I've proved this result for myself, but I have doubt in my proof whether it is true :
Let $E/K$ be a field extension and $G(E/K)$ its Galois group. Suppose $E^{G(E/K)}$ is its Fixed field, i.e. $\{x\in E\,|\:\forall\,\sigma\in G(E/K):\sigma(x)=x\}$. Is it true that: $$G(E/K)=G(E/E^{G(E/K)}) \:?$$
Let $K'=E^{G(E/K)}$. Here is my proof :
First, It is obvious that $K\subseteq K'$ and then $G(E/K')\subseteq G(E/K) \tag{*}$
Now, suppose $\phi\in G(E/K)$ and it's sufficient to prove $\phi\in G(E/K')$ or equivalently $\phi$ fixes all members of $K'$. And this is also clear as for all $y\in K'$, we have $\sigma(x)=x$, for every $\sigma\in G(E/K)$, especially for $\phi$.
This result, if it's correct, states a nice property that for any arbitrary extension of finite degree $E/K$, there exists an intermediate subfield, like $K'$, that : $$\left\{\begin{array}{ll} (i)&K\subseteq K'\subseteq E\\ (ii)&\text{$E/K'$ is a Galois extension (See [Proposition 6.1.3][1])}\\ (iii)& \:\:G(E/K)=G(E/K') \end{array}\right.$$
Or for any Galois group, there's a Galois Extension belongs to it !
# EDIT
The main problem I was working with was the following problem :
Let $E/K$ a field extension of finite degree. Prove that every subgroup $G$ of $G(E/K)$ is finite and $|G|$ divides $[E:K]$.