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

Solution 1:

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 $K^\times$.

Solution 2:

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.]

Solution 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