# A generalization of an old group problem [closed]

Here is an old exercise in group theory: (1) If $G$ is a group of order $2n$ with $n$ odd then $G$ is not simple and in fact $G$ has a normal subgroup of order $n$. I am going for one straight generalization: (2) If $G$ is a group of order $2^kn$ with $k\geq 1$ and $n$ odd then $G$ is not simple and has a normal subgroup of arder $n$. The proof of (1) would use the fact that $G$ must have an element of order $2$. So I don't think that (2) is true (even I have no counterexample yet). Now in (2) lets assume that $G$ has an element of order $2^k$. Does it make (the new) (2) a ture statement ?!

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## closed as off-topic by Mark Sapir, Ramiro de la Vega, YCor, Chris Godsil, David WhiteAug 31 '13 at 13:55

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Any finite group of even order has order $2^kn$ with $k\ge 1$ and $n$ odd, so of course any non-abelian simple group is a counterexample to your "straight generalization". Boris' reply is an answer to the last question (with the additional assumption of existence of an element of order $2^k$). – YCor Aug 31 '13 at 10:20
See math.stackexchange.com/questions/55964 – Derek Holt Aug 31 '13 at 17:27

Do you know any example to show that "having an element of order $2^k$" is necessary ? – user39125 Aug 31 '13 at 10:21