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Prove that inv(G) is a group. G is a monoid

  1. Apr 12, 2005 #1
    G is a monoid. Inv(G) = {a E G, exists b so that a b = b a = 1}

    prove that Inv(G) is a group. it's pretty obvious that inv(G) is a group. a monoid is a set with a law of composition which is associative and has a unit element. so inv(G) is clearly a group, because for all a in inv(G) there is an inverse element. but how do i prove this?

    i guess i have to show that for each x in inv(G), the inverse of x is also in G. but how do i do that?
     
    Last edited: Apr 12, 2005
  2. jcsd
  3. Apr 12, 2005 #2
    Given x in inv(G), you know there is a y in G so that xy=yx=1.
    Let's look at y. Guess what? yx=xy=1. But then y is in inv(G) by definition...
    1 is in inv(G), * is associative... did I forget something?
     
  4. Apr 12, 2005 #3
    this is what i have so far. e is the identity element.

    there is an element c such that c a = e.
    a=ea=cba=cbae=eae=ae
    further
    cba=ce=c
    and
    cba=ea=a

    have
    ab=ba=e
     
  5. Apr 12, 2005 #4
    Excuse me- what's the problem with my proof?
     
  6. Apr 12, 2005 #5

    Hurkyl

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    It's not his proof. When doing homework, one should present one's own work, not copy someone else's.
     
  7. Apr 12, 2005 #6
    i don't know. it looked kind of "sloppy". :rofl: i guess it is ok, but i wanted something a bit more "structured"...
     
  8. Apr 12, 2005 #7
    I agree with you both- but that's why I gave it in a sloppy way. It's not hard to formalize it, but it's your job... :wink:
     
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