I have field extension $\mathbb{Q}\left(\sqrt{\sqrt{2}+\sqrt{-2}}\,\right)$. I can prove that it is the field of decomposition of the polynomial $x^8+16$, and I can prove it's irreducible. This means that the Galois group of it has order $8$. But there are $5$ non-isomorphic groups of order $8$. How can I find exactly the Galois group of this extension?
Maple told me, that this is $\mathbb{Z}_4 \times \mathbb{Z}_2$, but I don't know how to prove it.
We know $\mathbb Q(\alpha)$ is the splitting field of $x^8 + 16$, every root can be expressed in terms of $\alpha$.
$$\alpha = \sqrt{\sqrt{2}+\sqrt{-2}} = \sqrt[4]{2}\sqrt{1 + i} = \sqrt{2}\frac{\sqrt{1 + i}}{\sqrt[4]{2}} = \sqrt{2} \zeta$$ where $\zeta = \zeta_{16}$ is a root of $\Phi_{16}(z) = z^8 + 1$.
$$\text{Gal}(\mathbb Q(\zeta) / \mathbb Q) = C_4 \times C_2.$$
We can recover $\sqrt{2}$ from $\zeta$ via $\sqrt{2} = \zeta^2 + 1/\zeta^2$ so $\alpha = \zeta^3 - \zeta^7$. $\mathbb Q(\alpha)$ is just a subfield of $\mathbb Q(\zeta)$ so its automorphisms are just a subgroup of the automorphisms of $\mathbb Q(\zeta)$, but there are 8 of them so we keep all of them and have the same Galois group.
