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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...

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

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

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