When can an infinite abelian group be embedded in the multiplicative group of a field?

Question

This question comes from this question by user72870. That question would easily be answered if we know the cyclicity of the group in question, but, as the OP appears to be trying to prove that the group is cyclic by embedding it into the multiplicative group of a field, we cannot work under the cyclic hypothesis.
So I loosed the condition and only assumed that it is abelian, but I could not find an appropriate field to embed it: finite fields are definitely out of options in reason of orders, and the ideal field must contain some sort of "cyclic algebras" for this embedding to happen.
The obstacle that I emcountered is that, in the decomposition of an abelian group into cyclic ones, the orders of different summands might not be relatively prime, so that the approach to embed an anelian group into the multiplicative group of a field by embedding respectfully its cyclic factors does not work.

Edit: Thanks to the reminder of @user72870, I realised that it is impossible to embed a finite abelian group into the multiplicative group of a field, unless the group is cyclic...I don't know what I was thinking at that moment, trying to embed a general abelian group into the multiplicative group of a field.
Hence I change the question to embedding a general infinite group into the multiplicative group of a (infinite, of course) field.
Any help or hint will be greatly appreciated; thanks in advance.

Answer 1

Let $\mathcal A$ be the class of those abelian groups embeddable in the multiplicative group of some field. Let $\mathcal B$ be the class of those abelian groups whose finite subgroups are cyclic. I claim that $\mathcal A=\mathcal B$. From this it follows that the answer to the question

When can an infinite abelian group be embedded in the multiplicative group of a field?

is

Those infinite abelian groups whose finite subgroups are cyclic.

This class of groups is axiomatized by universally quantified first-order sentences. The sentences needed are (i) some universally quantified sentences axiomatizing the class of abelian groups, together with (ii) for each $n$, a sentence $\sigma_n$ 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 $x$. (I am considering abelian groups as multiplicative groups.)

To be clear here, the sentence $\sigma_3$ is the following universally quantified first-order sentence:

$$ \forall x\forall y(((x^3=1)\wedge (y^3=1))\to ((x=1)\vee(x=y)\vee(x=y^2)\vee(y=1)\vee(y=x)\vee(y=x^2)) $$

---

To see that $\mathcal A=\mathcal B$, observe that

1. $\mathcal A$ is a class of first-order structures that is closed under the formation of substructures and ultraproducts. [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.]

(The preceding observation implies that $\mathcal A$ is axiomatizable by universally quantified first-order sentences.)

2. $\mathcal B$ is a class of first-order structures that is axiomatizable by universally quantified first-order sentences. [Reason: The sentences I mentioned above work. Recall that these sentences are the axioms for abelian groups together with all $\sigma_n$.]

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 classes that are axiomatizable by universally quantified first-order sentences 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.]

Answer 2

You can do it if and only if the torsion subgroup is a subgroup of $\mathbb{Q} / \mathbb{Z}$: i.e. it has at most one subgroup of order $n$ for every finite $n$.

If this is the case, then you can write the group as a product $T \times F$ where $T$ is torsion and $F$ is torsion-free.

Let $F \otimes \mathbb{Q} \cong \mathbb{Q}^\alpha$.

Then if $K$ is an algebraiaclly closed field of transcendence degree $\alpha$ over $\mathbb{Q}$, then you can embed $T$ in the roots of unity of $K$, and you can pick $\alpha$ multiplicatively independent elements to map a basis of $\mathbb{Q}^\alpha$ to. Since $F \subseteq F \otimes \mathbb{Q}$, these together allow you to embed your group in $K^\times$.

Answer 3

Nope! Any finite multiplicative group of a field is cyclic. See here, for example http://www.math.upenn.edu/~ted/203S06/References/multsg.pdf