# Cancellation laws and group

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.

• Looks good to me. Nov 23 '17 at 7:11
• 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 $\{ax:x\in S\}=\{ya:y\in S\}=S.$ Nov 23 '17 at 8:47

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.
• 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.
• @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 $G \text{ group} \implies \text{cancellation}$ Nov 23 '17 at 7:53
You have proved via using left and right cancellation in given equalities. By equality, it means a common term in a table. So, $$ae'=e'a, be'=e'b$$ leads to contradiction as $$a,b$$ are distinct. Although, your interpretation is correct as mean $$e'=a$$, hence $$aa=a$$; also $$e'=b$$, hence $$bb=b$$. But, still there seems flaw to me, as identity $$e'$$ should, by defn., mean $$e'a =a$$ & $$e'b =b$$. So, $$ae'=e'a\implies ae' = a, be'=e'b\implies be'=b$$. So, a better way to use left & right cancellation should be used, for the purpose of the proof. But, am unsure if can find one.