It is easy to prove: in a finite semigroup if for all $a$ and $b$, $ax=b$ and $ya=b$ has unique solution. then it is group. But if in a finite semigroup, if for all $a$ and $b$, $ax=b$ and $ya=b$ has solution(not necessarily unique given) would it be a group?

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    $\begingroup$ So for fixed $a$ and variable $x$, the elements $ax$ would have to be all distinct in order to include every element $b$, so the solution would have to be unique., and it would be a group. $\endgroup$
    – Derek Holt
    Jan 23, 2015 at 15:31
  • $\begingroup$ @DerekHolt Thanks may I ask one more question here itself. Can we find example of infinite group such that both above equation has solution but not unique $\endgroup$
    – Sushil
    Jan 23, 2015 at 15:36
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    $\begingroup$ Maybe you mean to have each of $ax=b$ and $ya=b$ to have unique solutions $x$ and $y$ respectively. Because in the semigroup for which the multiplication $xy$ always returns $y$ the equation $ax=b$ reduces to $x=b$ (unique solution) and the same goes for $ay=b$. $\endgroup$
    – coffeemath
    Jan 23, 2015 at 16:01
  • $\begingroup$ Assuming that you meant $ya=b$, then I think even for an infinite semigroup, if solutions for all $a,b$, then it is a group. $\endgroup$
    – Derek Holt
    Jan 23, 2015 at 17:42
  • $\begingroup$ Please give a correct version of your question, because $ax = b$ and $ay = b$ are just twice the same equation with a different name for the unknown. If you mean $by = a$, the answer is no and the minimal counterexample is the semigroup $S = \{a, b\}$ with the following multiplication table: $aa = ba = a$, $ab = bb = b$. $\endgroup$
    – J.-E. Pin
    Jan 24, 2015 at 17:33

2 Answers 2


Here is a short self-contained proof.

Suppose $S$ is a semigroup (finite or infinite, but of course nonempty) such that, for any $a,b\in S$, each of the equations $ax=b$ and $ya=b$ has at least one solution in $S$. We will show that $S$ is a group. (In fact this is how groups are defined in some textbooks. If memory serves, this is how they are defined in a book by Kurosh.)

Choose elements $a\in S$ and $e\in S$ such that $ae=a$.

Consider any $x\in S$. Choose $y\in S$ so that $ya=x$. Then $$xe=yae=ya=x.$$ Thus $e$ is a right identity element.

For any element $x\in S$ there is an element $x'\in S$ such that $xx'=e$, and an element $x''\in S$ such that $x'x''=e$. Then $$x'x=x'xe=x'xx'x''=x'ex''=x'x''=e,$$ that is, $x'$ is a two-sided inverse of $x$ with respect to $e$.

Finally, for any $x\in S$ we have $$ex=xx'x=xe=x,$$ so $e$ is a two-sided identity element. This completes the proof that $S$ is a group.


The answer is yes and relies on the properties of Green's relations and Green's theorem, which states that if $H$ is an $\mathcal{H}$-class of a semigroup $S$, then either $H^2 \cap H = \emptyset$ or $H$ is a subgroup of $S$. In particular, the $\mathcal{H}$-class of an idempotent is a subgroup of S.

Let $S$ be a nonempty finite semigroup. Then $S$ contains an idempotent $e$. Now, for all $a \in S$, there exists a pair $(x,y) \in S^2$ such that $ax = e$ and $ya = e$ and another pair $(s,t) \in S^2$ such that $es = a$ and $te = a$. This implies that, for all $a \in S$, $e \mathrel{\mathcal{R}} a$ and $e \mathrel{\mathcal{L}} a$, whence $e \mathrel{\mathcal{H}} a$. Therefore, $S$ is equal to the $\mathcal{H}$-class of $e$, which is a group, as any $\mathcal{H}$-class containing an idempotent.

  • $\begingroup$ Is there a more elementary explanation? I follow up to $es = a$ and $te = a$. $\endgroup$ Sep 16, 2015 at 19:38
  • $\begingroup$ The existence of an idempotent is proved at math.stackexchange.com/questions/353028/…. $\endgroup$ Sep 16, 2015 at 19:38
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    $\begingroup$ @6005 I have added references to Green's relations and stated Green's theorem. I am a bit reluctant to include a proof of Green's theorem, which is nontrivial but can be found in any book of semigroup theory. $\endgroup$
    – J.-E. Pin
    Sep 16, 2015 at 20:31
  • $\begingroup$ Thanks for the edit, I understand the proof now. $\endgroup$ Sep 16, 2015 at 22:56
  • $\begingroup$ From Wikipedia it doesn't seem like Green's theorem requires the finiteness assumption. So is it true that: a semigroup S is a group iff (1) it has an idempotent, and (2) $ax = b$ and $ya = b$ have at least one solution for any $a,b$? $\endgroup$ Sep 16, 2015 at 22:57

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