Simple proof involving groups

  1. Jun 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Let G be a set with an operation * such that:
    1. G is closed under *.
    2. * is associative.
    3. There exists an element e in G such that e*x = x.
    4. Given x in G, there exists a y in G such that y*x = e.

    Prove that G is a group.


    2. Relevant equations
    I need to prove that x*e = x and x*y = e, with x, y, e as given above.

    3. The attempt at a solution
    I dunno, I'm stumped. I've tried finding some sort of clever identity without any success.
     
  2. jcsd
  3. Jun 4, 2007 #2

    matt grime

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    I'm just playing around here to reduce the number of things we have to prove. We know x has a left inverse y. What is xyx? It is x(yx)=x, and it is (xy)x, thus if we could show that the e in 3 was unique, we'd have xy=e, and x would have a right inverse.

    So we just have to show that fx=x implies f=e, to obtain the existence of right inverses.

    Now, supposing that we have inverses on both sides, what can we say? Well, x*e=x(x^-1x)=(xx^-1)x=e*x, so we get a unique two sided identity.

    Putting that together, all I need to show is that fx=x implies f=e. Is that any easier? (it is implied by the existence of right inverses, but that is circular logic, so careful how you try to prove it).
     
    Last edited: Jun 4, 2007
  4. Jun 4, 2007 #3
    I'm still not making any progress. :(
     
  5. Jun 5, 2007 #4

    NateTG

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    Here's a start on one of them:
    Let's say we have some [itex]i[/itex] (not necessarily [itex]e[/itex]) with
    [tex]i \times x = x[/tex]
    then
    [tex]i \times i = i[/tex]
    [tex]i^{-1} \times (i \times i) =i^{-1} \times i[/tex]
    .
    .
    .
     
    Last edited: Jun 5, 2007
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