 Solution

Let $E=\mathbb{Q}(\alpha ,\beta )$. Sincepart (b) gives $[E:\mathbb{Q}]=8$. The roots of $X^4-6X^2+6$ are $\pm \alpha ,\pm \beta$, all in $E$, so $E$ is its splitting field and is Galois. Its action on the four roots is transitive; the order-eight entry in the list from part (a) is $D_8$. Therefore

For an explicit dihedral Galois action on four radical roots, defineThen $r^4=s^2=1$ and $srs=r^{-1}$. The subgroup lattice is determined byHere $⟨r^2⟩$ lies in all three order-four subgroups; $⟨s⟩,⟨r^{2}s⟩$ lie in $⟨r^2,s⟩$; and $⟨rs⟩,⟨r^{3}s⟩$ lie in $⟨r^2,rs⟩$.

Reversing inclusions under the Galois correspondence gives the complete field latticeThe incidence is likewise reversed: $\mathbb{Q}(\sqrt{2})$ lies in $\mathbb{Q}(\sqrt{2},\sqrt{3})$; $\mathbb{Q}(\sqrt{3})$ lies in that field, $\mathbb{Q}(\alpha )$, and $\mathbb{Q}(\beta )$; and $\mathbb{Q}(\sqrt{6})$ lies in that field, $\mathbb{Q}(\alpha +\beta )$, and $\mathbb{Q}(\alpha -\beta )$. The hint verifies the last two quartic fields through $\alpha +\beta =\sqrt{2}\gamma$ and $\alpha -\beta =\sqrt{2}\delta$.