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.