Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Show that if both cancellation laws i.e $w.a = w.b \implies a = b$ and $a.w = b.w \implies a = b$ holds then a finite semi-group (a finite set with associative binary operation) is a group.

I have seen some proofs which uses the alternative definition of group to prove it i.e. $a.x = b$ and $y.a =b$ have unique solutions for $x$ and $y$. I am not interested in such proofs.

How to prove this statement starting with cancellation laws and then showing that all axioms of group can be derived from them?

EDIT : As pointed out in one of the answer. This is only true when underlying set is finite. Edited accordingly.

share|improve this question
    
See Cancellable Finite Semigroup is Group at ProofWiki. –  Martin Sleziak May 8 '12 at 10:16
    
@MartinSleziak Thanks. This is what I have been looking for. –  Dilawar May 8 '12 at 10:21
    
Thanks, Martin. I did not know ProofWiki exists. You make my day. –  scaaahu May 8 '12 at 10:24
2  
I posted a proof of this here –  user23211 May 8 '12 at 10:34
    
Thanks, @ymar. I like that proof. You make my day even brighter - I have something to think about what cancellation means. –  scaaahu May 8 '12 at 11:02
show 1 more comment

2 Answers

up vote 4 down vote accepted

Hint $\rm\ \ \ell_a(x) = a\:\!x$ is $1\!-\!1$ so onto. So $\rm\:a\to \ell_a\:$ represents S as a subsemigroup of the finite group of permutations on S, which is necessarily a group, since every element has finite order.

Remark $\ $ Notice how conceptual the proof becomes using this regular representation (Cayley). Exploiting these structural insights reveals the essence of the matter with minimal calculation.

share|improve this answer
    
When I tried proving the same when only one cancellation law holds, it all came to me as 'Buddhist surge of consciousness' ... –  Dilawar May 8 '12 at 20:48
add comment

This is not true. The (strictly) positive integers under multiplication form a cancellative semigroup, but not a group.

share|improve this answer
    
Oops! I did not thought it through. I should have said 'finite set with ..'. I will now edit it. –  Dilawar May 8 '12 at 10:09
    
Indeed, please see the definition and the proof that a finite cancellative semigroup is a group here in Wikipediacancellative semigroup –  scaaahu May 8 '12 at 10:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.