Question: Let $E$ be a splitting field of $x^4-2$. Let $\sigma\in{\rm Gal}(E/\mathbb{Q})$ such that $\sigma(\alpha)=\bar\alpha$ for all $\alpha\in E$. Find $\mathrm{Inv}(\langle\sigma\rangle)=\{\alpha\in E:\sigma(\alpha)=\alpha\}$.
Thoughts: We can determine the splitting field $E$ to be $E=\mathbb{Q}(\sqrt[4]{2},i)$ (moreover, we can find that the order of this Galois group has to be $8$ and that the Galois group is iomorphic to the dihedral group of order $8$). Now, all elements of $E$ are of the form $\alpha_1+\alpha_2\sqrt[4]{2}+\alpha_3\sqrt[4]{4}+\alpha_2\sqrt[4]{8}+\alpha_5i$ (I may have this wrong.). So, an element of $E$ is in Inv$(\langle\sigma\rangle)$ whenever $\alpha_5=0$, thus Inv$(\langle\sigma\rangle)=\mathbb{Q}(\sqrt[4]{2})$, right? Moeover, could we determine whether or not $\langle\sigma\rangle$ is a normal subgroup of ${\rm Gal}(E/\mathbb{Q})$?