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I do not know if this correct. So I ask.

Given a number $n\geq 2$, can we find a Galois extension of $\mathbb Q$ such that the group has order $n$? Similarly, given $n\in \mathbb N$ can we find a totally imaginary number field that is a Galois extension of $\mathbb Q$ and for which the order of its Galois group is $2n$?

quantum
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    This seems like a weaker form to me. – Arthur Dec 01 '17 at 12:32
  • Oh is it. Ok I'm sorry. Feel free to correct if you know better. I was trying to figure if it should be "stronger" or "weaker". I am not confident. So please correct if you think the title of the question needs to be corrected. Thanks in advance. – quantum Dec 01 '17 at 12:38
  • It seems you are correct: https://math.stackexchange.com/questions/53708/precise-definition-of-weaker-and-stronger – quantum Dec 01 '17 at 12:39
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    According to this question and answer, there is always a real and cyclic extension of $\mathbb{Q}$ of degree $n$ (for any $n\geq 2$). To get a totally imaginary extension of degree $2n$, simply add a square root of $-1$. – Pierre-Guy Plamondon Dec 01 '17 at 12:55
  • The inverse Galois problem for abelian groups/extensions is much easier because we can deal with cyclotomic extensions. – reuns Dec 01 '17 at 16:09
  • Pierre-Guy I think your answer is valid and I would gladly mark it as an answer if you put it in as answer instead of comment! Thanks! – quantum Dec 09 '17 at 15:41
  • @Arthur I changed the title from "Stronger" to "Weaker". I think you are right. Thanks – quantum Dec 09 '17 at 15:46

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