Multiplicative Group of a Field The multiplicative group $F^{\times}=F\setminus \{0\}$ of a field is abelian, and it may contain torsion elements, may contain torsion free elements, or both may occur, as can be seen from the examples of any finite field, $\mathbb{Q},\mathbb{R}$ and $\mathbb{C}$. The Prüfer $p$-group is a (proper) subgroup of $\mathbb{C}^{\times}$. The question I would like to ask is
Question: Is there an infinite field $F$ such that $F^{\times}$ is isomorphic to the Prüfer $p$-group?
 A: Nice question! The answer is no. 
If $F$ has characteristic zero it contains a copy of $\mathbb{Q}$, so $F^{\times}$ has torsion-free elements; hence $F$ has positive characteristic $p$. If $F^{\times}$ is isomorphic to a Prüfer $\ell$-group for some prime $\ell$, then $F$ contains all $\ell$-power roots of unity, and $\mathbb{F}_p^{\times}$ must consist of $\ell$-power roots of unity, so in particular $p \neq \ell$. 
Let $k$ be a positive integer. Since $F$ contains all $\ell^k$-th roots of unity, $F$ contains $\mathbb{F}_{p^n}$ where $n$ is the smallest positive integer such that $\ell^k | p^n - 1$. Then $F^{\times}$ has a subgroup of order $p^n - 1$, which is necessarily a power of $\ell$. By choosing $k$ large enough, this is impossible by Zsigmondy's theorem. 
A: The following more general statement is also true.
There is no commutative ring $R$ whose unit group $R^*$ is the Prüfer $p$-group.
For a proof, see  https://arxiv.org/pdf/1505.03508.pdf
In this paper, Keir Lockride and I  gave a complete classification of all indecomposable abelian groups which occur as the group of  units of a commutative ring. Note that a Prüfer $p$-group is indecomposable and it does not appear in our list. 
