Galois group and intermediate fields for splitting field of $ x^3 -7 $

Question

I'm trying to do the following exercise:

find the Galois group $ G(E/\mathbb{Q}) $, where $ E $ is the splitting field of $ x^3 - 7 $, all its subgroups and the intermediate subfields $ E^H $ (subfields of $ E $ which are fixed by $ H \subset G $).

Of course $ E = \mathbb{Q}(\sqrt[3]{7}, \varepsilon_3)$, where $ \varepsilon_3 $ denotes the primitive root of unity of degree $ 3 $

The degree of this extension is $ 6 $ and I am able to find two automorphisms of $ E/\mathbb{Q} $ which do not commute, hence $ G \simeq S_3 $, since there are only two groups of order $6 $.

I know the subgroup structure of $ S_3 $ - all its proper subgroups are cyclic.

Is there a more elegant way to find all the intermediate subfields other than just checking which elements are fixed by every automorphism?

I'd appreciate some help with that.

Answer 1

You know that $x^{3}-7$ is irreducible over $\mathbf{Q}$ so the Galois group acts transitively on the set of roots, so $\mathbf{Q}(\sqrt[3]{7})$,$\mathbf{Q}(\zeta_{3}\sqrt[3]{7})$, $\mathbf{Q}(\zeta_{3}^{2}\sqrt[3]{7})$ are subfields of degree 3. And you know that these correspond to the subgroups of order 2 of $S_{3}$, and $S_{3}$ has exactly three such subgroups. Now $\mathbf{Q}(\zeta_{3})$ is a subfield of degree 2, and hence corresponds to a subgroup of order 3 in $S_{3}$, and $S_{3}$ has a unique such subgroup.