Does there exist a finite abelian group $G$, such that for no field $F$ is it the case that $G$ is isomorphic to either the additive group of $F$ nor the multiplicative group of non-zero elements of $F$? I would prefer a counterexample of minimum cardinality.
Is there a finite abelian group which is not isomorphic to either the additive or multiplicative group of a field
abelian-groupsfield-theoryfinite-groupsgroup-theory
Related Solutions
A pair of fields like this exists.
As noted in the question, one may assume that $E$ is a field of characteristic zero. Therefore $E^+$ is an infinite, torsion-free, divisible, abelian group; i.e. an infinite rational vector space. Moreover, any infinite rational vector space is isomorphic to the additive group of some field of characteristic zero. Thus the question reduces to this: is there an infinite field $F$ whose multiplicative group of units is a rational vector space?
Such a field is constructed in
Adler, Allan On the multiplicative semigroups of rings. Comm. Algebra 6 (1978), no. 17, 1751-1753.
Here is the first line of George Bergman's Math Review of this paper:
``Let $\mathbb Q$ denote the additive group of the rational numbers. It is shown that there exists a field whose multiplicative group is isomorphic to $\mathbb Q^{\aleph_0}$, but none whose multiplicative group is isomorphic to $\mathbb Q$.''
EDIT: 8/21/20
In response to the August 18 question of mr_e_man below, I asked my library to get a scan of Adler's paper. Adler's construction of a field whose multiplicative group is isomorphic to the direct power $\mathbb Q^{\aleph_0}$ goes like this:
Let $\mathbb F_{2^n}$ be Galois field of order $2^n$. Let $P$ be the set of primes. The desired field can be taken to be any field $\mathbb K$ that is an ultraproduct of the form $\prod_{\mathcal U} \mathbb F_{2^p}$ over a nonprincipal ultrafilter $\mathcal U$ defined on $P$. Let's show that the multiplicative group of $\mathbb K$ is isomorphic to $\mathbb Q^{\aleph_0}$.
Up to isomorphism, $\mathbb Q^{\aleph_0}$ is the unique torsion-free divisible abelian group of cardinality continuum. The multiplicative group $\mathbb K^{\times}$ is isomorphic to the abelian group $\prod_{\mathcal U} \mathbb F_{2^p}^{\times}$. The factors in this ultrproduct have sizes $2^2-1< 2^3-1 < 2^5-1 < \cdots$, so the sizes are a strictly increasing countable sequence. This is enough to prove that the ultraproduct has size continuum. (I am asserting that any nonprincipal ultraproduct of countably many finite sets of strictly increasing size has cardinality continuum.)
If $p\neq q$, then $\gcd(|\mathbb F_{2^p}^{\times}|,|\mathbb F_{2^q}^{\times}|)=\gcd(2^p-1,2^q-1)=1$. Thus, for any $m\leq \textrm{min}(2^p-1, 2^q-1)$, at most one of the groups $\mathbb F_{2^p}^{\times}$ has $m$-torsion. In fact, the map $x\mapsto x^m$ must be a bijection on all $\mathbb F_{2^p}^{\times}$ except possibly finitely many of them. In the ultraproduct $\prod_{\mathcal U} \mathbb F_{2^p}^{\times}=\mathbb K^{\times}$ the maps $x\mapsto x^m$, $m\in\mathbb Z^+$, must all be bijections. This is enough to show that $\mathbb K^{\times}$ is torsion-free and divisible.
Of course.
There trivial groups $\{0\}, +$ and $\{1\}, \times$ but we could have general cyclic groups with the same order.
Example in $\mathbb R$ then $<1,+> = \mathbb Z$ is isomorphic to $\{q^n|n\in \mathbb Z\}, \times$ (assuming $q>0; q\ne 1$).
Best Answer
The two results being used below are that the multiplicative group of a finite field is cyclic and the classification of finite abelian groups. For the former, see this MO thread for a discussion of some accessible proofs.
I made a really bad mistake in the first version of this answer! Since this answer's been accepted I can't delete it, but the fix below is due to Fishbane's comment so I've made this CW.
$C_2\times C_4$ is not the additive or multiplicative group of any field. Meanwhile, it's easy to check that every abelian group of size $<8$ is either the additive or multiplicative group of a field:
For each prime $p$ (here $p=2,3,5,7$), $C_p$ is the additive group of the field $\mathbb{F}_p$.
The only other abelian groups are the trivial group (= multiplicative group of $\mathbb{F}_2$), the Klein four group (= additive group of $\mathbb{F}_4$), and $C_6$ (= multiplicative group of $\mathbb{F}_7$).