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Finding an operation that makes a group

  1. Jun 26, 2006 #1

    learningphysics

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    Hi. I'm working through the book "Elements of Abstract Algebra" by Allan Clark. This question has stumped me... find an operation on (0,1) (set of reals x such that 0<x<1 ) that makes (0,1) a group and makes the inverse of x, 1-x.

    I'd appreciate any help. Thanks.
     
    Last edited: Jun 26, 2006
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  3. Jun 26, 2006 #2

    AKG

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    Do you know of an operation on R that makes it into a group with -x the inverse of x? Think of a way to make (0,1) correspond to R such that you can define your operation on (0,1) in terms of the very familiar operation on R.
     
  4. Jun 26, 2006 #3

    NateTG

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    Another thing you might want to think about is what the identity of the group will be.
     
  5. Jun 26, 2006 #4

    learningphysics

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    Thanks AKG and Nate. I believe I've found the answer but I'm not sure why the method I used "works"...

    I looked for a way to make (0,1) correspond to R as AKG said. I used the function: [tex]\frac{1}{2}tanh(u) + \frac{1}{2}[/tex]

    From here I decided to multiply out [tex]\frac{1}{2}tanh(u + v) + \frac{1}{2}[/tex]

    Then after setting [tex]x = \frac{1}{2}tanh(u) + \frac{1}{2}[/tex] and [tex]y = \frac{1}{2}tanh(v) + \frac{1}{2}[/tex], I rewrote the above as:

    [tex]\frac{xy}{1+2xy-x-y}[/tex] which is the product.

    (I got this idea because the previous problem had the function [tex]\frac{x+y}{1+xy} [/tex] which you can get from substiting [tex]x=tanh(u)[/tex] and [tex]y=tanh(v)[/tex] into [tex]tanh(x+y)[/tex] but I really don't understand why it works)

    Why did this method work? Thanks again for all your help.
     
    Last edited: Jun 26, 2006
  6. Jun 26, 2006 #5

    Hurkyl

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    Because the target of a group isomorphism is a group. You just constructed the isomorphism before you figured out what the group was.
     
  7. Jun 26, 2006 #6

    mathwonk

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    that is a wonderful book. hang in there and you will learn a lot.
     
  8. Jun 26, 2006 #7

    learningphysics

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    Thanks Hurkyl and mathwonk. The book hasn't covered group isomorphisms yet, but I looked it up and everything makes perfect sense.
     
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