# 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?

-

## 1 Answer

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.

-

I know the Galois group is isomorphic to the quasidihedral group of order 16, and the other automorphism is $$\sigma\colon{ \sqrt[8]{2}\mapsto\zeta_8\sqrt[8]{2}\atop i\mapsto i}$$ 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
– Zev Chonoles♦ Nov 24 '11 at 10:56
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
show 1 more comment
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.950846254825592, "perplexity_flag": "head"}