Timeline for Equational theories determined by "identities without variables"
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 14, 2023 at 8:40 | comment | added | Keith Kearnes | @მამუკაჯიბლაძე: A finite group generates a variety satisfying (C1) iff its Sylow subgroups are abelian. | |
| Mar 14, 2023 at 7:49 | comment | added | მამუკა ჯიბლაძე | One more question, if I may. Seems like just having a diamond in the lattice of congruences suffices for their construction, while in the event of congruence distributivity C1 trivially holds. This makes me wonder - are there any interesting cases of C1 without congruence distributivity? | |
| Mar 14, 2023 at 7:42 | comment | added | მამუკა ჯიბლაძე | Freese and McKenzie actually pick an element $0\in I$ and take instead of $K$ the subgroup of those $f$ with $f(0)=\prod_{i\ne0}f(i)$, but I believe in our case this is just the same, right? | |
| Mar 14, 2023 at 7:33 | comment | added | მამუკა ჯიბლაძე | So in fact when $I$ is finite this only works right away when its cardinality is odd | |
| Mar 14, 2023 at 7:24 | comment | added | მამუკა ჯიბლაძე | Absolutely, many thanks for this correction! | |
| Mar 14, 2023 at 7:23 | comment | added | Keith Kearnes | @მამუკაჯიბლაძე: If you are trying to build large subdirectly irreducible algebras in the variety $\mathscr{V}(G_G)$ (instead of in $\mathscr{V}(G)$), then your construction $G^I/K$ works, but you have to be more careful about the choice of $\hat{\prod}$. You can take $\hat{\prod}$ to be any extension of $\prod$ to $C^I$ provided that $\hat{\prod}(\{z\}^I) =z$, where $z\in C-\{1\}$. | |
| Mar 14, 2023 at 6:54 | comment | added | მამუკა ჯიბლაძე | Well more precisely $G^{(I)}\subset G^I$ is not a $G_G$-subalgebra. Maybe it still is in $\mathsf{HSP}(G_G)$, I don't know | |
| Mar 14, 2023 at 6:48 | comment | added | მამუკა ჯიბლაძე | Oh no $G^{(I)}$ is not in the variety of $G_G$ unless $I$ is finite! Maybe $G^{(I)}/\operatorname{Ker}(\prod)$ still is?... | |
| Mar 14, 2023 at 5:26 | comment | added | მამუკა ჯიბლაძე | Thank you. In fact there seems to be a simpler way - just take $G^{(I)}/\operatorname{Ker}(\prod)$, it has cardinality at least that of $I$ anyway. Maybe it is worth adding this as an explicit example to the final part of this answer? | |
| Mar 14, 2023 at 3:16 | comment | added | Keith Kearnes | @მამუკაჯიბლაძე: I think that your second construction of arbitrarily large SI's in the variety generated by the quaternion group is correct. The existence of $\hat{\prod}$ follows from the fact that $C^I$ is elementary abelian, hence may be viewed as a vector space. $G^I/K$ is large since it has a map onto $(G/C)^I$, which is large if $I$ is large. The center of $G^I/K$ can be seen to be $C^I/K(\cong C)$, which has size $2$. These observations are enough to see that your construction produces arbitrarily large SI groups in the variety. | |
| Mar 13, 2023 at 19:15 | comment | added | მამუკა ჯიბლაძე | Too bad I erased my previous attempt, I should probably wait for your reaction first. | |
| Mar 13, 2023 at 19:08 | comment | added | მამუკა ჯიბლაძე | Yes, thank you, and sorry. Could you please have a look at another attempt: choose any homomorphism ${\hat\prod}:C^I\to C$ that extends the homomorphism ${\prod}:C^{(I)}\to C$ from finitely supported $C$-valued functions to $C$, with ${\prod}(f):=\prod_{i\in I}f(i)$, and let $K$ be the kernel of ${\hat\prod}$. Then $G^I/K$ seems to work as expected. | |
| Mar 13, 2023 at 18:41 | comment | added | Keith Kearnes | I don't see why the center of $G^I/C_{\iota}$ has two elements. Suppose $g\in G-C$ and $\{u\}\notin \iota$. Let $f\in G^I$ be the function defined by $f(u)=g$ and $f(x)=1$ if $x\neq u$. Then $f\notin C_{\iota}$ since $f\notin C^I$, but $[f,h]\in C_{\iota}$ for any $h\in G^I$. Therefore $f/C_{\iota}$ is a nontrivial central element in $G^I/C_{\iota}$. This method should produce a lot of central elements. | |
| Mar 13, 2023 at 9:30 | comment | added | Keith Kearnes | The paper ''Freese, Ralph; McKenzie, Ralph Residually small varieties with modular congruence lattices. Trans. Amer. Math. Soc. 264 (1981), no. 2, 419-430'' proves that a finite algebra in a congruence modular variety that fails the commutator condition that is now called (C1) generates a residually large variety. Any constant expansion of a finite nilpotent group will fail (C1). If you look at page 422, line 17, set $\mu=1$ and $\nu$= the center, you will see how the condition fails. | |
| Mar 13, 2023 at 7:45 | comment | added | მამუკა ჯიბლაძე | Sorry, could you say a little bit more about your second example? What I figured out is that subdirectly irreducibles are the ones with the center cyclic of prime power order. But why is there a proper class of pairwise nonisomorphic such in $\mathsf{HSP}(G_G)$? | |
| Mar 12, 2023 at 18:50 | history | edited | Keith Kearnes | CC BY-SA 4.0 |
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| Mar 12, 2023 at 18:37 | history | edited | Keith Kearnes | CC BY-SA 4.0 |
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| Mar 12, 2023 at 18:07 | history | answered | Keith Kearnes | CC BY-SA 4.0 |