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I am reading in Dummit and Foote, page $579$. We have the following group:

$$G = \langle \tau,\sigma\mid \sigma^8,\tau^2,\sigma \tau = \tau \sigma^3\rangle = \{\sigma^k,\tau\sigma^k\mid 0\le k\le 7\}$$

which is just $Gal\left( \mathbb Q(i,\sqrt[8]{2})/\mathbb Q\right)$.

It is stated that: "determining the subgroups of this group is a straightforward problem".

How so? Is there an easy way to determine the subgroups? Here are the subgroups:

subgroups

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  • $\begingroup$ Just many calculations.. $\endgroup$ – Yanko Aug 10 '17 at 14:09
  • $\begingroup$ @yanko it's really a lot. It can take up to an hour! There must be some way around it. $\endgroup$ – Cauchy Aug 10 '17 at 14:23
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The Galois group of $K/\mathbb{Q}$ with $K:=\mathbb{Q}(\sqrt[8]{2},\zeta)$ has order $16$ and is an extension of $(\mathbb Z_8)^\times$ by $\mathbb Z_4$. For all groups of order $16$, in particular for this Galois group, the subgroups have been computed, see here, or here and similar references. One can use the classification of groups of order $1,2,4,8$ and other facts about $2$-groups. One also can use GAP. Doing the computations by hand may take some time.

For the Galois group see also this MSE-question: Finding the Galois group over $\Bbb{Q}$.

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