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			|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.