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March 2nd, 2010, 01:40 AM   #1
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normal subgroup

show that an intersection of normal subgroups of a group G is again a normal subgroup of G.

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March 2nd, 2010, 06:59 AM   #2
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Re: normal subgroup

What's the definition of a normal subgroup? What have you tried so far?
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March 3rd, 2010, 07:27 PM   #3
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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?
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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.
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March 3rd, 2010, 09:15 PM   #5
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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).
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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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