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This question already has an answer here:

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 $2n$?

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marked as duplicate by Pierre-Guy Plamondon, Namaste group-theory Dec 2 '17 at 1:45

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    $\begingroup$ This seems like a weaker form to me. $\endgroup$ – Arthur Dec 1 '17 at 12:32
  • $\begingroup$ 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. $\endgroup$ – quantum Dec 1 '17 at 12:38
  • $\begingroup$ It seems you are correct: math.stackexchange.com/questions/53708/… $\endgroup$ – quantum Dec 1 '17 at 12:39
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    $\begingroup$ 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 $-1$. $\endgroup$ – Pierre-Guy Plamondon Dec 1 '17 at 12:55
  • $\begingroup$ The inverse Galois problem for abelian groups/extensions is much easier because we can deal with cyclotomic extensions. $\endgroup$ – reuns Dec 1 '17 at 16:09