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.
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.