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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.2012-07-22
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    @colge Have you computed the Galois group of the big extension ?2012-07-22
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    @BenjaLim I have tried this (see my edit of the quesion) but I didn't get exact representation of this group.2012-07-22
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    @colge It's the Frobenius group $F_{20}$.2012-07-22
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    You should be able to find a relation involving $\sigma$ and $\tau$, decide whether there are any elements of order 10, etc.2012-07-22
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    I'm not sure what's left to do here, since you've asked whether $H$ has to be abelian and then you've determined that it isn't. You might want to show that $\sigma$ and $\tau^2$ are in $H$, but $\tau$ isn't. Do you know that there is a simple theorem that completely classifies the groups of order $2p$ for prime $p$?2012-07-23
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    Can someone please explain why $H$ is abelian imply $\mathbb{Q}(3^{1/5})/\mathbb{Q}$ is Galois ?2012-07-28

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