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 offtopic by Mark Sapir, Ramiro de la Vega, Yves Cornulier, Chris Godsil, David White Aug 31 '13 at 13:55
This question appears to be offtopic. The users who voted to close gave this specific reason:
 "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ramiro de la Vega, Yves Cornulier, David White
See a proof in: http://ysharifi.wordpress.com/2011/01/30/groupswithacyclicsylowpsubgroup2/ A more general situation see in: Wong, W.J. On finite groups whose 2Sylow subgroups have cyclic subgroups of index 2. J. Aust. Math. Soc. 4, 90112 (1964). 

