This question has an answer here but It's not clear to me. computing the orbits for a group action

So, I am asking it again.

Let G be the galois group of a field with 9 elementsover its subfield with 3 elements. Then the number of orbits for the action of G on the field with 9 elements is?

I want to ask which result in Galois theory should be used. I have studied galois theory but couldn't solve problems due to lack of time. SO, Assume knowledge of graduate text of galois theory while answering it and kindly provide a bit detailed reasoning.

I am really struck so I am asking . Thanks a lot for your help!!


(i) $|\mathbb{F}_9:\mathbb{F}_3|=2$ so by the Fundamental Theorem of Galois Theory [but we scarcely need it] the Galois Group $G$ has order $2$.

(ii) $\pi: x\mapsto x^3$ is an automorphism of order $2$, so the Galois group is $\langle \pi\rangle$.

(iii) The multiplicative group of $\mathbb{F}_9$ is cyclic of order $8$. [Clearly the multiplicative group of a field can't have finite non-cyclic subgroups, or we'd contradict the fact that the polynomial ring is a UFD.] Let $\theta$ be a generator.

(iv) Calculate the orbits: they are $\{0\}$, $\{1\}$,$\{-1\}$, $\{\theta, \theta^3 \}$, $\{\theta^{-1}, \theta^{-3}\}$, $\{\theta^{2},\theta^{-2}\}$.

  • 2
    $\begingroup$ Using very slightly more Galois theory, we also know that the elements of the base field are precisely the fixed points, so all other orbits much have order 2. $\endgroup$ Oct 17 '20 at 15:45
  • $\begingroup$ @ancientmathematician It may sound very easy easy to you but can you please tell which result you are using to compute $|\mathbb{F}_9:\mathbb{F}_3|=2$? I can't recollect how to find it and don;t wanna memorize it? $\endgroup$
    – No -One
    Oct 22 '20 at 16:58
  • $\begingroup$ A vector space of dimension $k$ over $\mathbb{F}_p$ clearly has $p^k$ elements. $\endgroup$ Oct 22 '20 at 19:13
  • $\begingroup$ @ancientmathematician how does π has order 2? $\endgroup$
    – No -One
    Nov 4 '20 at 16:43
  • $\begingroup$ Oh come on, because $\pi^2(x)=x$. $\endgroup$ Nov 4 '20 at 16:45

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