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Math Help - Normal subgroups

  1. #1
    Senior Member sfspitfire23's Avatar
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    Normal subgroups

    Prove: If K is normal to H and H is normal to G, show that aKa^{-1} is normal to H for all a\in G


    Would I say:

    hKh^{-1}=k and aHa^{-1}=h then try to get H=K and subsitute the K for H? Or is that way off?

    Thanks guys!
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  2. #2
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    Let h\in H,\,aka^{-1}\in aKa^{-1}. Then

    h(aka^{-1})h^{-1}\ =\ a\underbrace{\overbrace{(a^{-1}ha)}^{\in H}k\overbrace{(a^{-1}h^{-1}a)}^{\in H}}_{\in K}a^{-1}
    Last edited by proscientia; October 31st 2009 at 03:41 AM. Reason: Typo
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  3. #3
    Senior Member sfspitfire23's Avatar
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    and because the normal subgroup is abelian i can write the haka^{-1}h^{-1} as ahkh^{-1}a^{-1} as you did correct?
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    Quote Originally Posted by sfspitfire23 View Post
    and because the normal subgroup is abelian i can write the haka^{-1}h^{-1} as ahkh^{-1}a^{-1} as you did correct?

    No! Read again, very carefully and writing down the proof ina sheet of paper, the proof: in no place he did write anything from where one can deduce the normal sbgp. is abelian which, of course, is false in general...and he didn't write what you say he did.

    Tonio
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  5. #5
    Senior Member sfspitfire23's Avatar
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    I guess what I dont understandis how the a and a^{-1} get pushed outside.

    Heres what I have so far.

    We have to show h(aka^{-1})h^{-1} is true. Then, aha^{-1}=h and I can write the thing as aha^{-1}(aka^{-1})a^{-1}h^{-1}a then I get stuck
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  6. #6
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    Quote Originally Posted by sfspitfire23 View Post
    I guess what I dont understandis how the a and a^{-1} get pushed outside.

    Heres what I have so far.

    We have to show h(aka^{-1})h^{-1} is true. Then, aha^{-1}=h and I can write the thing as aha^{-1}(aka^{-1})a^{-1}h^{-1}a then I get stuck

    Take h(aka^{-1})h^{-1} and multiply it both from the left and from the right by 1=aa^{-1}, and then arrange parentheses accordingly (using associativity) and then use normality of H in G and normality of K in H. That's all the trick.

    Tonio
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