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It is well known that every finite multiplicative subgroup of a field is cyclic. I somehow got interested in a possible reverse implication:

Assume we have an abelian group $G$ whose every finite subgroup is cyclic.

Is $G$ necessarily the multiplicative group of all nonzero elements of some field $F$?

(i.e I ask whether we can "add" to $G$ a "zero" element, and define addition operation $+$ making $G\cup\{0\}$ a field w.r.t $+$ and the multiplication in $G$).

Of course for finite cyclic groups the answer is trivial. (A group $G$ can be completed to a field iff $|G|=p^n-1$ for some prime $p$).

Next, I considered the infinite group of all complex roots of unity (of all orders). I once proved to myself that it cannot be completed into a field, but I do not remember exactly how I did this.

I wonder if there are "nice" necessary & sufficient conditions which ensure $G$ can be completed.

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    $\begingroup$ I think $\mathbb{Z}$ cannot be isomorphic to a multiplicative group. If the characteristic is not 2, then the multiplicative group contains a cyclic group $\{1, -1\}$ of order 2, which $\mathbb{Z}$ does not. If char(k) = 2, then we easily rule out transcendence degree 0 over $\mathbb{Z}/2$, but then transcendence degree $\geq 1$ means $\mathbb{Z}/2(x)$ is a subfield, whose multiplicative group is not cyclic (not iso to $\mathbb{Z}$). $\endgroup$
    – Todd Trimble
    Commented Apr 28, 2015 at 11:30
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    $\begingroup$ A torsion group such as $\mathbb{Q}/\mathbb{Z}$ cannot be a subgroup of the multiplicative group of any field of positive characteristic $p$, because the group contains nontrivial elements of order $p$. The multiplicative group of any characteristic $0$ field contains elements of infinite order, e.g., $2$. Thus $\mathbb{Q}/\mathbb{Z}$ cannot be the group of units of a field. $\endgroup$
    – Jason Starr
    Commented Apr 28, 2015 at 11:30
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    $\begingroup$ The paper plms.oxfordjournals.org/content/s3-18/1/114 seems relevant. $\endgroup$
    – Tobias Kildetoft
    Commented Apr 28, 2015 at 11:37
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    $\begingroup$ If $F$ is an infinite field, then $F^\times$ isn't finitely generated. The proof is easy: if $\operatorname{char} F=0$, then $\mathbb{Q}^\times$ is a subgroup of $F^\times$; if $\operatorname{char} F=p>0$, then $\mathbb{F}_p(x)^\times$ is a subgroup of $F^\times$ (as explained in Todd Trimble's comment). Neither of these is finitely generated, hence the result. $\endgroup$
    – Jarek Kuben
    Commented Apr 28, 2015 at 14:05
  • $\begingroup$ @JarekKuben: it is not true that if $\operatorname{char}F>0$ and $F$ is infinite, then $F$ contains $\mathbb{F}_p(x)$ as a subfield. For example, take $F$ to be $\overline{\mathbb{F}}_p$, or any infinite subfield of this. However, in this case, any finite subset $S \subset F^\times$ will generate a finite extension over $\mathbb{F}_p$ in the sense of field extensions, and a fortiori will never generate $F^\times$ as a group. $\endgroup$
    – R.P.
    Commented Apr 28, 2015 at 21:56

2 Answers 2

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There are various constraints.

One of those is: the $\mathbf{Q}$-rank is either 0 or infinite. (Thus $F^*$ cannot be isomorphic to $\mathbf{Z}\times \mathbf{Q}/\mathbf{Z}$, among others).

Indeed, if $F$ has characteristic zero, then the primes form an infinite $\mathbf{Z}$-free family. If $F$ has characteristic $\ell>0$ and $F^*$ is not torsion, then there exists a transcendental element $x$, and then the $(x^p-1)$, where $p$ ranges over primes $\neq\ell$, form a $\mathbf{Z}$-free family. Indeed otherwise we could find a positive integer $k$, a non-negative integer $n$, and primes $p_1,\dots,p_k,q_1,\dots,q_n$, all distinct and distinct from $\ell$, and positive integers $t_1,\dots,t_k,m_1,\dots,m_n$, such that $\prod_i (x^{p_i}-1)^{t_i}=\prod_j (x^{q_j}-1)^{m_j}$. Since $x$ is transcendental, this equality still holds in the polynomial ring $\mathbf{F}_p[x]$. Evaluation at a primitive $p_1$-root of unity (in a suitable extension of $\mathbf{F}_\ell$) then yields a contradiction.

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Restricting to $\mathbb{Q}$-rank zero, there are more complicated torsion examples, such as $G=\mathbb{Z}[1/5]/\mathbb{Z}$. If $K^\times=G$, then $K$ must be algebraic over a finite field $k=\mathbb{F}_p$ for some prime $p>0$, since $G$ is torsion. But then $k^\times$ is a subgroup of $G$, and so we have that $\# k^\times$ is $5^n$ for some integer $n$. But we also have $\# k = p$, hence $$ p = 5^n + 1, $$ which is clearly impossible unless $p=2$. So assume $p=2$ now. In that case, $K$ contains an ascending chain of field extensions of $k=\mathbb{F}_2$, giving infinitely many positive integer solutions to $$ 2^m=5^n+1, $$ which is a contradiction, for instance with Catalan's conjecture (proven by Mihailescu), although a variety of more elementary arguments are possible (for which see the comments).

(With some more work, I think one can rule out $G=\bigoplus_\ell \mathbb{Z}[1/\ell]/\mathbb{Z}$, where the sum ranges over any finite set of primes $\ell$.)

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  • $\begingroup$ Isn't there any simpler argument to rule out $\mathbf{Z}[1/q]/\mathbf{Z}$ for prime $q$? Indeed, then $F$ has to be algebraic over $F_p$ for some prime $p\neq q$. Then for some increasing sequence $(n_i)$, we have $p^{n_i}-1$ equal to some power $q^{m_i}$ of $q$. I'm not a number theorist but this sounds unlikely... $\endgroup$
    – YCor
    Commented Apr 28, 2015 at 12:41
  • $\begingroup$ I absolutely agree that Catalan's conjecture is kind of a sledge-hammer here: it would suffice to know that $x^m−y^n=1$ has only finitely many integer solutions with $m,n>1$, and this was proven way before Mihailescu (by Rob Tijdeman). What you are saying is that, in my case, $x$ and $y$ are fixed as well, and one could presumably make use of this fact. Indeed I do think that applying some results from Diophantine approximation would probably do the trick (and prove the stronger result as well), but on the other hand those results have the disadvantage of being less well-known. $\endgroup$
    – R.P.
    Commented Apr 28, 2015 at 12:58
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    $\begingroup$ Well, if $p>2$ and $p^n-1=q^m$ with $m\ge 2$ and $(p^n,q^m)\neq (9,8)$, let's reach a contradiction. Then $p-1$ divides $p^n-1$, hence is a power of $q$, and is not 1 because $p\neq 2$. If $r$ is a prime divisor of $n$, then $p^r-1$ divides $p^n-1$, hence is a power of $q$, and so is $p^{r-1}+\dots+p+1>1$, so the latter is a power of $q$, but since $p$ is 1 mod $q$, the latter is equal to $r$ modulo $q$. Hence $r=q$. If $p-1\neq q$, actually $p$ equals 1 modulo $q^2$ and we get the previous equality mod $q^2$, so $p^{r-1}+\dots+p+1=q<p-1$, a contradiction. This finishes unless $(p,q)=(3,2)$... $\endgroup$
    – YCor
    Commented Apr 28, 2015 at 13:07
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    $\begingroup$ When $(p,q)=(3,2)$, the argument shows that $n$ is a power of 2, but since $p^{q^2}-1=80$ is not a power of $q=2$, we deduce $n=2$ and get $(9,8)$. It remains to consider $p=2$. $\endgroup$
    – YCor
    Commented Apr 28, 2015 at 13:10
  • $\begingroup$ That's very nice! $\endgroup$
    – R.P.
    Commented Apr 28, 2015 at 13:15

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