# If cancellation laws hold, then a finite semi-group is a group [duplicate]

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, MacavityJun 5 '14 at 10:03

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
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

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.