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

  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. 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. 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. Excuse me- what's the problem with my proof?
     
  6. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    It's not his proof. When doing homework, one should present one's own work, not copy someone else's.
     
  7. i don't know. it looked kind of "sloppy". :rofl: i guess it is ok, but i wanted something a bit more "structured"...
     
  8. 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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