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

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.

• You can show that the Galois group G is $S_3$. It is easy to see that it has degree 6($Q(\sqrt[3]7)$ is real) and $G$ embeds into $S_3$ since it permutes the $3$ roots.. – Asvin Jun 1 '15 at 20:58

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.
• Thank you. I'm wondering what happens if I substitute $\mathbb{Q}$ with $\mathbb{Q}(i)$. Then the polynomial $f$ is still irreducible and I think that this (my and your) argument still holds. Am I right or does something crucial change? – Jytug Jun 1 '15 at 21:24
• Jup, I still think so, but the subfields will be $\mathbf{Q}(i)(\zeta_{3}) \cdots$ – mich95 Jun 1 '15 at 21:30