Show that $\mathrm{Gal}(E/F)$ is cyclic 
For $E=\mathbb{Q}(i,2^{1/8})$ and $F=\mathbb{Q}(i)$, show that $\mathrm{Gal}(E/F)$ is cyclic.  

I know is that $\mathrm{Gal}(E/F)$ is of order 8, since $x^8-2$ is the minimal polynomial  of $2^{1/8}$. I can see that the roots of the minimal polynomial are $2^{1/8}$,$2^{1/8}\xi$, $2^{1/8}\xi^2$, $2^{1/8}\xi^3$, $2^{1/8}\xi^4...$ where $\xi$ is the 8th root of unity. From there, how can I proceed?  
Thanks
 A: First, choose a distinguished primitive $8$th root of unity by setting $\xi = \frac{\sqrt{2}}{2}(1+i)$. (This choice won't affect our computation of $G := \mathrm{Gal}(E/F)$, but rather how we label the roots of $X^{8}-2$.)
We can be very explicit in computing the $G$ by using the following facts/observations:
(1) $E = F(\sqrt[8]{2})$; hence, any field automorphism of $E$ which fixes $F$ is determined by where it sends $\sqrt[8]{2}$. 
(2) $G$ acts transitively on the roots of $X^{8}-2$; hence, for each $i = 0, 1, \ldots, 7$, there is an element $\sigma_{i} \in G$ which sends $\sqrt[8]{2}$ to $\xi^{i}\sqrt[8]{2}$. Moreover, each $\sigma_{i}$ is a distinct element of $G$, since $\sigma_{i}(\sqrt[8]{2}) \neq \sigma_{j}(\sqrt[8]{2})$ for distinct $i, j \in \{0, 1, \ldots, 7\}$.
(3) $|G| = [E:F] = 8$, so in fact, $G = \{\sigma_{0}, \sigma_{1}, \ldots, \sigma_{7}\}$ by observation (2).
Hence, we have computed the elements of $G$. Now we need to show that $G$ is cyclic. Indeed, it suffices to show that $G$ contains an element of order $8$. I claim that $\sigma_{1}$ has order $8$. The key thing to do is to determine what $\sigma_{1}$ does to $\xi$, so let's do that and see how it helps. 
Since $\sigma_{1}$ fixes $F$, all we need to do is check what $\sigma_{1}(\sqrt{2})$ is. This is easy enough: we have
$$\sigma_{1}(\sqrt{2}) = \sigma_{1}((\sqrt[8]{2})^{4}) = \sigma_{1}(\sqrt[8]{2})^{4} = (\xi \sqrt[8]{2})^{4} = \xi^{4}\sqrt{2} = -\sqrt{2}.$$
Hence, $\sigma_{1}(\xi) = -\xi$. You can then check that 
$$\sigma_{1}^{2}(\sqrt[8]{2}) = \sigma_{1}(\xi\sqrt[8]{2}) = \sigma_{1}(\xi)\sigma_{1}(\sqrt[8]{2}) = -\xi^{2}\sqrt[8]{2}$$
I leave the computation that $\sigma_{1}^{4}(\sqrt[8]{2}) = -\sqrt[8]{2}$ to you; it is similar.
By Lagrange, $\sigma_{1}$ has order $1, 2, 4$ or $8$. Since we see that $\sigma_{1}$ cannot have order $1, 2, $ or $4$, it must have order $8$ as desired. (You can explicitly compute the powers of $\sigma_{1}$ if you like, too.)
