If cancellation laws hold, then a finite semi-group is a group

Question

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.

@MartinSleziak Thanks. This is what I have been looking for. — Dilawar, May 08 '12 at 10:21
Thanks, Martin. I did not know ProofWiki exists. You make my day. — Nobody, May 08 '12 at 10:24
Thanks, @ymar. I like that proof. You make my day even brighter - I have something to think about what cancellation means. — Nobody, May 08 '12 at 11:02

score 6 · Accepted Answer · answered May 08 '12 at 14:40

6

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.

answered May 08 '12 at 14:40

Bill Dubuque

282,220

When I tried proving the same when only one cancellation law holds, it all came to me as 'Buddhist surge of consciousness' ... – Dilawar May 08 '12 at 20:48
Can I get an example for non associative structure(loop), with cancellation law holds – sabeelmsk Oct 04 '19 at 05:06

score 4 · Answer 2 · edited Jul 09 '25 at 02:23

4

This is not true [in the original version of the question where the set is not assumed finite]. The (strictly) positive integers under multiplication form a cancellative semigroup, but not a group.

edited Jul 09 '25 at 02:23

Bill Dubuque

282,220

answered May 08 '12 at 10:05

Chris Eagle

34,035

Oops! I did not thought it through. I should have said 'finite set with ..'. I will now edit it. – Dilawar May 08 '12 at 10:09
Indeed, please see the definition and the proof that a finite cancellative semigroup is a group here in Wikipediacancellative semigroup – Nobody May 08 '12 at 10:10

If cancellation laws hold, then a finite semi-group is a group

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