For an abelian torsion group of finite exponent, i.e. there is an integer $n$ such that $g^n=1$ for all $g\in G$, its theory appears to be $\aleph_0$-categorical by the theorem of Engeler, Ryll-Nardzewski and Svenonius. I want to confirm this fact.
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2$\begingroup$ Just to clarify: for you, "totient group" is a synonym for "finite abelian group"? The complete theory of any finite structure has only one model up to isomorphism. In particular, it has no infinite models. So no, it is not $\aleph_0$-categorical. $\endgroup$– Alex KruckmanCommented Jul 14, 2020 at 22:55
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2$\begingroup$ @AlexKruckman I added an answer assuming "finite order" meant "finite/bounded exponent". But perhaps it would have been good to clarify the terminology first since of course what you've said is correct! $\endgroup$– Gabe ConantCommented Jul 14, 2020 at 22:57
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3$\begingroup$ Why do you edit without addressing the comments? what do you mean by "finite order"? $\endgroup$– YCorCommented Jul 15, 2020 at 7:20
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2$\begingroup$ Perhaps “totient” here is a mangling of “torsion”? (That’s a weaker property, but related.) $\endgroup$– Emil JeřábekCommented Jul 15, 2020 at 7:22
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3$\begingroup$ It is correct that such a group is not considered to be a group of bounded order, but it could still be a torsion group. I say all of this because you have written "A torsion group is an abelian group of bounded order." But torsion groups need not have bounded order (or be abelian). $\endgroup$– Gabe ConantCommented Jul 15, 2020 at 17:56
1 Answer
This is a theorem of Rosenstein from the paper $\aleph_0$-categoricity of groups. But note that he uses the terminology bounded order rather than finite exponent.
I say that a group is $\aleph_0$-categorical if its complete theory in the language of groups is $\aleph_0$-categorical (i.e., has a unique countable model).
It is not hard to show that an $\aleph_0$-categorical group has finite exponent. This is also Theorem 1 of Rosenstein's paper. Theorem 2 of his paper provides a converse for abelian groups. So we have:
Theorem. (Rosenstein 1971)
An infinite abelian group is $\aleph_0$-categorical if and only if it has finite exponent.
Appendix. The discussion below has inspired me to add some discussion of the proof of theorems, and some other remarks. The theorem above comes from two results in Rosenstein's paper.
Theorem 1. An $\aleph_0$-categorical group has finite exponent.
Proof. Rosenstein uses the following fact, due to Ryll-Nardzewski/Enegler/Svenonius independently.
A countably infinite structure $M$ is $\aleph_0$-categorical if and only if for all $m>0$, the action of $\operatorname{Aut}(M)$ on $M^m$ has finitely many orbits.
Now suppose $\operatorname{Th}(G)$ is $\aleph_0$-categorical. We may assume $G$ is countable. Since elements of distinct orders are in distinct orbits of the action of $\operatorname{Aut}(G)$ on $G$, there is a uniform bound on the orders of torsion elements in $G$. Moreover, $G$ has no elements of infinite order since, if $g$ were such then $(g,g^n)$ for varying $n$ would be in distinct orbits in the action on $G^2$.
Theorem 2. An infinite abelian group of finite exponent is $\aleph_0$-categorical.
See Rosenstein's paper. The main tool is a structure theorem for abelian groups of finite exponent as direct sums of cyclic groups. This is apparently called Prufer's First Theorem.
Some remarks:
By YCor's comment about $\mathbb{Q}^{(\omega)}$, one cannot prove Theorem 1 by considering only orbits of singletons. ($\mathbb{Q}$ works also.)
It is easier to prove that an $\aleph_0$-categorical torsion group $G$ has finite exponent. Indeed, if not then by compactness/DLS there is a countable model of $\operatorname{Th}(G)$ with an element of infinite order, which cannot be $G$. But if $G$ is not a torsion group, I don't see a quick way to avoid Ryll-Nardzewski or Omitting Types of some kind (although one can substitute other facts that use these results, e.g., in an $\aleph_0$-categorical structure, the algebraic closure of a finite set is finite).
An $\aleph_0$-categorical group need not be abelian. An example is the countably infinite extraspecial $p$-group (see Definition 5.15 here). On the other hand, there are many results in model theory along the lines of "$\aleph_0$-categorical plus some model-theoretic property" implies some abelian-like structure. For example, Bauer, Cherlin, and Macintyre showed that an $\aleph_0$-categorical superstable group is abelian-by-finite.
An infinite group of finite exponent need not be $\aleph_0$-categorical. For example, take an infinite finitely generated group of finite exponent (like a Tarski monster). Indeed, by the more general result about algebraic closure stated above, if $G$ is $\aleph_0$-categorical then any finite subset of $G$ generates a finite subgroup.
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2$\begingroup$ In fact, a google search for "totient group" turns up this question as the only usage of that phrase in a mathematical context. $\endgroup$ Commented Jul 14, 2020 at 23:03
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1$\begingroup$ 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. $\endgroup$ Commented Jul 15, 2020 at 0:53
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1$\begingroup$ @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. $\endgroup$ Commented Jul 15, 2020 at 11:00
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1$\begingroup$ @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.” $\endgroup$ Commented Jul 15, 2020 at 11:12
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1$\begingroup$ @Emil Your argument is doesn't cover the case that $G$ is torsion-free. $\endgroup$ Commented Jul 15, 2020 at 11:42