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Maybe all that I am writing below is already exists at forum but I would like to know is my reasoning correct? So please do not duplicate my question. I am in the begining of group theory and it is very important to me to know that my reasoning is correct.

We know the following fact from group theory: Suppose a finite set $G$ is closed under an associative product and that both cancellation laws hold in $G$ $\Rightarrow$ $G$ must be a group.

However, if suppose that only one of the cancellation laws hold then $G$ need not to be a group.

Let $G$ be the set with three elements, i.e. $G=\{a,b,e\}$ and we define the product $x\cdot y=x$ for any $x,y\in G$. We see that this product is binary and associative since $(x\cdot y)\cdot z=x\cdot y=x$ and $x\cdot (y\cdot z)=x$ and we see that indeed associativity holds in $G$.

If $x\cdot z=y\cdot z$ then $x=y$ by the definition of product $\cdot$ hence right cancellation law holds.

However, from $a\cdot a=a\cdot b=a$ does not follow that $a=b$. So left cancellation law fails in $G$.

It's easy to verify that the set $G$ is not group since if we suppose that $G$ is group then $\exists e'\in G$ such that $e'\cdot a=a\cdot e'=a$ and we conclude that $e'=a$. If we apply this identity element for $b\in G$ we get $b\cdot e'=e'\cdot b=b$ and we get that $e'=b$ thus $a=b$ which is contradiction.

Would be grateful for verification.

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    $\begingroup$ Looks good to me. $\endgroup$ – Arthur Nov 23 '17 at 7:11
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    $\begingroup$ The shortest definition of a group that I have seen is a set $S$ with an associative binary operation (written as multiplication) such that for all $a,b \in S$ there is a unique $x\in S$ and unique $y\in S$ with $ax=ya=b. $(This implies existence of a unique two-sided identity, and unique two-sided inverses.)..... If $S$ is finite then right -&-left cancellation laws imply the existence of unique $x,y$ such that $ax=ya=b$ because $ \{ax:x\in S\}=\{ya:y\in S\}=S.$ $\endgroup$ – DanielWainfleet Nov 23 '17 at 8:47
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Your reasoning is very clear and correct.

Your final paragraph showing that $G$ is not a group is not really necessary. You have already proved that a cancellation law does not hold and therefore $G$ cannot be a group.

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  • $\begingroup$ We know the following: cancellation laws hold $\Rightarrow$ $G$-group. But are you suggesting that if cancellation laws do not hold $\Rightarrow$ $G$-not group? Am I right? It sounds bit false. $\endgroup$ – ZFR Nov 23 '17 at 7:20
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    $\begingroup$ @RFZ $\text{associative and cancellation} \implies G \text{ group}$. And $G \text{ group} \implies \text{cancellation}$. For let $G$ be a group and $ax = ay$. Then multiplying by $a^{-1}$ gives $a^{-1}ax = ex = x = y = ey = a^{-1}ay$ so $x = y$. Therefore $G \text{ group} \implies \text{cancellation}$ $\endgroup$ – eepperly16 Nov 23 '17 at 7:53
  • $\begingroup$ Yes Daniel. It is very important to realise that the cancellation laws hold in every group. As you become more experienced with groups you will find yourself cancelling on the left and right automatically. $\endgroup$ – S. Dolan Nov 23 '17 at 14:51

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