Fastest way to compute subfields of $\mathbb{Q}(\sqrt{2},i)$ which are Galois over $\mathbb{Q}$?

I have the lattice of subfields of the splitting field $\mathbb{Q}(\sqrt{2},i)$ 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?

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.
• I know the Galois group is isomorphic to the quasidihedral group of order 16, and the other automorphism is $$\sigma\colon{ \sqrt{2}\mapsto\zeta_8\sqrt{2}\atop i\mapsto i}$$ and the Galois group is defined by the relations $\sigma^8=\rho^2=1$ and $\sigma\rho=\rho\sigma^3$. I've written out the whole lattice of subgroups of $G$ in terms of their generators, and I've found 3 groups of order 8 (hence normal with index 2), so their corresponding subfields are Galois. I've also found 5 subgroups of order 4, and 5 subgroups of order 2. Do I really have to brute force compute it all? – Evariste Nov 24 '11 at 10:36
• I guess it boils down to how can I effeciently tell which of those 10 subgroups are normal in $G$? – Evariste Nov 24 '11 at 10:36
• I think the advice in the linked question is probably as good as we can get in general - you could compute the conjugacy classes of the group, and use that $H<G$ is normal iff it is a union of conjugacy classes, but it is not much faster, I think. One freebie is that the center is normal, but the rest we just have to brute force. – Zev Chonoles Nov 24 '11 at 10:54
• Thanks for the link, I might just trust it. There's no clever way to compute the conjugacy classes other than to just conjugate each element of the group by every element of the group for $16^2$ little computations, right? – Evariste Nov 24 '11 at 11:02