# Determine the lattice of subgroups of gal$(\mathbb Q(w):\mathbb Q)$ and of intermediate fields in $\mathbb Q(w)\supseteq\mathbb Q$

Let $w = e^{2\pi i / 7}$. Determine the lattice of all the subgroups of gal$(\mathbb Q(w):\mathbb Q)$ and the lattice of all the intermediate fields in the extension $\mathbb Q(w)\supseteq \mathbb Q$.

I have started the problem but I am a bit lost on the details.

We know, from different exercises and theorems, that $\mathbb Q(w)$ is a Galois extension of $\mathbb Q$ and that $G =\text{gal}(\mathbb Q(w): \mathbb Q) \simeq C_6$.

$\mathbb Z_7^* = \langle3\rangle$ so $G = \langle\sigma\rangle$ where $\sigma(w) = w^3$.

I am not really sure how to proceed with this information, though. I am pretty sure that $C_6 \simeq C_2 \times C_3$, so the subgroups are $C_2, C_3,$ and the identity ($\epsilon$).

I am not sure how to put this in the language of Galois groups and field extensions. Nor am I sure how to create the lattice of subfields.

According to the Galois correspondence, each subgroup corresponds to an intermediate field, the fixed field of that subgroup. The lattice of intermediate fields has exactly the same shape as the lattice of subgroups, except inclusions go in the opposite direction (because if $H \subset G$ as groups, and $L$ and $K$ are the fixed fields of $H$ and $G$ respectively, then everything fixed by $G$ is fixed by $H$, so $K \subset L$ as fields).
It is clear that $(\epsilon)$ fixes the entire field $\mathbb{Q}(\omega)$, and the whole group $C_6$ can fix only $\mathbb Q$. It remains to investigate $C_2$ and $C_3$.
$C_2$ is the group $\{\sigma^3, \epsilon\}$, and $\sigma^3$ sends $\omega$ to $\omega^{3^3} = \omega^{27} = \omega^{-1}$. So $\sigma^3$ fixes $\omega+\omega^{-1}$, and since $\omega+\omega^{-1} = 2\cos\frac{2\pi}7 \notin \mathbb Q$, the fixed intermediate field is $\mathbb Q(\omega+\omega^{-1})$. (You might notice that $\sigma^3$ is complex conjugation, so the fixed field must be $\mathbb R \cap \mathbb Q(\omega)$).
$C_3$ is the group $\{\sigma^2, \sigma^4, \epsilon\}$. By similar argument to $C_2$, we can find that $C_3$ fixes $\omega+\omega^2+\omega^4$, so the intermediate field is $\mathbb Q(\omega+\omega^2+\omega^4)$.