As pointed out by Mark Sapir in [his answer](https://mathoverflow.net/questions/99235/a-semigroup-with-the-property-that-xn-a-has-at-least-one-solution/99236#99236) 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?



Idempotent semigroups: Are they all residually finite?
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. 


