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As is well known, if $S$ is a semigroup in which the equations $a=bx$ and $a=yb$ have solutions for all $a$ and $b$, then $S$ is a group. This question arose when someone misunderstood the conditions as requiring that the solution to both equations be the same element of $S$. He suggested instead replacing one of the equations with a cancellation condition (he was thinking along the lines of trying to specify that the Cayley table would be a latin square). It is easy to see that there are semigroups that are not groups in which every equation of the form $a=xb$ has a solution and you can cancel on the right (the standard example that sets $ab=a$ for all $a,b$ works). What is not clear to me is what happens if the equations and cancellations are on the same side. That is:

Suppose $S$ is a semigroup in which the following two conditions hold:
  1. For all $a,b\in S$ there exists $x$ such that $a=xb$.
  2. For all $a,b,c\in S$, if $ab=ac$ then $b=c$.
Is $S$ a group?

It is easy to see that if $S$ contains an idempotent, then $S$ will be a group: if $e^2=e$, then for all $a\in S$ we have $e^2a=ea$, so $ea=a$ for all $a$; then solving $e=xa$ shows $S$ has a left identity and left inverses, hence is a group. In particular, $S$ will be a group if at least one cyclic subsemigroup of $S$ is finite, and also in particular if $S$ is finite.

I suspect that the answer will be "no" in full generality (that is, there are examples of semigroups $S$ that satisfy 1 and 2 above but are not groups), but I have not been able to construct one. Does any one have an example, a proof that $S$ will always be a group under these conditions, or a reference?

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If $S$ is finite, there exists $e$ s.t. $e^n=e$ for some $n>1$ (by iterating $x \mapsto x^2$), so $e^{n-1}a=a$ for every $a$ in $S$, and you can finish in the same way. So if a counterexample exists, it has to be infinite... –  Homology Jun 10 '10 at 17:56
    
@Homology - the original question observes that any counterexample, if it exists, would have to be infinite –  Yemon Choi Jun 10 '10 at 18:14
    
oops, sorry then! –  Homology Jun 10 '10 at 18:18
    
The cex would be not just infinite, but torsion-free. This sounds vaguely like a homework problem in Hungerford or Herstein. Does anyone know the difference between those homework problems and this formulation? Gerhard "Ask Me About System Design" Paseman, 2010.06.10 –  Gerhard Paseman Jun 10 '10 at 18:20
    
In Herstein's "Topics"(in Spanish, 1988); the exercises in Section 2.3 include: prove that a semigroup with right identity and right inverses is a group (problem 12); prove that right identity and left inverses do not yield a group (problem 13); prove that a finite semigroup with both cancellation laws is a group (problem 14); show that a single cancellation law does not suffice (problem 16); prove that if we drop "finite" in problem 14 then the cancellation laws do not suffice (problem 17). Hungerford (section I.2) probem 15 is the same as 14 and 17 from Herstein. No thers I can see. –  Arturo Magidin Jun 10 '10 at 18:39

1 Answer 1

up vote 13 down vote accepted

I am a victim of timing... I had asked this of a colleague a few days ago and had received no answers, but today at lunch he gave me a counterexample and reference (Clifford and Preston's The Algebraic Theory of Semigroups, volume II, pp. 82-86). The example is the Baer-Levi semigroup: the semigroup of all one-to-one mappings $\alpha$ of a countable set $X$ into itself such that $X\setminus \alpha(X)$ is not finite.

Left cancellation follows trivially; if $a$ and $b$ are such mappings, then to construct $c\in S$ such that $a=cb$ proceed as follows: if $y=b(x)\in b(X)$, set $c(y)=c(b(x))=a(x)$. Now let $\delta$ be a one-to-one map from $X\setminus b(X)$ to $X\setminus a(X)$ with $(X\setminus a(X))\setminus\delta(X\setminus b(X))$ infinite. Then define $c(y)$ for $y\notin b(X)$ by $c(y)=\delta(y)$.

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You can accept your own answer, that way people will know that the question has been resolved. –  Victor Protsak Jun 11 '10 at 2:05
    
When I try, it tells me I need to wait 38 hours. –  Arturo Magidin Jun 11 '10 at 2:36

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