In this paper, the author states in the first sentence:

Among the homomorphic images of a semigroup (= a set closed with respect to an associative binary operation) there is at least one group, namely the unit group $I$.

How is this meant, in what sense arises the unit group as a homomorphic image?

If $I$ is the group of invertible elements, if $S - I$ is an ideal, even the Rees factor semigroup introduces a zero element in the image, hence it could not be a group. So how does $I$ arises as a homomorphic image?

  • 1
    $\begingroup$ For each $s \in S \setminus I$ your semigroup define $f: S \to S$ to be $f: s \mapsto 1$. This effectively removes the non-invertibles, in other words the hom image $f(S) = G$ a group. Define $f$ to be identity on $I$. There is no restriction on the definition of semigroup homomorphis by the "the zero element" in a multiplicative semigroup. In other words $f(1) = 1$ is required by def of hom, but $f(0)$ we can set to anything, so set it to $1$ if present. $\endgroup$ – Shine On You Crazy Diamond Jun 5 '18 at 17:00
  • $\begingroup$ @EnjoysMath Thank you, but please make this into an answer, so I can accept this question! $\endgroup$ – StefanH Jun 5 '18 at 17:07

In this paper, the unit group is understood as the trivial group with one element, say $G = \{1\}$. Then $G$ is clearly a quotient of $S$. The paper you are referring to is devoted to find the maximal group image of a semigroup, when it exists. Note also that the paper deals with semigroups, which are not necessarily monoids and may have no unit at all.

  • $\begingroup$ Thanks for your answer on this one. Because I know you are an expert in tcs, may I point you to another recent question of mine on the algebraic approach to formal language theory: cstheory.stackexchange.com/questions/40920/… $\endgroup$ – StefanH Jun 6 '18 at 12:13

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.