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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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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 $2^k$). –  Yves Cornulier Aug 31 '13 at 10:20
    
See math.stackexchange.com/questions/55964 –  Derek Holt Aug 31 '13 at 17:27
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closed as off-topic by Mark Sapir, Ramiro de la Vega, Yves Cornulier, Chris Godsil, David White Aug 31 '13 at 13:55

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1 Answer

See a proof in: http://ysharifi.wordpress.com/2011/01/30/groups-with-a-cyclic-sylow-p-subgroup-2/

A more general situation see in:

Wong, W.J. On finite groups whose 2-Sylow subgroups have cyclic subgroups of index 2. J. Aust. Math. Soc. 4, 90-112 (1964).

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Do you know any example to show that "having an element of order $2^k$" is necessary ? –  user39125 Aug 31 '13 at 10:21
    
No, I don't know. I think it is not necessary. –  Boris Novikov Aug 31 '13 at 10:24
    
But both solutions that you have mentioned above use this assumption very essentially. –  user39125 Aug 31 '13 at 10:51
    
Yes, however... –  Boris Novikov Aug 31 '13 at 10:54
    
@nadal, (2) is not true (without additional assumptions); see Yves's comment. –  Anton Klyachko Aug 31 '13 at 11:38
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