Abelian groups whose finite subgroups are cyclic

If $(F,+,\times)$ is any field, then the abelian group $(F-\{0\},\times)$ has property that every finite subgroup of it is cyclic.

Question: If $G$ is an abelian group such that every finite subgroup of $G$ is cyclic, then can $G$ be embedded in the multiplicative group of some field?

It should be noted that if $G$ is an abelian group such that every finite subgroup of $G$ is cyclic, then $G$ may not be isomorphic to the multiplicative group of some field. For example, if $G$ is cyclic group of order $5$ then $G$ can not be isomorphic to $(F-\{0\},\times)$ for any field, because, then $|F|$ will be $6$, impossible.

The point to say here is that in question, I am stressing on embedding of $G$ rather than isomorphism of $G$ with $(F-\{0\},\times)$.

• I bet $\mathbb{C}^\times$ would do a job. – Orat Jun 6 '16 at 8:06
• Perhaps, if $G$ is finite abelian with property in question, then you are right. – p Groups Jun 6 '16 at 8:08
• As a special case you could ask whether every torsion free abelian group embeds in the multiplicative group of a field. – Derek Holt Jun 6 '16 at 13:34
• I think you can. See this question, but I would have to think more to actually understand it (I am not familiar with some of the concepts). Pretty sure the torsion subgroup of such groups is a subgroup of $\mathbb{Q}/\mathbb{Z}$ (the torsion subgroup is countable since if they were not there would be two cyclic subgroups $\langle a \rangle, \langle b \rangle$ of the same order but $\langle a,b \rangle$ would not be cyclic). – Paul Plummer Jun 21 '16 at 19:58

The answer to the question is Yes.

A universally axiomatizable class of structures is called a universal class. Such classes have been characterized as classes closed under the formation of substructures and ultraproducts.

Let $\mathcal A$ be the class of those abelian groups $A$ embeddable in the multiplicative group of some field. Let $\mathcal B$ be the class of those abelian groups $B$ whose finite subgroups are cyclic. The following are true.

1. $\mathcal A$ is a universal class. [Reason: it is clear that $\mathcal A$ is closed under the formation of substructures. If $\{A_i\;|\;i\in I\}\subseteq \mathcal A$ is a class of abelian groups, each embeddable in the multiplicative group of a field, then each ultraproduct that can be formed from this set is embeddable in the multiplicative group of the corresponding ultraproduct of fields. This ultraproduct is itself a field.]

2. $\mathcal B$ is a universal class. [Reason: one can write down the universal sentences that axiomatize $\mathcal B$. Beyond the axioms that say, I am an abelian group, one should include, for each $n$, a first-order axiom that says, for all $x, y$, if $x$ and $y$ both have exponent $n$, then $x$ is a power of $y$ or $y$ is a power of $x$. Here I am considering abelian groups as multiplicative groups.]

3. $\mathcal A\subseteq \mathcal B$. [Reason: it is well known that a finite subgroup of the multiplicative group of a field is cyclic.]

4. $\mathcal B\subseteq \mathcal A$. [Reason: It is a general fact about universal classes that if $\mathcal B\not\subseteq \mathcal A$, then there is a finitely generated $B\in \mathcal B-\mathcal A$. A finitely generated member of $\mathcal B$ has the form $\mathbb Z^k\oplus \mathbb Z_m$. So to establish this claim it suffices to show that groups of this form are embeddable in multiplicative groups of fields. To embed this group, choose algebraically independent elements $\alpha_1,\ldots, \alpha_k\in\mathbb C$ and let $\zeta\in\mathbb C$ be a primitive $m$-th root of unity. The multiplicative subgroup of $\mathbb C$ generated by $\{\alpha_1,\ldots,\alpha_k,\zeta\}$ is isomorphic to $\mathbb Z^k\oplus \mathbb Z_m$, so we are done.]