MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?!

share|cite|improve this question

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

This question appears to be off-topic. 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, YCor, David White
If this question can be reworded to fit the rules in the help center, please edit the question.

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 – Derek Holt Aug 31 '13 at 17:27

See a proof in:

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).

share|cite|improve this answer
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