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Proof of commutativity

  1. Jan 11, 2010 #1
    let G be a group following that whenever a, b and c belong to G and ab = ca, then b = c. prove that G is Abelian.

    here is what i have for the proof:

    (ab)c = c(ab)
    let c = aba^-1 (trying to find a c which which allows for commutativity)

    so (ab)aba^-1 = aba^-1(ab)
    (ab)aba^-1 = ab(a^-1 a)b
    then we see that aba^-1 = b (b = c)

    a on both sides on the right: aba^-1 a = ba

    then we can see that ab = ba which proves commutativity.

    I am not capable of using Latex at the moment, so a^-1 means the inverse of a.

    How is this? Where does b = c fit into all of this? I sort of came upon it through trying to prove commutativity and figured that that is logical. So I am not really all the comfortable with it, so I am looking for maybe some insight on how to better approach a problem like this. Thanks for any help.
  2. jcsd
  3. Jan 11, 2010 #2
    Are you assuming that
    is true? If so, how do you know this?

    I would have started my proof as follows:
    Let a,b be elements of G. So a has an inverse a-1. Clearly, b=(a-1a)b and b=b(aa-1).
  4. Jan 11, 2010 #3
    I'm sorry but i do not follow your proof there. I can see that it shows commutativity, for something, but I just don't see how it connects to the problem at hand. My professor told be to start off with something like (ab)c = c(ab) because that is the definition of commutativity (or something or other). So I am a little lost here. Perhaps it could be showing that substituting in the expression for c into the definition of commutativity and showing that it complies with the hypothesis is the right way to go here? I don't know. Any more suggestions?
  5. Jan 11, 2010 #4


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    Your suggestion to put c=aba^(-1) is a good idea. You've probably figured out that in that case ab=ca, right? So you can conclude b=c. What does that tell you? As iomtt6067 points out, you can't really start with something like (ab)c=c(ab).
  6. Jan 11, 2010 #5
    Well, we are being asked to show that G is Abelian, i.e. any two elements commute. So what I was trying to say is that if a and b are any two elements, we have

    [tex] (a^{-1}a)b=b(aa^{-1}) \Rightarrow a^{-1}(ab)=(ba)a^{-1} [/tex]

    which implies ab=ba.
  7. Jan 11, 2010 #6
    If I can't use (ab)c = c(ab), then what do I plug the equation for c into? And I think I see that if b = c, then the given ab = ca is the same as ab = ba which show commutativity. Right? But still I don't know where to begin now...
  8. Jan 11, 2010 #7


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    Plug c=aba^(-1) into ab=ca and show it's true. This really isn't as complicated as you think. Just stop thinking about (ab)c=c(ab).
  9. Jan 11, 2010 #8
    Hopefully this is the last one... c = a b a^-1 ==> b = a b a^-1 ==> ba = ab
  10. Jan 11, 2010 #9


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    Probably almost the last one. But can you state that in the form of a full proof?
  11. Jan 11, 2010 #10
    You mean write the whole thing from start to finish?
  12. Jan 11, 2010 #11


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    Um, yes? That's not so hard is it?
  13. Jan 11, 2010 #12
    ab = ca
    c = aba^-1

    ==> ab = aba^-1 a ==> ab = ab e = ab (this part I'm not sure what you were really saying, but anyhow, it works, obviously)

    so c = aba^-1 and b = c

    ==> b = aba^-1 ==> ba = aba^-1 a ==> ba = ab e = ab (commutativity so the group is Abelian)

    (e is the identity element)
  14. Jan 11, 2010 #13


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    Yes, that's it exactly. Just checking. Thanks. iomtt6076's version is also pretty nice.
  15. Jan 11, 2010 #14
    thanks for all the help.
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