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Abelian math help

  1. Sep 29, 2004 #1
    I want to show that if G is a group where there exists an [tex] x [/tex] which is it's own inverse then G is abelian. Ie [tex] x * x = e [/tex]. I get the hint that let [tex] x = ab [/tex]. So we have abab=e, I'm not sure how to continue from this. But I think I should try something like this

    [tex] (1) \quad a*abab = a [/tex]
    [tex] (2) \quad abab*a = a [/tex]
    [tex] (3) \quad b*abab = b [/tex]
    [tex] (4) \quad abab*b = b [/tex]

    But I'm not sure how to continue ... please give me help but don't spoil it :)
     
  2. jcsd
  3. Sep 29, 2004 #2

    jcsd

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    I'm slightly puzzled as all groups contain an idenityt so, every group has an elemnt such that x*x = e, namely e.




    I can also think of Abelian and non-Abelian groups that contain elements such that x*x = e were x is not equal to e (for example muplication in the rationals and matrix mutplication).
     
  4. Sep 29, 2004 #3
    Hmm I might have understood it wrong. But it say

    Prove that in a group G where [tex] x \ast x = e [/tex] for all [tex] x \in G [/tex] is abelian. Hint: Look at [tex] (a \ast b) \ast (a \ast b) [/tex]. Maybe I misunderstood the question ...

    Edit: I think I can prove it now, I MUST have misunderstood. We have
    [tex] x \ast x = e [/tex] let the x:s be [tex] x_1 [/tex] and [tex] x_2 [/tex] with [tex] x_1 = x_2 [/tex] (Just to make it easier to see the difference).

    [tex] x_1 \ast x_2 = e [/tex]
    [tex] x_1 \ast x_2 \ast x_2^{-1} = x_2^{-1} [/tex]
    [tex] x_1 \ast e = x_2^{-1} [/tex]
    [tex] x_2 \ast x_1 = x_2 \ast x_2^{-1} [/tex]
    [tex] x_2 \ast x_1 = e [/tex]
    And since [tex] x_1 \ast x_2 = x_2 \ast x_1 [/tex] the group G must be abelian?
     
    Last edited: Sep 29, 2004
  5. Sep 29, 2004 #4
    Yes, there is a difference between "there exists an x such that x*x = e" (which you said in your first post) and "for all x, x*x = e".

    Another hint is to consider (ab)^-1 (there is a formula for that expression in terms of the inverses of a and b).
     
  6. Sep 29, 2004 #5

    jcsd

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    I'm still a little confused tho', does that also implicit that x is not equal to e?
     
  7. Sep 29, 2004 #6
    I'm not sure what you mean. Are you asking if x^2 = e for all x in G => x != e...? x could of course be e.
     
  8. Sep 29, 2004 #7

    jcsd

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    Right now, sorry I relaize now, the orginal quetsion wasn't the the question that he wanted answered.
     
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