Prove that if every element in the monoid is right invertible, then every element has exactly one right inverse.
That is in the monoid $(M,\circ )$ with the identity element e,
$$\forall a\in M \; \exists a^\prime \in M\; : \; a\circ a^\prime=e \quad \Rightarrow \quad \forall a\in M \; \exists! a^\prime \in M\; : \; a\circ a^\prime=e $$
Right invertible elements in a monoid.
abstract-algebramonoid
Best Answer
Ok, so there is a standard argument for this kind of questions (unit of a group is unique; the unit element of a ring is unique etc). The proof is basically the same as below:
For $\forall a \in M$, assume we have two right inverses $b_1$ and $b_2$, then we have $ab_1=ab_2=e$. By hypothesis, we knew there are right inverses for both $b_1$ and $b_2$, say they are $c_1$ and $c_2$ respectively. Then multiply $c_1$ to the equation $ab_1=e$ from right, we have $ab_1c_1=c_1$, then since $b_1c_1=e$, we have $a=c_1$. Similarly, multiplying $c_2$ to the equation $ab_2=e$ from right, we get $a=c_2$. Therefore we showed that $b_1$ and $b_2$ have the same right inverse, namely just $a$. So $b_1a=b_2a=e$. Multiplying $b_1$ to the equation $b_2a=e$ from right, we have $b_2=b_2ab_1=b_1$. Therefore there exists exactly one right inverse.
Actually, this proof gets you a stronger that, this monoid is actually a group.