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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
See…. – 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…. – rschwieb May 15 '13 at 15:05
up vote 3 down vote accepted

"The answer" for classifying which finite groups are subgroups of units of division rings was given by Amitsur in Finite subgroups of division rings. Trans. Amer. Math. Soc. 80 (1955), 361–386. This is generally considered to be the analogue of the "finite subgroups cyclic" theorem for the unit groups of fields.

(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 $\Bbb H^\ast$.)

This clearly acts as "local" restrictions on elements of finite order.

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