Suppose a finite set $G$ is closed under an associative product and that both cancellation laws hold in $G$. Prove that $G$ must be a group.

The cancellation laws are:

  1. $a\cdot$$u=a\cdot$$w$ implies $u=w$

  2. $u\cdot$$a=w\cdot$$a$ implies $u=w$


The cancellation laws imply that the map $f_a: G \to G$ given by $f_a(b)=ab$ is a permutation. Each permutation is identity when raised to an appropriate power. So there is an element $e$ that acts as identity. So you have a left unit and a left inverse (because a map given by right multiplication is also a permutation). Thus it is a group.

Just saw, that on wiki page for group there is no definition with only one-sided units and inverses. So I will continue: We know for all $a$ $ea=a$. If $ae=b$ then $a^{-1}a=e=a^{-1}ae=a^{-1}b$ so $b=a$. if $aa^{-1}=b$ then $a^{-1}e=a^{-1}=a^{-1}aa^{-1}=a^{-1}b$ thus $b=e$.

  • $\begingroup$ By being a permutation, are you saying that $G$ is abelian? $\endgroup$ – user88923 Jan 13 '14 at 8:08
  • 1
    $\begingroup$ No, I am just saying that $f_a$ is bijective $\endgroup$ – user68061 Jan 13 '14 at 8:24

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