
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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! 

Tags 
containts, group, order, proof, subgroup 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Proof conjecture about the order of group  jonbryan80  Abstract Algebra  6  October 21st, 2013 12:50 AM 
Order of a Subgroup M  flostamjets  Abstract Algebra  1  December 24th, 2012 01:48 AM 
Subgroup of Group with Order = Prime^2  Relmiw  Abstract Algebra  9  April 10th, 2011 10:50 AM 
Normal sylow subgroup of a group of order p^aq^b  Spartan Math  Abstract Algebra  6  September 21st, 2009 09:52 PM 
Find a abelian subgroup of S5 of order 6  shari2004  Abstract Algebra  1  February 16th, 2009 07:05 AM 