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This question already has an answer here:

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.

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marked as duplicate by user1729, Najib Idrissi, Davide Giraudo, Norbert, Macavity Jun 5 '14 at 10:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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  • $\begingroup$ When I tried proving the same when only one cancellation law holds, it all came to me as 'Buddhist surge of consciousness' ... $\endgroup$ – Dilawar May 8 '12 at 20:48
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This is not true. The (strictly) positive integers under multiplication form a cancellative semigroup, but not a group.

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  • $\begingroup$ Oops! I did not thought it through. I should have said 'finite set with ..'. I will now edit it. $\endgroup$ – Dilawar May 8 '12 at 10:09
  • $\begingroup$ Indeed, please see the definition and the proof that a finite cancellative semigroup is a group here in Wikipediacancellative semigroup $\endgroup$ – scaaahu May 8 '12 at 10:10

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