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Homework Help: Question about isomorphic direct products of groups and isomorphic factors.

  1. Dec 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose G and F are groups and GxF is isomorphic to G'xF', if G is isomorphic to G', can we conclude that F is isomorphic to F'?

    2. Relevant equations



    3. The attempt at a solution
    I'm trying to give a proof using the first isomorphism theorem (using that GxF/Gx(e) is isomorphic to F, and that G'xF'/G'x(e) is isomorphic to F'), but I can't find an isomorphism between the quotients. I also can't find a counter example of the statement, so any help or suggestions would be appreciated.
     
  2. jcsd
  3. Dec 5, 2012 #2
    Did you know that [itex]\mathbb{R}^2[/itex] and [itex]\mathbb{R}[/itex] are isomorphic as groups?? Try to prove this.
     
  4. Dec 5, 2012 #3
    I didn't know that, thanks a lot that solves my problem.

    The statement of the problem is false then. Because if RxR is isomorphic to R, then it's also isomorphic to Rx(e), and the statement of the problem would imply that R is isomorphic to the trivial group, which is false.

    Thanks : )
     
  5. Dec 5, 2012 #4
    Yeah, but I think you still need to prove that [itex]\mathbb{R}^2[/itex] is isomorphic to [itex]\mathbb{R}[/itex]. This is not trivial.
     
  6. Dec 5, 2012 #5
    Yeah, I still need to prove that, but at least I know that the initial statement is wrong.
     
  7. Dec 5, 2012 #6

    Dick

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    If you want an example that's a little more manageable then the R^2, R thing, try taking G to be an infinite direct product of factors of Z (the integers). Or any other group you like.
     
  8. Dec 5, 2012 #7
    That one is actually a really nice counterexample, since it generalizes to other fields of mathematics as well. A similar example works in topology, for example. The [itex]\mathbb{R}^2[/itex] thing does not.
     
  9. Dec 5, 2012 #8

    Dick

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    And you can actually write down what the isomorphism is explicitly.
     
  10. Dec 5, 2012 #9
    Thanks a lot, that works.
     
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