I've computed the splitting field of $x^8-3$ over $\mathbb{Q}$ to be $\mathbb{Q}(\sqrt[8]{3},\zeta_8)=\mathbb{Q}(\sqrt[8]{3},\sqrt{2},i)$, which is of degree 32 over $\mathbb{Q}$.

The possible automorphisms are the maps fixing $\mathbb{Q}$ of form $$ \sqrt[8]{3}\mapsto \zeta_8^i\sqrt[8]{3}\quad (0\leq i\leq 7),\qquad \sqrt{2}\mapsto\pm\sqrt{2},\qquad i\mapsto\pm i. $$ There are 32 automorphisms, and thus these are all automorphisms. So I have an explicit description of the automorphisms in the Galois group $G$, but if I wanted to actually say what $G$ is isomorphic to, how do I find that? I looked on groupprops subwiki, and there seem to be 51 groups of order 32 up to isomorphism, at least.

I made some little observations, like that there are 7 elements of order 2, but not sure how to actually classify the Galois group.

I've also noticed that the maps fixing $\sqrt[8]{3}$ will form a subgroup isomorphic to the Klein-4 group, and the maps fixing $\sqrt{2}$ and $i$ will form a cyclic subgroup of order 8. Does this narrow it down?


Your description of $G$ is perfectly fine as it is.

But maybe a representation of $G$ in $GL_2(\mathbb{Z}/8\mathbb{Z})$, would be more to your taste : if $\sigma \in G$ satisfies $\sigma(\zeta_8) = \zeta_8^a$ and $\sigma(\sqrt[8]{3}) = \zeta_8^b \sqrt[8]{3}$, it is represented by the matrix $ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}$

You can observe that the automorphisms fixing $\sqrt[8]{3}$ are those with $b=0$, and they form a (non normal) subgroup $H$ isomorphic to $(\mathbb{Z}/8\mathbb{Z})^*$. The automorphisms fixing $\zeta_8$ are those with $a=1$, and they form a normal subgroup $N$ isomorphic to $\mathbb{Z}/8\mathbb{Z}$, and $G$ can also be seen as a semi-direct product of those two groups :

In the exact sequence $0 \rightarrow N = Gal_{\mathbb{Q}(\zeta_8)}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) \rightarrow G = Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) \rightarrow G/N = Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8)) \rightarrow 0$

there is a section $s : Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8)) \rightarrow H \subset Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8,\sqrt[8]{3})) $, defined simply by extending an automorphism $\sigma$ with $s(\sigma)(\sqrt[8]{3}) = \sqrt[8]{3}$.

$Gal_{\mathbb{Q}}(\mathbb{Q}(\zeta_8))$ is isomorphic to $(\mathbb{Z}/8\mathbb{Z})^*$, so that the canonical surjection $f$ is simply picking $f(\sigma) = a$, and the section $s : (\mathbb{Z}/8\mathbb{Z})^* \rightarrow H$, is $s(a) = \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}$

This gives an isomorphism between $G/N$ and $H$ and shows that $G$ is the semidirect product of $H$ acting on $N$

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    $\begingroup$ Ok, I think I understand most of this. $(\mathbb{Z}/8\mathbb{Z})^*$ and $\mathbb{Z}/8\mathbb{Z}$ are the two groups I mentioned also. Now I just want to know why $G$ is the semi-direct product of them, can you explain why that is? $\endgroup$ – Manny Quinn Nov 25 '11 at 9:34
  • $\begingroup$ @MannyQuinn Any split short exact sequence of groups gives rise to a semi-direct product. I don't know of a specific reference, but any book on homological algebra should have it. $\endgroup$ – Potato May 8 '13 at 21:09
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    $\begingroup$ @Potato the Wikipedia article talks about this (in case you still don't have a specific reference!) en.wikipedia.org/wiki/Semidirect_product $\endgroup$ – Cauchy Aug 12 '17 at 6:09

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