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Prove that inv(G) is a group. G is a monoid... 
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#1
Apr1205, 09:24 AM

P: 39

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? 


#2
Apr1205, 10:10 AM

P: 266

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? 


#3
Apr1205, 12:54 PM

P: 39

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 


#4
Apr1205, 02:53 PM

P: 266

Prove that inv(G) is a group. G is a monoid...
Excuse me what's the problem with my proof?



#5
Apr1205, 03:07 PM

Emeritus
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PF Gold
P: 16,091

It's not his proof. When doing homework, one should present one's own work, not copy someone else's.



#6
Apr1205, 03:36 PM

P: 39




#7
Apr1205, 03:56 PM

P: 266

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...



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