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I have been asked to determine the lattice of subgroups of $G=\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ where $\zeta$ is a primitive $24^{th}$ root of unity. I am then asked to determine their corresponding fixed fields.

We know that the automorphisms of a primitive $n^{th}$ root of unity $\alpha$ are given by $a^k$ such that $k$ and $n$ are coprime. I have therefore determined the elements of $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ to be:

$$\phi_k(\zeta)=\zeta^k \ \ \ \ \ \ \ k=1,5,7,11,13,17,19,23.$$ And I know that these elements determine the whole group since $[\mathbb{Q}(\zeta):\mathbb{Q}]=|\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})|=8.$ I also know that this Galois group will be isomorphic to a product of cyclic groups of prime power order, and I believe this to be $\mathbb{Z}/\mathbb{Z}4\times\mathbb{Z}/\mathbb{Z}2$ but am not exactly sure how to show this. I know that once I've shown this I can determine the subgroups of $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ by examining the subgroups of the isomorphic product of cyclic groups (the one I believe to be $\mathbb{Z}/\mathbb{Z}4\times\mathbb{Z}/\mathbb{Z}2$). I am unsure how to then determine their corresponding fixed fields, can anyone give me guidance?

Thanks in advance.

UPDATE: I have determined the isomorphic group to actually be $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ by the following. $\Gamma(\mathbb{Q}(\zeta):\mathbb{Q})$ is abelian and of order 8. Thus it is isomorphic to one of: $\mathbb{Z}_8$, $\mathbb{Z}_4\times\mathbb{Z}_2$ or $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$. We have that $\phi_k^2=\phi_1=id$ for $k=1,5,7,11,13,17,19,23$ and thus every non identity element has order 2, meaning we have isomorphism to $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$. How do I proceed?

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    $\begingroup$ $\mathbb{Z}_4 \times \mathbb{Z}_4$ has $16$ elements. However, since the Galois group is abelian and of order $8$, there are only $2$ possibilities for it. $\endgroup$ – Kaj Hansen Apr 2 '17 at 16:52
  • $\begingroup$ @KajHansen You mean $3$ possibilities... $\endgroup$ – peter a g Apr 2 '17 at 17:25
  • $\begingroup$ Apologies for the typo, I meant $\mathbb{Z}_4\times\mathbb{Z}_2$, I have edited the original post. $\endgroup$ – maccamaths976 Apr 2 '17 at 17:26
  • $\begingroup$ Oops, yes @peterag. Forgot about $(\mathbb{Z}_2)^3$. $\endgroup$ – Kaj Hansen Apr 2 '17 at 17:44
  • $\begingroup$ I have found the isomorphic group to be $(\mathbb{Z}_2)^3$ as can be seen in the updated post. I am still unsure how to proceed. I know I can now find the subgroups of $(\mathbb{Z}_2)^3$ and relate this back to the galois group, but I'm unsure how to identify the elements of the galois group with the elements of $(\mathbb{Z}_2)^3$. Can anyone give me some advice? $\endgroup$ – maccamaths976 Apr 2 '17 at 18:28

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