December 10th, 2008, 11:30 AM  #1 
Newbie Joined: Dec 2008 Posts: 1 Thanks: 0  Normal Groups
Let G a group , let Define Please explain me the proof that N(a) is a normal subgruop in G. Thanks 
December 11th, 2008, 05:27 PM  #2 
Senior Member Joined: Dec 2008 Posts: 160 Thanks: 0  Re: Normal Groups
By the definition of normal subgroup: For all a in G, aN(~a) = N: if xa = ax, (~a) is an a inverse, so xa(~a) = ax(~a) => N  normal.

December 23rd, 2008, 07:36 AM  #3 
Member Joined: Jul 2008 From: Minnesota, USA Posts: 52 Thanks: 0  Re: Normal Groups
I really don't think this is true. In fact, I'm almost certain this is not true. Counter examples aren't much fun in group theory. Hmm.... This N(a) is known as the CENTRALIZER of a in G, by the way. I would try letting a = (12345) in the permutation group S_5. The other powers of a give you that the subgroup N(a) has at least order 5. Yet, N(a) is definately not the whole group, just try conjugating a by any transposition. I could be wrong of course, but this just seems like a ridiculously strong statement. 
December 23rd, 2008, 11:54 AM  #4  
Newbie Joined: Dec 2008 Posts: 2 Thanks: 0  Re: Normal Groups Quote:
NClement is right, but here is a more precise reason why. Lets restrict ourselves to A_5. With a = (12345), we know that N(a) contains more than the identity as NClement pointed out. So if N(a) were normal then A_5 would have a normal subgroup contradicting simplicity of A_5. And just to note: if a is in G then {gag~: g is in G} is a normal subgroup in G. However, in my counterexample above this set would exapnd to be all of A_5.  

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