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Let $K$ be the splitting field of $x^5-3 \in \mathbb{Q}[x]$.

We can see $K = \mathbb{Q}(3^{1/5}, \zeta_5)$ where $\zeta_5 = e^{2 \pi i/5}$, and $[K: \mathbb{Q}] = 20$. It's easy to see $\sqrt{5} \in \mathbb{Q}(\zeta_5)\subseteq K$.

Let $H= \operatorname{Gal}(K/\mathbb{Q}(\sqrt{5})$). Does $H$ have to be abelian?

Edit: The Galois group of the big extension $K/\mathbb{Q}$ is the group generated by $\sigma$ (order 5 element) and $\tau$ (order 4 element) defined by $\sigma: 3^{1/5} \mapsto 3^{1/5} \zeta_5$ and $\tau: \zeta_5 \mapsto \zeta_5^2$.

I don't know what this group is, but it's definitely not abelian since if it's abelian the extension $\mathbb{Q}(3^{1/5})/\mathbb{Q}$ is Galois.

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You should be able to find the group of $K$ over $\bf Q$, and then find $H$ as a subgroup of that group. – Gerry Myerson Jul 22 '12 at 11:33
@colge Have you computed the Galois group of the big extension ? – user38268 Jul 22 '12 at 11:35
@BenjaLim I have tried this (see my edit of the quesion) but I didn't get exact representation of this group. – colge Jul 22 '12 at 12:01
@colge It's the Frobenius group $F_{20}$. – Cocopuffs Jul 22 '12 at 12:09
You should be able to find a relation involving $\sigma$ and $\tau$, decide whether there are any elements of order 10, etc. – Gerry Myerson Jul 22 '12 at 12:25

The Galois group is of order 20, generated by $x$ of order 5 any $y$ of order 4, where $yxy^{-1}=x^{3}$. Its the Frobenius group of order 20. The subgroup of order 10 is generated by $x$ and $y^2$. It is a dihedral group.

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If the group is abelian, how do you explain the fact that $\mathbb{Q}[3^{1/5}]/\mathbb{Q}$ is not an abelian extension, as the OP mentionned? – M Turgeon Jul 22 '12 at 15:57
I think you've got the dicyclic group of order 20; the Frobenius group has relation $yxy^{-1}=x^3$. – Gerry Myerson Jul 23 '12 at 4:31
@GerryMyerson-you are right. So the subgroup of order 10 is dihedral, not abelian. – i. m. soloveichik Jul 28 '12 at 17:57

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