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


by grimster
Tags: invg, monoid, prove
grimster
grimster is offline
#1
Apr12-05, 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?
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Palindrom
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#2
Apr12-05, 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?
grimster
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#3
Apr12-05, 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

Palindrom
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#4
Apr12-05, 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?
Hurkyl
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#5
Apr12-05, 03:07 PM
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It's not his proof. When doing homework, one should present one's own work, not copy someone else's.
grimster
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#6
Apr12-05, 03:36 PM
P: 39
Quote Quote by Palindrom
Excuse me- what's the problem with my proof?
i don't know. it looked kind of "sloppy". i guess it is ok, but i wanted something a bit more "structured"...
Palindrom
Palindrom is offline
#7
Apr12-05, 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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