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Show that a group has exactly one idempotent element

  1. Feb 20, 2017 #1
    1. The problem statement, all variables and given/known data
    Prove that a group has exactly one idempotent element.

    2. Relevant equations


    3. The attempt at a solution
    So we need to show that the identity element is the unique idempotent element in a group.

    First, we know that by definition of a group there is at least one element, e, such that ##e * e = e##.

    Second, we need to show that there is at most one idempotent element. We do this by showing that if ##x*x=x## and ##y*y=y## then ##x=y##... This is as far as I get. Am I on the right track?
     
  2. jcsd
  3. Feb 20, 2017 #2

    fresh_42

    Staff: Mentor

    What does it mean for ##x\cdot x = x^2=x\,##? Any idea to get rid of one ##x\,##?
     
  4. Feb 20, 2017 #3
    I can use the cancellation law. But how does that logically show that e is the unique element in the group that is idempotent?
     
  5. Feb 20, 2017 #4

    fresh_42

    Staff: Mentor

    If ##x## is any idempotent element, i.e. ##x^2=x## then you can multiply (as in school on both sides) the whole equation with ##x^{-1}##, which is probably what you meant by cancellation. You can do this, because all elements of a group have an inverse. Then - if you want to be very rigorous and pedantic - you can apply associativity and the existence and definition of ##e##. Write it down and see what it says.
     
  6. Feb 20, 2017 #5

    FactChecker

    User Avatar
    Science Advisor
    Gold Member

    You know more than that about e. You know x*e=x. Use that. The fact that e*e = e should tell you that you need to prove that x=e.
     
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