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 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 = {h|h?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
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Re: normal subgroup

Quote:
 Originally Posted by tinynerdi Define H?K = {h|h?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 .
Isn't this if and only if? I.e. (forall h?H) ghg^-1?H iff H is normal? Then there's your proof.
(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:
 => 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).
h', of course, comes from H?K. Also, I think you meant. gh'?(H?K)g

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
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Re: normal subgroup

Quote:
 Originally Posted by tinynerdi 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).
Sorry, I meant at the end when you say gh'?(H?K), which it is not

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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