Multiplicative group of a field implies every finite subgroup is cyclic

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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)
View all group property implications | View all group property non-implications
Get more facts about multiplicative group of a field|Get more facts about group in which every finite subgroup is cyclic

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

For commutative rings that are not fields

Noncommutative analogues

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).