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I am exploring the following question: Does there exist an associative binary operation (i.e., a semigroup) on a set ( G ) that has at least a right identity and such that every element has at least one left inverse, but where the left inverses are not necessarily unique?

I initially considered the operation defined by $x*y = x + x^2y$ on $ \mathbb{C} $.

However, I found that this operation is not associative. I attempted to modify it (or consider similar variants) to achieve associativity while breaking the uniqueness of left inverses.

I am unsure whether there exists a standard theorem stating Edit to clarify: In any semigroup with at least one right identity (I fix one, which I name $e$), if every element has at least one left inverse with respect to $e$, then this inverse must be unique with respect to $e$.

Could someone provide a rigorous explanation or reference confirming this result? Or, if possible, suggest a construction of an associative binary operation with a right identity where left inverses exist but are not necessarily unique?

Thanks in advance for your insights!

ADD Belkacem Abderrahmane's proof presents an awkwardness; I correct it: Denote by $e$ an right identity element of G ( not necessarily unique) . Suppose $x\in G$, have two left inverse $y, v\in G$ , i.e. $yx=vx=e$, . Put $t=xy$, then $t^2=t$, and let $s\in G$ be an left inverse of $t$ ( not necessarily unique). Then, for all $u\in G$, we have $ut=uet=u(st)t=ust^2=ust=ue=u$. Thus, $t$ is another right identity of $G$ and hence $y=yt=y(xy)=(yx)y=(vx)y=v(xy)=vt=v$.

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  • $\begingroup$ Is your right identity unique? $\endgroup$ Commented Mar 29 at 9:31
  • $\begingroup$ No, it's not necessarily unique $\endgroup$ Commented Mar 29 at 9:42
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    $\begingroup$ Then how do you define your left inverse? Is it $\forall x\exists y\forall z: zyx=yxz=z$? $\endgroup$ Commented Mar 29 at 9:51

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Denote by $e$ the right identity.let $x\in G$, by hypothesis, there is $y\in G$ such that: $yx=e$,so $yxyx=y(xy)x=yx$.let $t\in G$ be an idempotent (i.e $t^2=t$),and let $s\in G$ be a left inverse of $t$,then: $$ e=st=st^2=t $$ i.e the only idempotent of $G$ is $e$,(edit) indeed, let $f$ be a left inverse of $e$, then $$ e=fe=fet=t $$ so $yx=e$,that is $y$ is also a right inverse, hence every elemnt of $G$ admit an inverse.Now, let $a\in G$, then $$ ae=aa^{-1}a=a $$ So $e$ is also a left identity,So $G$ must be a group.

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  • $\begingroup$ @Ziad $yxyx=yxe=yx$ $\endgroup$ Commented Mar 28 at 18:00
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    $\begingroup$ e is right identity why st²=t, st²=(st)t=et why gives t $\endgroup$ Commented Mar 28 at 18:36
  • $\begingroup$ edit:using a left inverse of $e$ completes the proof. $\endgroup$ Commented Mar 29 at 12:56
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Usual definition of left inverse

The meaning of left inverse depend on the existence of an identity, which is then the only one right identity and the only one left identity. In that case, the assumption of the existence of a left inverse for all element implies that this left inverse is unique.

Proof: Let $x\in G$, note $y$ a left inverse of $x$ and note $z$ a left inverse of $y$. Then $x=(zy)x=z(yx)=z$, hence for all $y'$ another left inverse of $x$, we get $y'=y'(zy)=y'(xy)=(y'x)y=y$.

Unusual definition of left inverse

Given that Ziad writes that the right identity may be not unique in $G$, I wondered what happened if the the notion of left inverse was altered in the following sense:
$y$ is a left inverse for $x$ iff $yx$ is a right identity.

Then we have a counterexample.

Let $G=\{a,b\}$ a set with two elements and an internal law defined by $$\forall (x,y)\in G^2,\quad x.y=x$$ which is clearly associative.

Then $a$ and $b$ are distinct right identities and both are left inverses to both of them.

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    $\begingroup$ In your example, every element of $G$ is a right identity. We fix one, which I denote by $a$. Let $x \in G$, what are its right inverses with respect to $a$? The only obvious one is $a$, since $a \cdot x = a$. $\endgroup$ Commented Mar 29 at 11:10

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