# Given $G$, when can we find a division ring $R$ with $R^*=G$?

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!

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This is certainly not always possible. Recently a question was posed here to prove that there are no rings at all (let alone division rings) whose group of invertibles has order $5$. –  Marc van Leeuwen May 15 '13 at 14:40
@MarcvanLeeuwen I have the feeling that this is in general not possible. But I wonder whether there is some nice way to know when this is possible. –  Hui Yu May 15 '13 at 14:43
–  lhf May 15 '13 at 14:55
Pretty sure this question has been asked before. There is a positive answer possible under several restrictions described at math.stackexchange.com/questions/156363/…. –  rschwieb May 15 '13 at 15:05

(In Lam's paper on the quaternions, he describes an interesting lead-up to that discovery about Coexeter's work determining the finite subgroups of $\Bbb H^\ast$.)