Showing that the Class of Cyclic Groups Aren't Axiomatizable

Question

The class of finite cyclic groups are not axiomatizable, for if we supposed they were by some set of sentences $\Sigma$, then there would exist a model for $\Sigma$ of at least order $n$ for all $n \in \mathbb{N}$ (i.e., the model $Z_{n+1}$ for each $n$). Then we would have that there would also exist an infinite model as well by the compactness theorem (and hence we would have a contradiction).

But evidently the theory all cyclic groups (including the infinite ones) are also not axiomatizable. Why is this so?

Answer 1

Basically the same idea will work: By Upward Lowenheim-Skolem, there would be models of arbitrarily high (in particular, uncountable) cardinality. But an uncountable group cannot be cyclic.

Answer 2

Suppose there is a set $\Sigma$ of first-order sentences such that $\mathrm{Mod}(\Sigma)$ is the class of all cyclic groups $(A,+)$. For $m,n\in\mathbb Z\setminus\{0\}$, let $\varphi_{m,n}(x,y)$ be the formula $(x\ne0)\land(y\ne0)\land(mx\ne ny)$. Then $\Sigma\cup\{\varphi_{m,n}(x,y):m,n\in\mathbb Z\setminus\{0\}\}\ $ is not satisfiable, although every finite subset is satisfiable. This contradicts the compactness theorem.

Answer 3

André's answer provides an answer to your question, but the result can be strengthened: the cyclic groups are not axiomatisable even among countable groups.

This is, in part, because the theory of $({\bf Z},+)$ is not $\omega$-categorical. For instance ${\bf Z}\oplus {\bf Q}$ is elementarily equivalent to it, and is evidently not cyclic. In fact, the theory has $2^{\aleph_0}$ nonisomorphic countable models. To see this, notice that there are that many complete types over $\emptyset$.

On the other hand, using the classification theorem for finitely generated abelian groups, one can show that cyclic groups are elementary relative to the (clearly non-elementary) class of finitely generated groups.