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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?

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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