Fastest way to compute subfields of $\mathbb{Q}(\sqrt[8]{2},i)$ which are Galois over $\mathbb{Q}$? I have the lattice of subfields of the splitting field $\mathbb{Q}(\sqrt[8]{2},i)$ over $x^8-2$, and the corresponding lattice of subgroups of the Galois group $G$ of the splitting field. 
I'm now interested in the finding the subfields which are Galois over $\mathbb{Q}$. What's the fastest way to find them?
I know that a subfield will be Galois over $\mathbb{Q}$ iff the automorphisms in $G$ fixing the subfield form a normal subgroup, but it seems difficult to go through and actively find all the normal subgroups. Is there a faster way?
 A: On the contrary, (a small part of) the power of Galois theory is precisely that it reduces the difficult question of finding Galois subfields to the significantly easier question of finding normal subgroups. This is the faster way :)
You should figure out what the group $G=\text{Gal}(\mathbb{Q}(\sqrt[\large 8]{2},i)/\mathbb{Q})$ looks like - for starters, we know that
$$|G|=[\mathbb{Q}(\sqrt[\large 8]{2},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[\large 8]{2},i):\mathbb{Q}(\sqrt[\large 8]{2})][\mathbb{Q}(\sqrt[\large 8]{2}):\mathbb{Q}]=2\cdot8=16.$$
What else do we know - for example, can you think of some elements you know will be in $G$? One that we know will be there is complex conjugation, which I will denote $\rho$,
$$\rho:\mathbb{Q}(\sqrt[\large 8]{2},i)\to\mathbb{Q}(\sqrt[\large 8]{2},i),\qquad \rho:{\sqrt[\large 8]{2}\mapsto \sqrt[\large 8]{2}\atop i\mapsto -i}$$
Can you think of any others? Once we work out the elements and how they interact (i.e. the group structure), it actually isn't that bad of a problem to find the normal subgroups. This question may help you out, and if you're still having trouble, you could ask a separate question about how to find the normal subgroups of this particular group.
