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Let $T$ be a first-order theory of cyclic groups. Even if an abelian group $(G,+)$ satisfy $(G,+)\models T$ there is no reason that $(G,+)$ is a cyclic. (For example, by Löwenheim–Skolem theorem there is uncountable abelian group $G$ that satisfy $T$.)

I tried to find a first-order formula that is true for all cyclic groups, but is false for some abelian group. But I don't know how to find it. Thanks for any help.

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    $\begingroup$ How about for all $x,y$ in the group, at least one of $x$, $y$ and $xy$ has a square root. You can use this idea to show that the elementary theories of free abelian groups of rank $n$ are different for different $n$, in contrast to free groups. $\endgroup$
    – Derek Holt
    Oct 27, 2013 at 8:06
  • $\begingroup$ @DerekHolt It is not true in infinite cyclic group $\Bbb{Z}$. (e.g. $x=2$, $y=3$.) and it is true in $C_1$ and $C_2$. $\endgroup$
    – Hanul Jeon
    Oct 27, 2013 at 8:09
  • $\begingroup$ I meant square root in a multiplicative group. So the square root of 2 in $({\mathbb Z},+)$ is 1. $\endgroup$
    – Derek Holt
    Oct 27, 2013 at 8:19
  • $\begingroup$ @DerekHolt Oh, it is my mistake. $\endgroup$
    – Hanul Jeon
    Oct 27, 2013 at 8:19

2 Answers 2

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To make my comment more formal, the following statement (using multiplicative notation for groups) is true in all cyclic but not in all abelian groups. It is false, for example, in the free abelian group of rank 2, or in the Klein 4-group.

$\forall x,y \in G, \exists z \in G$ such that $z^2=x \vee z^2=y \vee z^2=xy.$

A statement that is true in all 2-generator but not in all 3-generator abelian groups is

$\forall x,y,z \in G, \exists w \in G$ such that $w^2=x \vee w^2=y \vee w^2=z \vee w^2=xy \vee w^2=xz \vee w^2=yz \vee w^2 = xyz.$

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The proof that the multiplicative group of a finite field is cyclic uses the property of a cyclic group $C$ that for every natural number $n > 0$, there are at most $n$ elements in $C$ of order $n$. This is an elementary (first-order) statement, for each $n$, which also holds for the infinite cyclic group (vacuously). For the finite cyclic groups, it is enough to say that for each prime $p$, there are at most $p$ elements of order $p$. This holds for the infinite cylic group $\Bbb{Z}$ vacuously, but it also holds for any free abelian group. So I think it is better to take the dual sentences, namely, that for every prime $p$, the quotient $A/pA$ has order at most. This is also an elementary statement, and implies the bound on the elements of order $p$ in case $A$ is finite. I'm pretty sure that this is an axiomatization of the cyclic groups in the language of abelian groups. An abelian group is called pseudo-cyclic if it is a model of these axioms. So an example of a pseudo cyclic abelian group would be the $p$-Prüfer group of the $p$-adic completion of $\Bbb{Z}$, or the direct sum of the two, but not a direct sum of two $p$-Prüfer groups or two $p$-adic completions. The rationals also seem to be pseudo cyclic.

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  • $\begingroup$ Interesting material here, but I think something got skipped in the sentence ending with "has order at most". Should be "at most $p$"? $\endgroup$
    – hardmath
    Mar 21, 2015 at 19:07

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