 Solution

The cyclotomic polynomial ${\Phi }_7=1+X+\cdots +X^6$ is irreducible: replacing $X$ by $X+1$ gives an Eisenstein polynomial at seven. Hence the extension has degree six, andThere is one subgroup of each order $1,2,3,6$, and therefore precisely four intermediate fields. The trivial subgroup fixes $\mathbb{Q}(\eta )$; the whole group fixes $\mathbb{Q}$. The order-two subgroup $\{1,6\}$ is complex conjugation and fixes $\mathbb{Q}(\eta +{\eta }^{-1})$, of degree three: $\eta$ satisfies $X^2-(\eta +{\eta }^{-1})X+1$ and is not real.

The order-three subgroup $\{1,2,4\}$ fixes $u=\eta +{\eta }^2+{\eta }^4$. Its other conjugate is $v={\eta }^3+{\eta }^5+{\eta }^6$. Expanding products and using ${\sum }_{j=1}^6{\eta }^j=-1$ gives $u+v=-1$ and $uv=2$. Thus $u^2+u+2=0$, and this fixed field is $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{-7})$, of degree two. These generators have the required degrees and are fixed by the indicated subgroups, so the Galois correspondence proves they are the full fixed fields.