Galois correspondence

For a finite Galois extension, subgroups correspond in reverse order to intermediate fields by taking fixed fields and field-fixing subgroups.

For a finite Galois extension $L/K$ with group $G$, an intermediate extension $E/K$ is normal exactly when its corresponding subgroup $\mathrm{Gal}(L/E)$ is a normal subgroup of $G$.

Every subgroup of an abelian group is normal. Hence every intermediate field of a finite abelian Galois extension is itself Galois over the base field.

For $r>1$, choose a prime $\ell \equiv 1(\mathrm{mod}r)$ by Dirichlet's theorem. The cyclic grouphas a subgroup of index $r$, whose fixed field is cyclic Galois of degree $r$ over $\mathbb{Q}$.

Let an automorphism $\sigma$ of an algebraically closed field $L$ have finite order and fixed field $K$. Artin's fixed-field theorem makes $L/K$ finite cyclic Galois. Every finite extension $M/K$ embeds in $L$: in characteristic zero, choose a primitive generator and send it to a root in $L$. Its image is an intermediate field of a cyclic extension and is therefore Galois and cyclic over $K$.

Put $\alpha =\sqrt[4]{7}$ and $L=\mathbb{Q}(\alpha ,i)$. The dihedral group $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ acts byThe ten subgroups and their fixed fields areThe normal subgroups are $D_8$, $⟨r⟩$, the two displayed Klein four-groups, $⟨r^2⟩$, and $\{1\}$.