Normal Basis for Galois extension $ \Bbb K / \Bbb F$ is defined as looking at $\Bbb K$ as a vector space over $\Bbb F $ with the basis of $(\sigma(a)| \sigma \in Gal(\Bbb K / \Bbb F))$ for $a \in \Bbb K$
I need to proof that $ \Bbb K=\Bbb F(a)$
I know that Galois group is acting transitively on the roots of the minimal separable and normal polynomial over $\Bbb K$: $f(x)=\prod_{i=1}^n (x-\sigma_i(a))$
How I proof that for every i, $\sigma_i(a)=a^k $ for some $k \in \Bbb N$?
If the conjugates of $\alpha$ form a basis for the extension, then the number of conjugates must equal the degree of the extension, so the degree of $\alpha$ must equal the degree of the extension, so then degree of $F(\alpha)$ over $F$ must equal the degree of the extension, so $F(\alpha)=K$.