Under what conditions does a Galois extension contain all the conjugates of its elements?

Given a finite Galois extension $\frac{E}{K}$ and an element $\alpha \in E \setminus K$ it is easy to proof that all the elements of $C = \Big\{\sigma(\alpha):\sigma \in Gal\Big(\frac{E}{K}\Big) \Big\}$ where $Gal\Big(\frac{E}{K}\Big)$ stands for the Galois group of the extension, are conjugates, that is, if we denote the minimal polynomial by $Irr$ then $Irr(\alpha,K) = Irr(\sigma(\alpha),K)$.

My question is now if $C$ contains all the conjugates of $\alpha$.

My thoughts

Using Artin's theorem I know that the degree of the extension is $degree(Irr(\alpha,K)) = [E:K] = \Big|Gal\Big(\frac{E}{K}\Big) \Big|$ and therefore there are as many conjugates as automorphisms in the group. But I have no guarantee that the automorphism do not coincide on $\alpha$...

My definition of Galois extension

My definition of a finite Galois extension is as follows

Given a finite extension $\frac{E}{K}$ we say it is a finite Galois extension if there exists a subgroup $G \le Aut(E). K = E^G$ where $E^G$ is the set $\{u \in E:\forall \sigma \in G. \sigma(u) = u \}$.

In fact, I have a couple of characterizations of this concept as:

1. A finite normal and separable extension.
2. The splitting field of a separable polynomial.
• The fixed field definition is the most useful, even if it is not immediate it implies $|Gal(E/K)| = [E:K]$. If $K = E^G$ then for any $\alpha \in E, f(x) = \prod_{\beta \in G \alpha} (x-\beta)=\sum_{n=0}^d a_n x^n \in E[x]$ and for $\sigma \in G$ : $\sum_{n=0}^d \sigma(a_n) x^n =\prod_{\beta \in G \alpha} (x-\sigma(\beta)) = f(x)$ thus $f \in E^G[x]= K[x]$. By definition the $K$-minimal polynomial of $\alpha$ divides $f$. Finally $f$ is irreducible because any root $\beta$ of the minimal polynomial induces some automorphism of $K(\alpha)^{gal}/K$ which extends to $E/K$ – reuns Nov 1 '17 at 19:21
• $G \alpha= \{ \sigma(\alpha),\sigma \in G\}$ the orbit of $\alpha$ under $G$'s action – reuns Nov 1 '17 at 19:32
• I meant $f$ is irreducible because $h(\alpha) = 0 \implies h(\sigma(\alpha))=0$ – reuns Nov 1 '17 at 19:42