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Proof showing group is abelian?

  1. Feb 1, 2012 #1
    Proof showing group is abelian???

    1. The problem statement, all variables and given/known data

    Show that every group G with identity e such that x*x=e for all x in G is abelian.





    3. The attempt at a solution

    I know that Ii have to show that it's commutative. I start by taking x,y in G and then xy is in G, so

    x*x=e
    y*y=e,
    so (x*y)*(x*y)=e

    I'm not sure where to go from here to show that it's commutative...any help is appreciated. Thank you.

    I'm not sure where to go from here
     
  2. jcsd
  3. Feb 1, 2012 #2

    jbunniii

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    Re: Proof showing group is abelian???

    So far so good. Now you want to perform some operations to both sides of the equation in order to end up with just x*y on the left hand side.
     
  4. Feb 1, 2012 #3
    Re: Proof showing group is abelian???

    @jbunniii- That's the part i'm a bit confused about, would I use the inverses in this case?
     
  5. Feb 1, 2012 #4

    jbunniii

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    Re: Proof showing group is abelian???

    Yes. You will also use a key fact about the inverses of elements in this group. What does [itex]x*x = e[/itex] imply?
     
  6. Feb 1, 2012 #5
    Re: Proof showing group is abelian???

    I see what you mean, so x=e? Thank you!
     
  7. Feb 1, 2012 #6

    jbunniii

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    Re: Proof showing group is abelian???

    No...

    x*x = e does not necessarily imply that x = e. If it did, then there would only be one element in this group, namely e.

    Think about inverses. If x * x = e, then what is the inverse of x?
     
  8. Feb 2, 2012 #7
    Re: Proof showing group is abelian???

    1/x?
     
  9. Feb 2, 2012 #8

    jbunniii

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    Re: Proof showing group is abelian???

    What's the definition of the inverse of an element?
     
  10. Feb 2, 2012 #9
    Re: Proof showing group is abelian???

    Suppose that x' denotes the inverse of x,

    then by definition of inverse x*x'=e. Are you implying that the binary operation is on x with its inverse? Or maybe x is equal to its own inverse? Maybe i'm off on some tangent here sorry...
     
  11. Feb 2, 2012 #10
    Re: Proof showing group is abelian???

    Use the fact that your group operation must be assosiative, then try and use the fact that x*x=e and y*y=e

    Do you know what it means for a group to be abelian?
     
  12. Feb 2, 2012 #11
    Re: Proof showing group is abelian???

    Basically x=x^-1 for all x in G, so xy=x^-1y^-1=(yx)^-1=yx
     
  13. Feb 2, 2012 #12

    jbunniii

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    Re: Proof showing group is abelian???

    Right, x*x = e means that x is its own inverse. This is true for every element in the group. So if

    (x * y) * (x * y) = e

    then multiplying each side on the left by x and applying associativity gives

    (x * x) * y * x * y = x * e

    and since x * x = e, the term in parentheses vanishes and you're left with

    y * x * y = x

    Now proceed to the next logical step.
     
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