March 2nd, 2010, 01:40 AM  #1 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  normal subgroup
show that an intersection of normal subgroups of a group G is again a normal subgroup of G. thanks 
March 2nd, 2010, 06:59 AM  #2 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: normal subgroup
What's the definition of a normal subgroup? What have you tried so far?

March 3rd, 2010, 07:27 PM  #3 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  Re: normal subgroup
Define H?K = {hh?H and h?K} let h?H?K, ie. h?H and h?K then ghg^1?H and ghg^1?K. Therefore we have ghg^1?H?K . We want to show that g(H?K) = (H?K)g for all g?G. => proof: for all g?G and for all h?(H?K) there exist h' such that g^1hg=h' => gg^1hg=gh' => hg=gh' =>hg?H?K . Hence (H?K)g ?g(H?K). <= proof: for all g?G and for all h?(H?K) there exist h' such that ghg^1=h' => ghg^1g=h'g => gh=h'g => gh?(H?K)g. Hence g(H?K) ?(H?K)g. is this right? 
March 3rd, 2010, 08:56 PM  #4  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: normal subgroup Quote:
(that condition is actually equivalent to the condition you prove below.) You should probably say where g comes from. For the sake of pedantry (and hopefully pedagogy) Quote:
Other than those two things, your proof is correct... but it's the proof that ghg^1?H => gH = Hg. The other direction is basically the same.  
March 3rd, 2010, 09:15 PM  #5 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  Re: normal subgroup
sorry but I am confuse when you say " Also, I think you meant. gh'?(H?K)g". is it because hg=gh' therefore gh'?(H?K)g. I thought gh'?g(H?K).

March 4th, 2010, 04:14 AM  #6  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: normal subgroup Quote:
Your proof is showing that g(H?K) ?(H?K)g and (H?K)g ? g(H?K); so you take an element of g(H?K) and show it's in (H?K)g (and vice versa).  

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