As pointed out by Mark Sapir in his answer to a related question, every residually finite divisible semigroup is idempotent (hence uniquely divisible). On another hand, it is not difficult to prove that any idempotent Abelian semigroup is residually finite. So it is natural to ask whether or not the same holds even in the case where the semigroup operation is not commutative. Does it happen to be a well-established result? Is there a trivial counterexample that I can't see?
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asked Jun 21, 2012 at 9:15
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2 Answers
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Idempotent semigroups in general are not residually finite. There are some examples in Golubov, È. A.; Sapir, M. V. Varieties of finitely approximable semigroups. Izv. Vyssh. Uchebn. Zaved. Mat. 1982, no. 11, 21–29. If you want an easy example, I can post it here.
Here is an example. Let $S=L\cup R\cup Z\cup \{0\}$ where $L=\{l_1,l_2,...\},R=\{r_1,r_2,...\}$ are infinite sets, $Z=\{x,y\}$. Define a product by $l_i^2=l_i, r_i^2=r_i, x^2=x,y^2=y$, $l_ir_j=x$ if $i=j$, $l_ir_j=y$ if $i\ne j$, $l_ix=x, yr_i=y$, all other products are 0 (in particular, $xy=0$ and $0$ plays the role of zero. The associativity is easy to check. So $S$ is an idempotent semigroup. Suppose that $S$ is residually finite and there exists a homomorphism $\phi$ onto a finite semigroup separating $x$ from $y$. Note that there exist $i\ne j$ such that $\phi(l_i)=\phi(l_j)$. Then $\phi(x)=\phi(l_ir_i)=\phi(l_jr_i)=\phi(y)$, a contradiction.
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The answer turns out to be no. For a counterexample see example 2 in "On the infinite subsemigroups of a compact semigroup" by R.P. Hunter and L.W. Anderson. On the other hand, proposition 12, says that idempotent semigroups satisfying a further restriction, $xyzx=xzyx$, are always residually finite.
answered Jun 21, 2012 at 9:43