS

I have been thinking that it may help proving that $G$ is isomorphic to one of the $5$ possible groups of order $20$. In that way, I may could prove the existence of such elements for (1) and (2). Would that be an option? Any hint? For (3) I do not know what to do. I have been thinking to use the following theorem

$G \cong G_1 \rtimes G_2$ (where $\rtimes$ is the semidirect product) if and only if $G$ has two subgroups $H$ and $K$, with $H \vartriangleleft G$, $H \cong G_1$, $K \cong G_2$, $H \cap K = \{1\}$ and $G = HK.$

Although I do not know if it will help me. Could you help me, please?

idriskameni
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    For your (1) and (2), an element of $G$ can be defined by its action on the generators (i.e., $\sqrt[5]{2}$ and $\xi$); what you have to show is that those are actually 'realizable' elements of the Galois group (i.e., that $\sigma$ and $\tau$ are in fact automorphisms.) – Steven Stadnicki Jan 07 '20 at 20:17

1 Answers1

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You know that the Galois group has 20 automorphism, which are determined by their actions on $\sqrt[5]{2}$ and $\zeta$.We also know that it must send $\sqrt[5]{2}$ to any root of $x^5-2$ and must send $\zeta$ to a root of $x^4+x^3+x^2+x+1$. This gives you the exact 20 possibilities! Hence, you can find such a $\tau$ and $\sigma$. Now, consider the subgroups $<\tau>$ and $<\sigma>$. You can show that one of them is normal using the fact that $\mathbb{Q}(\zeta)/\mathbb{Q}$ is a Galois extension (use the Galois correspondence).

mich95
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