Timeline for If a semigroup embeds into a group, then is it a subdirect product of groups?
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| Oct 10 at 7:46 | comment | added | Salvo Tringali | By the way, YCor's answer can be used to prove that a cancellative semigroup need not be a subdirect product of subdirectly irreducible cancellative semigroups, see mathoverflow.net/questions/480364#480364 | |
| Oct 10 at 5:31 | comment | added | Salvo Tringali | As noted by @PaceNielsen in the question linked in my previous msg, every cancellative duo embeds in a group (and hence in its universal group). | |
| Oct 10 at 2:56 | comment | added | Salvo Tringali | I have the impression that the "extension lemma" in this answer holds for every duo sgrp S that embeds into its own universal group G (regardless of whether G is totally orderable and S is the associated positive cone), one key point being that, by duoness, $g^{-1}hg\in S$ for all $g,h\in S$. Here, duo means that aS = Sa for every a ∈ S. This begs another question: mathoverflow.net/q/480390/16537 | |
| Oct 9 at 12:38 | comment | added | Salvo Tringali | 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.) | |
| Oct 9 at 12:14 | comment | added | YCor | @SalvoTringali I don't think of this in this way (I'm not sure how to define kernel of a semigroup homomorphism, so I wouldn't invoke kernels — quotient semigroups, in general, are by congruences). How I'd phrase it: the surjective homomorphism $S\to G_i$ is surjective, $S$ is not a group while $G_i$ is a group, so it cannot be bijective, hence $G_i$ is, by the assumption, abelian. | |
| Oct 9 at 12:10 | comment | added | Salvo Tringali | 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). | |
| Oct 8 at 23:54 | vote | accept | Salvo Tringali | ||
| Oct 8 at 18:08 | comment | added | YCor | PS in categorical terms, the lemma precisely says that for a bi-ordered group $G$, the embedding $G_+\to G$ is the universal group embedding. | |
| Oct 8 at 8:54 | history | edited | YCor | CC BY-SA 4.0 |
fixed typos
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| Oct 8 at 8:50 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Oct 8 at 8:13 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Oct 8 at 7:50 | history | answered | YCor | CC BY-SA 4.0 |