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 $(F-\{0\},\times)$.
 A: 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.


*

*$\mathcal A$ 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.]

*$\mathcal B$ 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.]

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

*$\mathcal B\subseteq \mathcal A$. [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.]
