0

If $K$ is a splitting field of $x^5-11$ over $\Bbb Q$. Find $Gal(K/\Bbb Q)$ and find all intermediate subfields.

Now $x^5-11$ is irreducibles and splitting field of $x^5-11$ is $\Bbb Q(\sqrt[5]11, a)$ where $a$ is $5$ -th root of unity. Again $[\Bbb Q(\sqrt[5]11, a): \Bbb Q]=20$. I can trace all the $20$ elements but can't see what is $Gal(K/\Bbb Q)$ and all the intermediate fields. So can anyone help me in this if I have not done anything wrong.

Ri-Li
  • 9,466
  • 4
  • 56
  • 104
  • If anyone request I can add the group elements but I have to write a lot then, so I have skipped. – Ri-Li Nov 01 '15 at 20:57
  • Have you heard about semidirect preoducts? See http://sierra.nmsu.edu/morandi/notes/semidirect.pdf and take a look at examples 6 – Klaramun Nov 01 '15 at 21:00
  • yeah i know semidirect product.. – Ri-Li Nov 01 '15 at 21:06
  • But what is the group and how to calculate it? – Ri-Li Nov 01 '15 at 21:34
  • Up to isomorphism there are only five groups of order $20$, and one can narrow down the possibilities with a little effort. http://groupprops.subwiki.org/wiki/Groups_of_order_20#The_list – Travis Willse Nov 01 '15 at 22:04
  • @Travis
    Which group is isomorphic then this group of galois?     – user425181 Nov 12 '17 at 20:58
  • @user425181 One can check readily that the Galois group is nonabelian and with a little more work that its center is trivial. Among the groups of order $20$ this leaves a single possibility, the Frobenius group of order 20, namely, $\operatorname{Aff}(\Bbb F_5) = \operatorname{GL}(\Bbb F_5) \ltimes \Bbb F_5 \cong \Bbb F_4 \ltimes \Bbb F_5$: https://groupprops.subwiki.org/wiki/General_affine_group:GA(1,5) – Travis Willse Nov 13 '17 at 10:57
  • Or one can see here, that the Galois group is isomorphic to the Frobenius group. – Dietrich Burde Jun 13 '25 at 18:11

0 Answers0