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Salvo Tringali's user avatar
Salvo Tringali's user avatar
Salvo Tringali
  • Member for 13 years, 5 months
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  • Shijiazhuang, Hebei, China
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If a semigroup embeds into a group, then is it a subdirect product of groups?
Concerning "I'm not sure how to define kernel of a semigroup homomorphism [...] — quotient semigroups, in general, are by congruences": the kernel of a sgrp homomorphism is a congruence (on the domain of the homomorphism); see, for instance, Prop 3.3 in the 1995 edition of Grillet's book on sgrps. That being said: with my previous msg, I mainly wanted to emphasize that each of the $G_i$'s is indeed a quotient of S, which is of course true of any factor in any subdirect rep of any algebra (by the general formulation of the 1st isomorphism thm for algebras.)
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If a semigroup embeds into a group, then is it a subdirect product of groups?
Thanks for the reference (and +1). I'm accepting YCor's answer because (i) it came first and (ii) I believe his construction can be used to give a negative answer to mathoverflow.net/questions/480202
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If a semigroup embeds into a group, then is it a subdirect product of groups?
Very nice. Just let me make one point slightly more explicit: If a sgrp S is a subdirect prod of a family $(G_i)_{i\in I}$ of groups, then, by def., S embeds into the direct prod of the $G_i$'s and, for each $j\in I$, there is a surjective sgrp hom $f_i:S\to G_i$. It follows, by the 1st isomorphism thm for sgrps, that $G_i$ is iso to the quotient sgrp $S/\ker(f_i)$, where $\ker(f_i)$ is the kernel of $f_i$. If S is not a group and each of its proper quotients is commutative, then $G_i$ is abelian and hence $S$ is commutative (as it's a subsgrp of the direct product of abelian groups).
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If a semigroup embeds into a group, then is it a subdirect product of groups?
@PeterTaylor The additive semigroup of positive integers embeds into the group of integers. However, it is neither a group nor the empty semigroup.
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If a semigroup embeds into a group, then is it a subdirect product of groups?
@HJRW As already explained by Carl-Fredrik Nyberg Brodda, cancellative is weaker than group-embeddable (you may want to have a look at the edit to the other question linked in the OP). That being said, a semigroup is embeddable in a group iff the canonical homomorphism from S to its universal group is injective (see, e.g., Exercise 4.11.1 in Bergman's An Invitation to General Algebra and Universal Constructions). So, your reference to the universal group was completely fine, after all.
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If a semigroup embeds into a group, then is it a subdirect product of groups?
@HJRW I'm slightly confused by your reference to the universal group of $S$: you mean that Kearnes' proof carries over to any semigroup that embeds into a residually finite group, don't you?
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If a semigroup embeds into a group, then is it a subdirect product of groups?
By definition, S is the quotient of the free semigroup on the two-element set $\{a, b\}$ by the smallest congruence $\theta$ containing the pair $(ab, baa)$. Let me ask a naive question: is there any "direct relation" between $\theta$ and the intersection of all proper congruences on S? Are they one and the same thing (up to the replacing of a and b with their $\theta$-classes)? If yes, this is probably a basic property of presentations that I should know but I don't.
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If a semigroup embeds into a group, then is it a subdirect product of groups?
@Carl-FredrikNybergBrodda Let's call S the semigroup you are suggesting to consider. By Adian's embedding theorem, S embeds into a group. What about the intersection of all proper congruences on S? If it's proper, then S is subdirectly irreducible and we have an example proving that the answer to the OP is no. Here (following Birkhoff), a congruence is proper if it's not the discrete congruence $\Delta := \{(x,x): x \in S\}$.
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