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Using one-sided axioms to show <G,*> is a group

  1. Sep 26, 2007 #1

    eq1

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

    Prove that a set G, together with a binary operation * on G, satisfying the
    following three axioms:
    A1) The binary operation * on G is associative
    A2) There exists a left identity element e in G such that e*x=x for all x in G
    A3) For each a in G, there exists a left inverse a' in G such that a'*a=e
    is a group

    2. Relevant equations

    Our definition of a group:

    A group is a set G, and a closed binary operation * on G, such that
    the following axioms are satisfied:
    G1) * is associative on G
    G2) There is an element e in G such that for all x in G
    x*e = e*x = x (existence of an identity element e for *)
    G3) Corresponding to each a in G there is an element a' in G such that
    a*a' = a'*a = e (existence of inverse a' of a)


    3. The attempt at a solution

    G1, is given by A1

    By A2 we know e*x=x for all x in G.
    We must show x*e=x for all x in G to finish proving G2.

    x*e=x, what is to be proved
    x'*(x*e)=x'*x, we know x' exists from A3, existence of inverse
    (x'*x)*e=e, using A1, associativity
    e*e=e, using A3, existence of inverse
    e=e, therefore x*e=x for x in G proving G2 holds.

    By A3 we know there exists x' in G such that x'*x=e for all x in G.
    We must show x*x'=e for all x in G to finish proving G3.

    x*x'=e, what is to be proved
    (x'*x)*x'=x'*e, using A1, associativity
    e*x'=x', using the above theorem
    x'=x', therefore x*x'=e for x in G proving G3 hold.

    Therefore the set G, together with a binary operation * on G, satisfying
    A1, A2 and A3 is a group.

    4. My questions

    Does ending with the e=e statement or x'=x' statement show the original statement holds? Since it ends up with LHS=RHS I think it does but maybe I should try to string statements in the form e*x = ... = x*e although I am not sure why that would be better. I guess it is just what I am used to seeing.

    Also, I skipped the part about the binary operation being closed for a group. Kind of important, I know. I think (hope) it was indirectly mentioned in the axioms when it was stated * in associative -on- G. Is this correct or is it something I will also need to prove? If so, any ideas?

    Thank you for taking the time to read all this.
     
  2. jcsd
  3. Sep 26, 2007 #2
    "Does ending with the e=e statement or x'=x' statement show the original statement holds?"

    No, you've only shown that e=e and x'=x'...profound statements but not what you set out to show. Here's a hint: x' is in G so x' also has a left inverse...what is it?

    As for the closure, I am unsure. I don't see how one could show the operation is closed without more information about the nature of the group.
     
  4. Sep 26, 2007 #3

    HallsofIvy

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    Doesn't the phrase "binary operation on G" (as opposed to "binary operation from G to H") imply the operation always gives a value in G? In other words, can't you take closure as "given" by the hypotheses?
     
  5. Sep 26, 2007 #4

    eq1

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    mcampbell,
    I think understand what you are trying to tell me and unfortunately I don't think I even showed e=e and x'=x' since I started from unproved statements. I'll work with your suggestion now and see what I can come up with.

    HallsofIvy,
    I was thinking that too but I wasn't positive. Specifically I wasn't sure if on and onto were equivalent. I kind of suspected they were as my text seems to use it that way, but it doesn't explicity state it.

    Thank you guys for the replies.
     
  6. Sep 26, 2007 #5

    quasar987

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    It's not hard, but you gotta start at the right place or you're gonna try for a long god damn time. The right place to start is first with the inverse business. You're gonna try to show that the left inverse of each element in also a right inverse. And for this, you're gonna start with the equation

    a*a'=a*e*a'

    Then showing the left identity is also a right identity is easy.
     
  7. Sep 27, 2007 #6

    eq1

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    Wow quasar987, you were not kidding about the long god damn time. I spun on this for almost two hours. But when I combined your approach and mcampbell's approach I think I have a proof that works.

    Proof of right inverse

    As x' is in G by A3 there exists an inverse, say x'', such that x''*x'=e

    Thus x'*x = e, and x''*x'=e
    setting them equal to eachother
    x' * x = x'' * x'
    x' * x = x'' * e * x'
    x' * x = x'' * x' * x * x'
    x' * x = e * x * x', by associativity and left identity
    x' * x = x * x' = e

    And the right identity follows easily.
    Thank you guys so much. This problem is going to give me nightmares. So simple but so hard at the same time.
     
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