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I would be grateful if someone pointed out an error in my answer.

Prove that a if finite set $G$ is closed under an associative product and that both cancellation laws hold in $G$, then it is a group. Also prove that if only one cancellation law holds, then $G$ need not be a group.

In my proof, I seem to be using only one of the cancellation laws. Hence it is a contradiction to both parts of the question.

Proof: Take an element $a\in G$. Find $a^n$ for all $n\in\Bbb{N}$. We know $G$ is closed under multiplication, and that it is also finite. Hence, there has to be some $a^x=a^y$. Using the left hand cancellation law, we get $a^{x-y+1}=a$, which implies $a^{x-y-1}$ is the inverse of $a$ ($a.a^{x-y}=a\implies a.a^{a-y-1}=e$).

Now we have to prove that the inverse of $a$ exists in $G$. Let us suppose it is $x$. Assumption: if $x$ is the inverse of $a$, then $axa=a$. Using the left cancellation law, suppose we find $a^m=a$. Then $x=a^{m-2}$. As $G$ is closed under multiplication, $x\in G$. Here I'm assuming $m\geq 3$.

Hence, $G$ is a group.

Are the assumptions (if $ax=a\implies x$ is the identity and $axa=a\implies x$ is the inverse) valid?

Thanks in advance!

  • You seem to be assuming the existence of an identity element in your proof, but that is not part of the requirement. – Tobias Kildetoft Oct 02 '13 at 09:08
  • I'm trying to prove the existence of an identity element in $G$ by proving that for every $a\in G$, there exists an element $x\in G$ such that $ax=a$. I don't think I have assumed there exists an identity element in $G$ already... –  Oct 02 '13 at 09:24
  • But the $x$ you find depends on $a$ (at least at first glance), and the identity element has to be the same for all elements. – Tobias Kildetoft Oct 02 '13 at 09:26
  • Ah I suppose you're correct. How do you propose I go about proving this? –  Oct 02 '13 at 09:32
  • Take the $x$ you have found and try to multiply it on some other element. Then multiply by $a$ and see what happens. Then use cancellation. – Tobias Kildetoft Oct 02 '13 at 09:34

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Take $G=\left\{ a,b\right\} $ with $a\neq b$ and multiplication $xy=x$. It is associative and has the righthand cancellation law. However, $G$ is not a group. Apply your 'proof' in this simple case and see what you did wrong.

drhab
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