$K=\mathbb{Q}(\sqrt[8]{2},i)$ and let $F=\mathbb{Q}(\sqrt{-2})$. Find the Galois Group $G(K,F)$

Question

Let $K=\mathbb{Q}(\sqrt[8]{2},i)$ and let $F=\mathbb{Q}(\sqrt{-2})$. Identify the $G(K,F)$ with a subgroup of permutations of the roots of $x^8-2$.

You have a guideline for the answer in here. But I would really appreciate some further clarification:

1. Why does it imply when the order of $\sigma\in G(K,F)$ is four, then that it is $Q_8$?

2. And I know that in a splitting field of a polynomial, the automorphisms are completely identified by it's action on the roots. But here why should $K$, be the splitting field of $x^8-2$? I can show that the polynomial splits in $K$ in to linear factors. But for it to be a splitting field shouldn't we need the additional condition that it is generated by the roots of the polynomial also. ( I understand that for any $\sigma\in G(K,F)$, $\sigma(i)\sigma(i)=\sigma(-1)=-1$. Thus $\sigma(i)=i$ or $-i$.)

3. Also if we are to draw the lattice structure of the intermediate fields, is there any specific details that I should look into

Appreciate your help

Answer 1

1. The The argument shows that every automorphism $\sigma\in\operatorname{Gal}(K/F)$ is of the form $$\sigma:\ \begin{array}{lcl} \sqrt[8]{2}&\ \longmapsto&\ \zeta_8^n\sqrt[8]{2}\\ i&\ \longmapsto&\ (-1)^ni\end{array}$$ for some integer $0\leq n\leq8$. From this representation it is easy to verify that the order of every automorphism divides $4$, and that there are precisely $2$ automorphisms of order $2$. The only group of order $8$ that has these properties is $Q_8$.
2. The field is generated by the roots of $x^8-2$. Clearly $\sqrt[8]{2}$ and $i\sqrt[8]{2}$ are roots, and their ratio is $i$.
3. The Galois group is isomorphic to $Q_8$, so the lattice of intermediate fields of $F\subset K$ is identical to the lattice of subgroups of $Q_8$. So this lattice of subgroups is worth looking into.