# Number of orbits for the action of G on the field with 9 elements

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!!

## 1 Answer

(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 $$\{0\}$$, $$\{1\}$$,$$\{-1\}$$, $$\{\theta, \theta^3 \}$$, $$\{\theta^{-1}, \theta^{-3}\}$$, $$\{\theta^{2},\theta^{-2}\}$$.

• 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. Oct 17 '20 at 15:45
• @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? Oct 22 '20 at 16:58
• A vector space of dimension $k$ over $\mathbb{F}_p$ clearly has $p^k$ elements. Oct 22 '20 at 19:13
• @ancientmathematician how does π has order 2? Nov 4 '20 at 16:43
• Oh come on, because $\pi^2(x)=x$. Nov 4 '20 at 16:45