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We know that the splitting field of $x^5 - 2 $ over $\mathbb Q$ is $\mathbb Q(2^{1/5}, \rho)$, where $\rho$ is a fifth root of unity.

Therefore, $\left[\mathbb Q(2^{1/5} , \rho) : \mathbb Q \right] = 20 $. Let $G$ be the Galois group of $\mathbb Q(2^{1/5} , \rho)$. Then $|G| = 20$.

How to find the Galois group $G$?

Can we generalize to the Galois group of the splitting field of $x^p -2$ over $\mathbb Q$?

user26857
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Struggler
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    Very closely related. Could actually be a duplicate, but for hopefully understandable reasons I'm a bit reluctant to use my AA-powervote. The answer there is more about hints, but the asker of the duplicate elaborates nicely IMHO. Anyway, you get $C_p\rtimes C_{p-1}$ as the answer. – Jyrki Lahtonen Dec 18 '14 at 05:02

2 Answers2

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By Sylow's Theorem, we have that $P_5\trianglelefteq G$ is normal, which we expect since $\Bbb Q(\rho)/\Bbb Q$ is Galois. Since we know that $\text{Gal}(\Bbb Q(\rho)/\Bbb Q)\cong \Bbb Z/4$ we see that there is a surjection $G\to\Bbb Z/4$, hence we see that we have the short, exact sequence:

$$1\to \Bbb Z/5\stackrel{i}{\longrightarrow} G\stackrel{p}{\longrightarrow} \Bbb Z/4\to 1.$$

Now since the map $G\to \Bbb Z/4$ is surjective, $G$ has an element of order $4$, hence we have a section of the surjection, i.e. a map $\psi: \Bbb Z/4\to G$ such that $\psi\circ p= \text{id}_{P_2}$, and we conclude $G$ is a semidirect product, but since $P_2$ is not normal--$\Bbb Q \left(\sqrt[5] 2\right) /\Bbb Q$ is not Galois after all--we have that the product is not direct so:

$$G\cong \Bbb Z/4\ltimes \Bbb Z/5$$

We can generalize as well: With $\zeta_p$ a primitive $p^{th}$ root of $1$, we have $\text{Gal}\left(\Bbb Q(\zeta_p)/\Bbb Q\right)$ is always isomorphic to $\left(\Bbb Z/p\right)^*$ which is cyclic when $p$ is prime, the sequence will always split by the surjectivity of the map from $G\to\Bbb Z/(p-1)$ and you will always get the semi-direct product from the lack of being abelian.

Adam Hughes
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It's the Frobenius group $C_5\rtimes C_4$.

Hint: Start with $$\sigma:\left\{\begin{array}{l}2^{1/5}\mapsto \rho 2^{1/5}\\ \rho\mapsto \rho\end{array}\right.\hspace{15pt}\text{and}\hspace{15pt}\tau:\left\{\begin{array}{l}2^{1/5}\mapsto 2^{1/5}\\ \rho \mapsto \rho^2\end{array}\right.$$ To generalize, take a look at the order $\bmod p$ of the power in $\tau$, and play around with the conjugation $\tau^{-1}\sigma\tau$ until your group comes out to match the degree of your extension.

Alexander Gruber
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