2
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ You should be able to find the group of $K$ over $\bf Q$, and then find $H$ as a subgroup of that group. $\endgroup$ Jul 22 '12 at 11:33
  • $\begingroup$ @colge Have you computed the Galois group of the big extension ? $\endgroup$
    – user38268
    Jul 22 '12 at 11:35
  • $\begingroup$ @BenjaLim I have tried this (see my edit of the quesion) but I didn't get exact representation of this group. $\endgroup$
    – colge
    Jul 22 '12 at 12:01
  • $\begingroup$ @colge It's the Frobenius group $F_{20}$. $\endgroup$
    – Cocopuffs
    Jul 22 '12 at 12:09
  • $\begingroup$ You should be able to find a relation involving $\sigma$ and $\tau$, decide whether there are any elements of order 10, etc. $\endgroup$ Jul 22 '12 at 12:25
1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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? $\endgroup$
    – M Turgeon
    Jul 22 '12 at 15:57
  • 1
    $\begingroup$ I think you've got the dicyclic group of order 20; the Frobenius group has relation $yxy^{-1}=x^3$. $\endgroup$ Jul 23 '12 at 4:31
  • $\begingroup$ @GerryMyerson-you are right. So the subgroup of order 10 is dihedral, not abelian. $\endgroup$ Jul 28 '12 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.