Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was asked to prove the galois group of a given normal extension is non-abelian. My original solution was to use isomorphism extension theorem but that was not taught in class. So, in my new attempt to solve it, I hit a road bum.

To show non-abelian, I need 2 automorphisms. I already know the 2 automorphisms exist but I don't know how to prove one of them exist.

Let $K$ be a normal extension of $\mathbb{Q(\sqrt[3]7)}$. How would I prove that given any element $a\in K$ and $b$, a conjugate of $a$, there exist an automorphism $\alpha\in Gal(K/\mathbb{Q})$ such that $\alpha(a)=b$? Let's assume $K$ is a finite dimension over $\mathbb{Q(\sqrt[3]7)}$ because infinite dimension was not taught in class.


EDIt: Added a few more details

share|cite|improve this question
@Giygas: I don't think your question is the same as the one being asked here, since they assume that the extension is normal and are specifically asking for an argument that doesn't use the fact that isomorphisms can be extended to normal extensions. I suggest you instead ask your question as a new question. – Eric Wofsey Sep 28 '15 at 19:09
I'm going to do that. Thank you for your suggestion. – Giygas Sep 28 '15 at 19:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.