This is motivated by a characterization of finite cyclic groups, in which one proves
Let $G$ be a finite group. If $\#\{g\in G\colon g^d=e\}$ is at most $d$, then $G$ is cyclic.
The proof is not actually difficult, but a unnecessarily complicated idea occurred to me (maybe because the first time I saw such a result, it was used to prove that finite subgroups of the multiplicative group of a field are cyclic).
If we can construct this $G$ as the multiplicative group of a certain division ring, that is, $R^*\colon=\{r\in R\colon r\neq 1\}$. Then we know $G$ is abelian since finite division rings are commutative, and the abelian case follows from a direct use of the structure theorem of finite abelian groups.
Of course such a proof is unnecessarily complicated and very likely results in some cyclic arguments since it uses two very big structure theorems. But I guess it would still be nice to know what kind of groups are multiplicative groups of division rings.
Thanks very much!