A group $G$ is residually finite if, for any two elements $g$ and $g^\prime$ in $G$, there is a finite group $G^\prime$ and a (group) homomorphism $f: G \rightarrow G^\prime$ such that $f(g)$ doesn't equal $f(g^\prime)$. The definition for a semigroup is analagous: just make $G$ and $G^\prime$ semigroups and make $f$ a semigroup homomorphism. I was wondering if there is a good reference which will answer questions like the following:

Is there a group $G$ which is not residually finite as a group but is residually finite as a semigroup (in other words there is a finite semigroup $S$ and a semigroup homomorphism from $G$ to $S$ which separates elements, but there is no finite group $G^\prime$ and a group homomorphism from $G$ to $G^\prime$ which separates elements)?

If $S$ is a residually finite semigroup and $G$ is a subgroup of $S$, then $G$ is residually finite as a semigroup. Is $G$ residually finite as a group?


  • 1
    $\begingroup$ I may be missing something, but isn’t it true that if φ:G→S is a semigroup homomorphism where G is a group, then the image φ(G) is a group and φ:G→φ(G) is a group homomorphism? If this is correct, a group which is residually finite as a semigroup is also residually finite as a group. $\endgroup$ Aug 2, 2010 at 18:47

1 Answer 1


I first posted this as a comment, but I guess that this is an answer.

If a group is residually finite as a semigroup, it is residually finite as a group. This is an immediate consequence of the following easy fact: if G is a group and φ:G→S is a semigroup homomorphism, then the image φ(G) is a group and φ is a group homomorphism from G to φ(G). I guess that the latter fact is in any textbook on semigroups, though I do not have one at hand.

  • $\begingroup$ I don't think a reference to the literature on semigroups is required here. By definition, a semigroup homomorphism is a map such that $\varphi(xy) = \varphi(x) \varphi(y)$ for all $x,y \in G$. But this is also the definition of a group homomorphism! The point is that, for groups, that this implies that the identity gets sent to the identity and that inverses get sent to inverses. If one has not done so before, it is a good exercise to think about why for ring homomorphisms, in contrast, one must require explicitly that $1 \mapsto 1$. $\endgroup$ Aug 2, 2010 at 22:21
  • $\begingroup$ Thanks for the comment! I agree that this fact is easy to prove, and there is no real need to consult a textbook on semigroups. However, because the questioner asked for a good reference, I wanted to make clear that I did not provide any reference in my answer. $\endgroup$ Aug 2, 2010 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.