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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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    @scaaahu Thank you, I'm glad my answer is useful.2012-05-08

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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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    When I tried proving the same when only one cancellation law holds, it all came to me as 'Buddhist surge of consciousness' ...2012-05-08
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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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    Indeed, please see the definition and the proof that a finite cancellative semigroup is a group here in Wikipedia[cancellative semigroup](http://en.wikipedia.org/wiki/Cancellative_semigroup)2012-05-08