Uniqueness of Galois group

Question

I'm wondering whether the Galois group is unique in the sense that:

Let $E/K$ be a field extension and let $K = E^G$ for $G = Aut_K(E)$. Is it possible that there exists a proper subgroup $H$ of $G$ such that $K=E^H$?

Answer 1

Let $\mathbb{F}_p$ be a finite field with $p$-elements. Consider its algebraic closure $\overline{\mathbb{F}_p}$ and let $\phi:\overline{\mathbb{F}_p}\rightarrow \overline{\mathbb{F}_p}$ be the Frobenius automorphism i.e. $\phi(x) = x^p$. $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)$ is a profinite group and hence it cannot be isomorphic to $\mathbb{Z}$. Thus $$\mathbb{Z} \cong \langle \phi\rangle\subsetneq Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)$$ On the other hand $$\overline{\mathbb{F}_p}^{\langle \phi\rangle} =\{x\in \overline{\mathbb{F}_p}\,|\,x^p=\phi(x)=x\} = \mathbb{F}_p$$

Remark.

The general Galois correspondence gives rise to a bijection between closed subgroups of the Galois groups and intermediate fields. One can also derive from it that every subgroup dense in the Galois group has as its fixed points the base field.

For instance, in the example above the closure of the subgroup generated by $\phi$ is equal to $Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)$.