Intermediate fields between $\mathbb{Q}(\zeta_7)$ and $\mathbb{Q}$. So I know $|\text{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q})| = \phi(7) = 6$
It is $\{1, \sigma_2, \dots,\sigma_6\}$ where $\sigma_a(\zeta_7) = \zeta_7^a$
Through brute force computation, I found that the subgroups are $H = \{1,\sigma_2,\sigma_2^2 \}$ and $K = \{1,\sigma_6\}$
Now what I don't get is in $H$, while the element $\alpha = \zeta_7 + \zeta_7^2 + \zeta_7^4$ is fixed by $H$, so is the element $\beta = \zeta_7^3 + \zeta_7^5 + \zeta_7^6.$ Since $3 = [\mathbb{Q}(\alpha): \mathbb{Q}] = [\mathbb{Q}(\beta): \mathbb{Q}]$, are the two fields $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ isomorphic?
For $K$, $\zeta_7 + \zeta_7^6$ is fixed. But so is $\zeta_7^2 + \zeta_7^3 + \zeta_7^4 + \zeta_7^5$. However $2 =[\mathbb{Q}(\zeta_7 + \zeta_7^6):\mathbb{Q}]\neq [\mathbb{Q}(\zeta_7^2 + \zeta_7^3 + \zeta_7^4 + \zeta_7^5):\mathbb{Q}] = 4$ and $\mathbb{Q}(\zeta_7^2 + \zeta_7^3 + \zeta_7^4 + \zeta_7^5)$ can't be an intermediate field.
 A: You have done very well calculating the subgroups.
In general we have $$\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) = (\mathbb Z/n \mathbb Z)^\times$$ which in this case is $C_6$, which you can more easily read off the fact there is 1 subgroup of index 2 and 1 subgroup of index 3.
It is useful to find a generator for $(\mathbb Z/7 \mathbb Z)^\times$. 2 usually works:

*

*powers of 2 mod 7: $1, 2, 4, 1, 2$ not in this case

*powers of 3 mod 7: $1, 3, 2, 6, 4, 5$. So 3 is a generator for this group.

This lets us find the "periods" that are fixed under subgroups of the Galois group more easily.
Let $\sigma$ be the generator of the Galois group corresponding to 3. i.e. $\sigma \zeta_7^r = \zeta_7^{3r}$. Then we have a subgroup of index 2: $\langle \sigma^3 \rangle$ and a subgroup of index 3: $\langle \sigma^2 \rangle$.
The period sums are invariant:

*

*$\zeta_7 + \zeta_7^2 + \zeta_7^4$ invariant under $\sigma^2$, lies in a subfield of degree 2.

*$\zeta_7 + \zeta_7^6$ invariant under $\sigma^3$ lies in a subfield of degree 3.

