# First order theory of abelian groups and first order theory of cyclic groups are coincide?

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.

• 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. Oct 27, 2013 at 8:06
• @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 $C_2$. Oct 27, 2013 at 8:09
• I meant square root in a multiplicative group. So the square root of 2 in $({\mathbb Z},+)$ is 1. Oct 27, 2013 at 8:19
• @DerekHolt Oh, it is my mistake. Oct 27, 2013 at 8:19

$\forall x,y \in G, \exists z \in G$ such that $z^2=x \vee z^2=y \vee z^2=xy.$
$\forall x,y,z \in G, \exists w \in G$ such that $w^2=x \vee w^2=y \vee w^2=z \vee w^2=xy \vee w^2=xz \vee w^2=yz \vee w^2 = xyz.$
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.
• Interesting material here, but I think something got skipped in the sentence ending with "has order at most". Should be "at most $p$"? Mar 21, 2015 at 19:07