Be $E\supset F$ galois extension and $\alpha \in E$, $O$ the orbit of $\alpha$ under $Aut(E/F)$. $|O|=|F(\alpha)/F|$?

Question

Be $E\supset F$ galois extension and $\alpha \in E$. Be $O$ the orbit of $\alpha$ under $Aut(E/F)$.

I know that $|O|=|F(\alpha)/F|$ if $F(\alpha)\supset F$ is a galois extension. My question is: there are some cases or hypotesis in which the equality is true but $F(\alpha)\supset F$ is not a galois?

Answer 1

The equation always holds if you consider the action of the galois group of the normal closure of the extension. A root of an irreducible polynomial must be mapped to a root of the same polynomial. (I intended the number in your question as the degree of the extension) if you don’t consider a galois group otherwise, you have not defined the automorphism and so you are not allowed to define actions.