# 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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## Contents

## 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

### 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

### Noncommutative analogues

## Facts used

## Proof

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