Timeline for Is an abelian group of bounded exponent $\aleph_0$-categorical
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2021 at 3:44 | vote | accept | Eugene Zhang | ||
| S Nov 23, 2021 at 3:01 | history | edited | Alex Kruckman | CC BY-SA 4.0 |
fix a grammatical error and change content
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| S Nov 23, 2021 at 3:01 | history | suggested | Eugene Zhang | CC BY-SA 4.0 |
fix a grammatical error and change content
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| Nov 22, 2021 at 20:11 | review | Suggested edits | |||
| S Nov 23, 2021 at 3:01 | |||||
| Jul 16, 2020 at 11:25 | history | edited | Gabe Conant | CC BY-SA 4.0 |
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| Jul 16, 2020 at 11:16 | comment | added | Gabe Conant | Thanks @Emil I will make some edits | |
| Jul 16, 2020 at 6:33 | comment | added | Emil Jeřábek | FWIW, the number (finite or infinite) $n$ of copies of $C_{p^k}$ in the decomposition of $G$ to prime power cyclic groups is given by $G_{p^k}/(G_{p^{k-1}}(G^p\cap G_{p^k}))\simeq C_p^{(n)}$ if I didn't make a mistake. | |
| Jul 16, 2020 at 6:19 | comment | added | Emil Jeřábek | I don't think Theorem B is correct as stated: $G=C_4\times C_2^{(\omega)}$ and $H=C_4^2\times C_2^{(\omega)}$ both have infinitely many elements of orders 2 and 4, but they are not isomorphic: $G/G_2\simeq C_2$ and $H/H_2\simeq C_2^2$, where $G_2=\{g\in G:g^2=1\}$. | |
| Jul 16, 2020 at 0:40 | comment | added | Eugene Zhang | @bof, I mean it is countable because $ℵ_0$-categorical is countable. | |
| Jul 16, 2020 at 0:25 | comment | added | bof | @hermes According to that definition, isn't the answer to your question trivially "no", since an Abelian group of bounded exponent may be uncountable or finite? Did you mean to say "is a countably infinite Abelian group of bounded exponent $\aleph_0$-categorical? | |
| Jul 15, 2020 at 21:20 | history | edited | Gabe Conant | CC BY-SA 4.0 |
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| Jul 15, 2020 at 18:39 | comment | added | Eugene Zhang | @bof, $\aleph_0$-categorical structure is defined as a structure of size $\aleph_0$ whose theory is $\aleph_0$-categorical . | |
| Jul 15, 2020 at 13:53 | comment | added | Emil Jeřábek | Fair enough. -- | |
| Jul 15, 2020 at 11:45 | history | edited | Gabe Conant | CC BY-SA 4.0 |
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| Jul 15, 2020 at 11:42 | comment | added | Gabe Conant | @Emil Your argument is doesn't cover the case that $G$ is torsion-free. | |
| Jul 15, 2020 at 11:12 | comment | added | Gabe Conant | @bof Yes you are correct it’s a property of the theory. But it’s common for people to say that a structure has a property when they mean its complete theory does. E.g. “$M$ is stable” means “$Th(M)$ is stable.” | |
| Jul 15, 2020 at 11:00 | comment | added | Gabe Conant | @YCor I’m talking about the action of $Aut(G)$ on $G^m$ for varying $m$. $\aleph_0$-categoricity is equivalent to those actions having finitely many orbits for all $m$. “Singleton” and “pair” refers to $m=1$ and $m=2$. In the first case elements of different orders must be in different orbits. One can also think of $m$-types over $\emptyset$ rather than orbits of $m$-tuples. | |
| Jul 15, 2020 at 7:51 | comment | added | bof | @YCor I never studied logic seriously so my knowledge of the subject is rather scattershot, but I think I know that one. A theory $T$ is $\kappa$-categorical if all models of $T$ of cardinality $\kappa$ are isomorphic. Examples: theory of plain sets with no additional structure; theory of dense linear orders with (or without) top and bottom elements; theory of atomless Boolean algebras. Hmm. I guess the theory of the random infinite graph is another one. | |
| Jul 15, 2020 at 7:22 | comment | added | YCor | @bof what's the meaning for a noncomplete theory to be $\aleph_0$-categorical? (I didn't find the answer) | |
| Jul 15, 2020 at 7:21 | comment | added | YCor | @GabeConant what's an orbit of singleton? for the group $\mathbf{Q}^{(\omega)}$ the automorphism group acts transitively on nonzero elements. | |
| Jul 15, 2020 at 6:13 | comment | added | bof | I thought categoricity was a property of theories, not structures. I'm guessing that people are saying "$\aleph_0$-categorical sructure" when then mean "structure whose elementary theory is $\aleph_0$-categorical". | |
| Jul 15, 2020 at 5:30 | vote | accept | Eugene Zhang | ||
| Nov 23, 2021 at 3:44 | |||||
| Jul 15, 2020 at 0:53 | comment | added | Gabe Conant | For the other direction: If $G$ has elements of arbitrarily large finite order then there are infinitely many orbits of singletons. If $G$ has an element $g$ of infinite order then $(g,g^n)$ for varying $n$ yield infinitely many orbits of pairs. | |
| Jul 14, 2020 at 23:20 | comment | added | Gabe Conant | @YCor His proof of Thm2 is definitely longer, for whatever that's worth. But yes I agree. Anyway, in light of all these comments I'm definitely less sure what the original intention of the question was. Hopefully it will be clarified. | |
| Jul 14, 2020 at 23:15 | comment | added | YCor | Actually, that every abelian group of bounded exponent is $\aleph_0$-categorical is pretty easy (and probably straightforward from Szmielew); I guess the bulk of Rosenstein's theorem is the reverse direction. | |
| Jul 14, 2020 at 23:12 | comment | added | Alex Kruckman | Haha. I found that paper too, but I considered "the totient group of $n$" to be a different usage than "totient group" as a general concept (certainly not every finite abelian group is the group of units mod $n$ for some $n$). | |
| Jul 14, 2020 at 23:10 | comment | added | Gabe Conant | @Alex On the page 2 "badlands" of Google searches I found it used in this paper. I think that there author actually does mean the group of units mod $n$, which supports your interpretation. I'm now also quite intrigued by this. | |
| Jul 14, 2020 at 23:03 | comment | added | Alex Kruckman | In fact, a google search for "totient group" turns up this question as the only usage of that phrase in a mathematical context. | |
| Jul 14, 2020 at 22:55 | history | answered | Gabe Conant | CC BY-SA 4.0 |