Timeline for If a semigroup embeds into a group, then is it a subdirect product of groups?
Current License: CC BY-SA 4.0
23 events
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| Oct 9 at 1:51 | answer | added | Benjamin Steinberg | timeline score: 4 | |
| Oct 8 at 23:54 | vote | accept | Salvo Tringali | ||
| Oct 8 at 18:10 | comment | added | YCor | To cover the residually finite and abelian case, observe that if a semigroup $S$ is embeddable in a residually torsion group, then it is a subdirect product of groups. | |
| Oct 8 at 11:31 | history | became hot network question | |||
| Oct 8 at 9:15 | comment | added | HJRW | @Carl-FredrikNybergBrodda, SalvoTringali, thanks for correcting my semi-group terminology. I think the basic point — that residual finiteness is what you need to make the argument go through — stands. I note that YCor’s counterexample is, indeed, not residually finite. | |
| Oct 8 at 8:05 | comment | added | Salvo Tringali | @PeterTaylor The additive semigroup of positive integers embeds into the group of integers. However, it is neither a group nor the empty semigroup. | |
| Oct 8 at 8:02 | comment | added | Peter Taylor | If a semigroup embeds into a group, doesn't that require it to be also a quasigroup, whence it is either empty or a group? | |
| Oct 8 at 7:55 | comment | added | Salvo Tringali | @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. | |
| Oct 8 at 7:50 | answer | added | YCor | timeline score: 9 | |
| Oct 8 at 7:43 | comment | added | Carl-Fredrik Nyberg Brodda | @HJRW That's usually called being "group-embeddable". Cancellativity just means we can cancel on the left and right (i.e. either $xy = xz$ resp. $yx = zx$ implies $y=z$, and There are cancellative semigroups which are not group-embeddable (Malcev constructed such examples), see this very readable article. | |
| Oct 8 at 7:38 | comment | added | HJRW | @Carl-FredrikNybergBrodda: Oh sorry, I miscounted! Yes, I think the $BS(2,3)$ example looks more promising. | |
| Oct 8 at 7:37 | comment | added | HJRW | @SalvoTringali: Yes, that’s what I mean! (I assume that, in this setting, the universal group is the subgroup generated by $S$. Sorry, I really do know nothing about semigroups. And I assume that “cancellative” means that the semigroup embeds into its universal group.) | |
| Oct 8 at 7:32 | comment | added | Salvo Tringali | @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? | |
| Oct 8 at 7:26 | comment | added | Carl-Fredrik Nyberg Brodda | @HJRW The $U$ for my example is $\operatorname{BS}(1,2)$ (it is just the group with the same presentation) which is residually finite. But we can easily take $\langle a, b \mid a^2 b = b a^3 \rangle$ which embeds into $\operatorname{BS}(2,3)$. | |
| Oct 8 at 7:17 | comment | added | HJRW | This makes @Carl-FredrikNybergBrodda’s suggestion a very good one, because it looks like the associated $U$ is the non-residually-finite Baumslag—Solitar group $BS(2,3)$. | |
| Oct 8 at 7:15 | comment | added | HJRW | I know nothing about semigroup, but… if I understand rightly, Kearnes’ proof goes through when the universal group $U$ of $S$ is residually finite. (Such a $U$ embeds into the product of all its finite quotients, and $S$ will surject each finite quotient.) | |
| Oct 8 at 7:10 | comment | added | Salvo Tringali | 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. | |
| Oct 8 at 6:47 | comment | added | Salvo Tringali | @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 $\Delta := \{(x,x): x \in S\}$. | |
| Oct 8 at 6:19 | comment | added | Carl-Fredrik Nyberg Brodda | I don't know much about subdirect products, but is the semigroup $\langle a, b \mid ab = baa \rangle$ isomorphic to a subdirect product of groups? What happens when we push this through the machinery of Birkhoff's theorem? | |
| Oct 8 at 4:11 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 8 at 4:00 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 8 at 3:40 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 8 at 3:29 | history | asked | Salvo Tringali | CC BY-SA 4.0 |