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Question: Let $K$ be a splitting field for $x^5-2$ over $\mathbb{Q}.$

$(1)$ Determine $[K:\mathbb{Q}].$

$(2)$ Show that $Gal(K/\mathbb{Q})$ is non-abelian.

For $(1),$ note that roots of $x^5-2$ are $$\sqrt[5]{2}, \sqrt[5]{2}\omega, \sqrt[5]{2}\omega^2, \sqrt[5]{2}\omega^3, \sqrt[5]{2}\omega^4$$ where $\omega$ is the $5$th root of unity, that is, $$\omega = e^{\frac{2\pi i}{5}}.$$ So we have $$K = \mathbb{Q}(\sqrt[5]{2}, \omega),$$ implying that $$[K:\mathbb{Q}] = [K:\mathbb{Q}(\omega)] \cdot [\mathbb{Q}(\omega):\mathbb{Q}] = 5 \times 4 = 20.$$

For $(2),$ I know that $Gal(K/\mathbb{Q})$ is isomorphic to the Dihedral group with $20$ elements. Since such Dihedral group is non-abelian, so is $Gal(K/\mathbb{Q}).$ However, I do now know how to prove it.

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Idonknow
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  • $|\Bbb Q(\omega):\Bbb Q|=4$. – Angina Seng Nov 24 '18 at 05:12
  • Consider the action of the Galois group as permutations of the roots. – Angina Seng Nov 24 '18 at 05:12
  • You can find two elements in the Galois group that don't commute. – Seewoo Lee Nov 24 '18 at 05:12
  • @LordSharktheUnknown Thanks for pointing out my mistake. Corrected it. – Idonknow Nov 24 '18 at 05:14
  • The Galois group is not dihedral. – Angina Seng Nov 24 '18 at 05:16
  • This Galois group is not dihedral. It is isomorphic to the holomorph of $C_5$, i.e. the semi-direct product $C_5\rtimes Aut(C_5)=C_5\rtimes C_4$. The dihedral group of order twenty has a non-trivial center but the center of this group is trivial. – Jyrki Lahtonen Nov 24 '18 at 06:29
  • A way forward is to observe that an automorphism $\sigma$ is fully determined once you know $\sigma(\omega)$ and $\sigma(\root5\of2)$. There are four (resp. five) choices for those two images. Because we know that there will be twenty automorphisms, all combinations must occur exactly once. After that point you can identify the Galois group as a subgroup of the permutations of the five roots of $x^5-2$. – Jyrki Lahtonen Nov 24 '18 at 06:34
  • A hint for a shorter way of settling part (2): If the Galois group were abelian, then all its subgroups would be _____. Consequently all the intermediate fields would be _____ extensions of the rationals. But the extension $\Bbb{Q}(\root5\of2)/\Bbb{Q}$ is not _____, so this cannot be the case. The same word goes into all the blank spaces :-) – Jyrki Lahtonen Nov 24 '18 at 06:36
  • @JyrkiLahtonen The missing word is normal. – Idonknow Nov 24 '18 at 06:44
  • Correct. Feel free to turn that into an answer :-) – Jyrki Lahtonen Nov 24 '18 at 06:57

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