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Are the following result is true?

Every group of order 2n (n odd) is solvable.

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1 
 Yes, but this is not really a research level question (even though the proof does use Feit-Thompson which is certainly not an easy theorem to prove) – Tobias Kildetoft Mar 31 2011 at 13:43
As Tobias said, the result is true. You can find the details of the proof (which is easy, if you accept Feit-Thompson theorem) in this paper: marathwadamathsociety.org/vol10-1/6-Salunke%20Gotmare.pdf – Francesco Polizzi Mar 31 2011 at 13:49
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 Note that what is entirely elementary is to show that no group of order congruent to $2$ mod $4$ is simple. This is in fact a standard exercise in introductory group theory texts, but one that many people seem to forget over the years. (I speak of course about myself...) – Pete L. Clark Mar 31 2011 at 14:32

closed as too localized by Pete L. Clark, Simon Thomas, Mark Sapir, Zev Chonoles, Andy Putman Mar 31 2011 at 16:47

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