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Math Help - How do you show that this is abelian?

  1. #1
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    Smile How do you show that this is abelian?

    Show that if H,K are normal to G and H intersect K = {1}, then hk=kh for all h in H, and k in K.

    I have tried playing around with this for a long time. I just can't quite get it right. I've used the facts of the normal subgroup tests, gHg^-1 in H and gKg^-1 in K, the fact that gH=Hg and gK=Kg. I've tried using kHk^-1 in H and hKh^-1 in K. I just can't quite get it to work. All I am able to achieve is getting hk=k2h and kh=h2k...

    Any tips would be greatly appreciated!!! I have tried searching around a bit and can't find anything.
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  2. #2
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    Quote Originally Posted by xoyankeegirlx3 View Post
    Show that if H,K are normal to G and H intersect K = {1}, then hk=kh for all h in H, and k in K.

    I have tried playing around with this for a long time. I just can't quite get it right. I've used the facts of the normal subgroup tests, gHg^-1 in H and gKg^-1 in K, the fact that gH=Hg and gK=Kg. I've tried using kHk^-1 in H and hKh^-1 in K. I just can't quite get it to work. All I am able to achieve is getting hk=k2h and kh=h2k...

    Any tips would be greatly appreciated!!! I have tried searching around a bit and can't find anything.
    h^{-1}k^{-1}hk=h^{-1}\left(k^{-1}hk\right)\in H , since H is normal, but also h^{-1}k^{-1}hk=\left(h^{-1}k^{-1}h\right)k\in K ,

    because K is normal, thus...

    Tonio
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xoyankeegirlx3 View Post
    Show that if H,K are normal to G and H intersect K = {1}, then hk=kh for all h in H, and k in K.

    I have tried playing around with this for a long time. I just can't quite get it right. I've used the facts of the normal subgroup tests, gHg^-1 in H and gKg^-1 in K, the fact that gH=Hg and gK=Kg. I've tried using kHk^-1 in H and hKh^-1 in K. I just can't quite get it to work. All I am able to achieve is getting hk=k2h and kh=h2k...

    Any tips would be greatly appreciated!!! I have tried searching around a bit and can't find anything.
    Also, one can show that h,k)\mapsto hk" alt="\phi:H\times K\to Gh,k)\mapsto hk" /> is an isomorphism--this tells you more than you need though.
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    Quote Originally Posted by Drexel28 View Post
    Also, one can show that h,k)\mapsto hk" alt="\phi:H\times K\to Gh,k)\mapsto hk" /> is an isomorphism--this tells you more than you need though.

    The above is an isomorphism iff G=HK , otherwise it is only a monomorphism.

    Tonio
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    The above is an isomorphism iff G=HK , otherwise it is only a monomorphism.

    Tonio
    Tonio,

    I did not see that this condition was not included. My mistake.
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    Tonio,

    I did not see that this condition was not included. My mistake.


    What?!!? You oversaw, or added on your own, a condition? Welcome to the club...

    Tonio
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