My Math Forum Proof that group of order 2k containts order k subgroup

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 April 22nd, 2012, 01:52 PM #1 Newbie   Joined: Feb 2012 Posts: 10 Thanks: 0 Proof that group of order 2k containts order k subgroup Hi everyone, I'm stuck on this proof. G is a finite group of order 2k, with k odd. Prove that G contains a normal subgroup of order k. My attempt thus far: I reduced the problem to finding a subgroup of order k, since that subgroup is automatically a normal subgroup, since the index is 2. I know that if I can prove that G contains an element of order k, the subgroup generated by that element will have order k. That would complete the proof. So the problem now is: prove that G contains an element of order k (k odd). I can't seem to figure this one out. If k is prime, or a product of distinct primes (ie squarefree), G must contain an element of order k. I cannot generalise my proof. Am I on the right track? Can anyone give me some hints? Help much appreciated!

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### a group of order 2k ( k is odd) has a normal subgroup

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