It is a well-known fact that a finite subgroup of the multiplicative group of a field is cyclic. We will prove this result shortly. We will also extend it to any division ring of non-zero characteristic. Note that this result is not necessarily true in a division ring of zero characteristic. For example, in the division ring of real quaternions, the subgroup is not even abelian let alone cyclic. We will also prove that finite abelian subgroups of the multiplicative group of any division ring are cyclic.

**Theorem**. Every finite subgroup of the multiplicative group of a field is cyclic.

*Proof*. Let be a field and let be a finite subgroup of Let and, for any divisor of let be the number of elements of of order Obviously

Let be the Euler’s totient function. Recall from number theory that

*Claim*. If and then

*Proof of the claim*. Since there exists such that Let and Then every element of is a root of But has at most roots in Thus is exactly the set of roots of Finally, the fact that an element has order if and only if implies

It now follows from and the claim that for all In particular, and so for some

**Corollary 1. **Every finite abelian subgroup of the multiplicative group of a division ring is cyclic.

*Proof*. Let be a division ring with the center Let be a finite abelian subgroup of and put It is obvious that is a commutative domain and Also, since is finite, is a finite dimensional vector space over and thus every element of is algebraic over Let and suppose that is the minimal polynomial of over Then and so Therefore is a field and we are done by the above theorem.

**Corollary 2**. Every finite subgroup of the multiplicative group of a division ring of non-zero characteristic is cyclic.

*Proof*. Let be a division ring with Let be the prime subfield of and suppose that is a finite subgroup of Let Clearly is a finite subring of and contains Let Since is a domain, is a domain too. Thus and so for some Therefore is a division ring. But, by Wedderburn’s little theorem, a finite division ring is a field. So is a field and we are done by the above theorem.