# Only one cancellation law? Then $G$ may not be a group…

Suppose that the following result is known:

"Let $G$ be a finite set, closed with respect to an associative product and that both of the cancellation laws are valid. Then $G$ is a group with respect to this product."

The question is: "Find a counter-example which shows that, if one supposes only one of the cancellation laws, then we can't to conclude that $G$ is a group."

Cancellation laws: $$ax=ay \implies x=y$$ $$xa=ya \implies x=y$$

Which is this counter-example?! I can't find it!! Need some help...

## 1 Answer

Define $xy=x$ for all $x,y\in G$. This operation is associative and satisfies the right cancellation law but not the left.

• Thank you very much @Brian M. Scott. You really saved me. ;) – Ders Apr 12 '15 at 13:58
• @Anderson: You’re welcome! – Brian M. Scott Apr 12 '15 at 13:59