this was an old homework problem that I didnt quite understand, could someone provide some insight to any of the parts?
We are supposed to prove that given $K= \mathbb{Q}[\sqrt[8]{2},i]$
a) $G(K/\mathbb{Q}[i])\cong \mathbb{Z}_8$
b) $G(K/\mathbb{Q}[\sqrt{2}])\cong D_4$
c) $G(K/\mathbb{Q}[i\sqrt{2}])\cong $ Quaternion Group
any help is appreciated, thanks.
$K$ is the splitting field over ${\mathbb Q}$ of $x^8-2$, which has roots $tw^i$ ($0 \le i \le 7$), where $t = \sqrt[8]{2}$ and $w = (1+i)/\sqrt{2}$ is a primitive $8$th root of $1$.
So we need to work out the Galois group of $K$ over ${\mathbb Q}$. This acts transitively on the roots, so there is an automorphism $\alpha$ withs $\alpha(t)= tw$. So $\alpha(\sqrt{2}) = \alpha(t^4) = t^4w^4 = -t^4 = -\sqrt{2}$. Another automorphism, $\beta$, is complex conjugation. By replacing $\alpha$ by $\alpha\beta$ if necessary, we can assume that $\alpha(i)=i$, and hence $\alpha(w) = -w = w^5$.
We can now calculate the action of $\alpha$ on all roots, and we find $\alpha$ is the $8$-cycle $(0,1,6,7,4,5,2,3)$, where I have just written $i$ for the root $tw^i$. We can also calculate $\beta = (1,7)(2,6)(3,5)$, and we find that $\beta\alpha\beta = \alpha^3$, so $\alpha$ and $\beta$ generate a group of order $16$, which is equal to the degree $|K:{\mathbb Q}|$, and hence is the full Galois group of the extension. (This group is sometimes known as the semidihedral group.)
The Galois groups you have been asked to calculate are the subgroups with the specified fixed fields. We chose $\alpha$ with $\alpha(i)=i$, so ${\mathbb Q}[i]$ is the fixed field of the subgroup $\langle \alpha \rangle$, which is cyclic of order $8$.
Similarly, you will find that the subgroups $\langle \alpha^2,\beta \rangle$ and $\langle \alpha^2,\alpha\beta \rangle$ fix $\sqrt{2}$ and $i\sqrt{2}$, respectively.
