Multiplicative group of a field implies every finite subgroup is cyclic

This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., multiplicative group of a field) must also satisfy the second group property (i.e., group in which every finite subgroup is cyclic)
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Statement

The multiplicative group of a field is a group in which every finite subgroup is cyclic: in other words, every finite subgroup of the multiplicative group of a field is a cyclic subgroup.

(Note that the result holds more generally for the multiplicative group of an integral domain).

Related facts

For finite fields

• Multiplicative group of a prime field is cyclic
• Multiplicative group of a finite field is cyclic

For infinite fields

• Classification of fields whose multiplicative group is locally cyclic
• Classification of fields whose multiplicative group is uniquely divisible

For commutative rings that are not fields

• Classification of natural numbers for which the multiplicative group is cyclic

Noncommutative analogues

• Multiplicative group of a skew field implies every finite abelian subgroup is cyclic

Facts used

1. Multiplicative group of a field implies at most n elements of order dividing n
2. At most n elements of order dividing n implies every finite subgroup is cyclic

Proof

The proof follows directly by piecing together facts (1) and (2).