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Mapping an isomorphism b/w 2 grps

  1. Feb 14, 2013 #1
    I googled this but couldn't find a clear answer.

    Is every invertible mapping an isomorphism b/w 2 grps or does it have to be linear?
     
  2. jcsd
  3. Feb 14, 2013 #2
    Re: Isomorphism

    also does an isomorphism maps connected (separated) sets to connected (separated) sets?
     
  4. Feb 14, 2013 #3

    jbunniii

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    Re: Isomorphism

    No!
    It has to be invertible AND a homomorphism, meaning it must satisfy ##\phi(ab) = \phi(a)\phi(b)##, where ##\phi## is the mapping and ##a,b## are arbitrary elements of the group. Here, the group operation is written multiplicatively. The additive version is ##\phi(a+b) = \phi(a) + \phi(b)##.

    By the way, one might think that it would also be necessary to stipulate that ##\phi^{-1}## is a homomorphism, but that turns out to be automatically true if ##\phi## is a bijection and a homomorphism.
     
  5. Feb 14, 2013 #4

    jbunniii

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    Re: Isomorphism

    Are we still talking about group isomorphisms? There is no notion of "connected" or "separated" for a general group. You need to impose some additional topological structure. So what kind of groups are you working with?
     
  6. Feb 14, 2013 #5
    Re: Isomorphism

    The whole question has to deal with analysis..

    if A is connected and we have T: A ---> B an isomorphism, can we say T(A) in B is connected?

    I guess one still have to show that a mapping is a homomorphism even in analysis. right?
     
  7. Feb 14, 2013 #6

    micromass

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    Re: Isomorphism

    OK, but you are clearly not working with just groups. What is the structure you are working with?? What are A and B?? What kind of map is T? It's an isomorphism of what?
     
  8. Feb 15, 2013 #7
    I think I was confusing the invertibilty of a Linear Mapping between 2 Vector Spaces with any function that has an inverse.

    I remember in my Lin. Alg. course, we learned that if a Linear Transformation T is invertible, then it is an isomorphism between the 2 VS.

    Clearly this is not the general case.
     
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