The question:
Let $F:=\mathbb{Q}, E:=\mathbb{Q}(\zeta)$ where $\zeta :=\exp({2\pi i\over 7})$. Find all the middle fields of the extension $E/F$.
My attempt:
The roots of $f(x):=\text{irr}(\zeta,F)=x^6+\dots+1$ are $\zeta^k$ with $1\leq k\leq 6$. Let $G:=\text{Aut}(E/F)$.
$f$ is separable so $|G|=[E:F]=\deg(f)=6$. Thus, $$ G=\{\sigma_k|1\leq k\leq 6,\sigma_k=(\zeta\mapsto\zeta^k)\} $$ Notice that$$ |\sigma_1|=1,|\sigma_6|=2,|\sigma_2|=|\sigma_4|=3,|\sigma_3|=|\sigma_5|=6$$ Thus $G\cong \mathbb{Z}_6$. Thus, the subgroups of $G$ are $$ H_1=\{\sigma_1,\sigma_6\},H_2=\{\sigma_1,\sigma_2,\sigma_4\}$$ From the fundamental theorem of Galois theory, the midterm fields of the extension are $K_1=E^{H_1}$ and $K_2=E^{H_2}$.
Consider $K_1$. Let $E\ni x=\sum_{i=0}^6 a_i\zeta ^i$. If $x\in K_1$ then $x=\sigma_6(x)$: $$ a_0+a_1\zeta+a_2\zeta^2+a_3\zeta^3+a_4\zeta^4+a_5\zeta^5+a_6\zeta^6= \\ a_0+a_1\zeta^6+a_2\zeta^5+a_3\zeta^4+a_4\zeta^3+a_5\zeta^2+a_6\zeta \\ \Rightarrow a_1=a_6,a_2=a_5,a_3=a_4 \\ \Rightarrow K_1=\mathbb{Q}(\cos({2\pi\over7})) $$
With similar argument we get that if $x\in K_2$ then $$a_1=a_2=a_4,\\a_3=a_5=a_6$$ and thus, $$x=\beta_0+\beta_1(\zeta+\zeta^2+\zeta^4)+\beta_2(\zeta^3+\zeta^5+\zeta^6)$$
Here I stuck, I dont know how to identify $K_2$ with that characterisation of $x$.