# Group with at most n elements of order dividing n

From Groupprops

This article defines a group property: a property that can be evaluated to true/false for any given group, invariant under isomorphism

View a complete list of group propertiesVIEW RELATED: Group property implications | Group property non-implications |Group metaproperty satisfactions | Group metaproperty dissatisfactions | Group property satisfactions | Group property dissatisfactions

## Contents

## Definition

A group with identity element is termed a **group with at most n elements of order dividing n** if the following is true for every natural number : the number of such that is at most .

## Relation with other properties

### Stronger properties

Property | Meaning | Proof of implication | Proof of strictness (reverse implication failure) | Intermediate notions |
---|---|---|---|---|

locally cyclic group | every finitely generated subgroup is cyclic | |FULL LIST, MORE INFO | ||

multiplicative group of a field | isomorphic to the multiplicative group of a field | Group with at most n nth roots for any element|FULL LIST, MORE INFO | ||

group with at most n nth roots for any element | for any element of the group and any positive integer , there are at most roots of that element | |FULL LIST, MORE INFO |

### Weaker properties

Property | Meaning | Proof of implication | Proof of strictness (reverse implication failure) | Intermediate notions |
---|---|---|---|---|

group in which every finite subgroup is cyclic | every finite subgroup is cyclic | at most n elements of order dividing n implies every finite subgroup is cyclic | Every finite subgroup is cyclic not implies at most n elements of order dividing n | |FULL LIST, MORE INFO |

group with at most n pairwise commuting elements of order dividing n | for any positive integer , there are at most pairwise commuting elements of order dividing | |FULL LIST, MORE INFO | ||

group with finitely many elements of order dividing n | for any positive integer , there are finitely many elements of order dividing | |FULL LIST, MORE INFO | ||

group with finitely many conjugacy classes of elements of order dividing n | |FULL LIST, MORE INFO |

## Facts

For a finite group, we have a theorem that the number of nth roots is a multiple of n for each dividing the order of the group. Thus, for a finite group, this condition would imply that there are *exactly* elements of order dividing . In fact, a finite group satisfying this condition is cyclic.

A more general question is the following: given any finite group and a natural number dividing the order of such that there are exactly elements whose order divides , do those elements form a subgroup? This is the Frobenius conjecture on nth roots.