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For a monoid $G$, consider the following conditions:

(i) $G$ is a group.

(ii) For $a, x, y ∈ G$, each of the equations $ax = ay$ and $xa = ya$ implies $x = y$.

Then (i) $\implies$ (ii). Show that the reverse implication holds for finite monoids $G$, but not for arbitrary monoids G.

I think that I need to prove this problem by proving that ever $a \in G$ has an inverse that is also in $G$. I also think that $x$ is the inverse of $a$ since one of the axioms in group theory states that for all $a\in G$, there exists another element $b\in G$ such that $ba=e=ab$. I've tried showing $ax=xa$ since I think that this would imply that the equation is equal to $e$ and thus $x$ is the inverse of $a$. I've done this by multiplying x to both sides of $ax=ay$ to get $x(ax)=x(ay)=(xa)y=(ya)y=(ya)x$ but I can't seem to deduce how this would lead to $ax=xa=e$.

J.-E. Pin
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  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Also, you may use site search to find your question is a duplicate. – Martin Brandenburg Jan 04 '25 at 03:33
  • To see the connection to the linked duplicate, notice that we may assume $M$ is commutative (in fact, cyclic!) by considering the submonoid $1,a,a^2,\dotsc$. So cancelling on the left or right is no different. Or just consider this answer, it applies here directly. – Martin Brandenburg Jan 04 '25 at 04:01

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