# Group with at most n elements of order dividing n

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

A group $G$ with identity element $e$ is termed a group with at most n elements of order dividing n if the following is true for every natural number $n$: the number of $g \in G$ such that $g^n = e$ is at most $n$.

## 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 $n$, there are at most $n$ $n^{th}$ 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 $n$, there are at most $n$ pairwise commuting elements of order dividing $n$ |FULL LIST, MORE INFO
group with finitely many elements of order dividing n for any positive integer $n$, there are finitely many elements of order dividing $n$ |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 $n$ dividing the order of the group. Thus, for a finite group, this condition would imply that there are exactly $n$ elements of order dividing $n$. In fact, a finite group satisfying this condition is cyclic.

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