math.stackexchange.com/questions/350421/finite-group-cyclic
math.stackexchange.com/questions/346936/finite-group-for-which-xxm-e-leq-m-for-all-m-is-cyclic
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY
Volume 32, Number 1, March 1972A CONDITION FOR A FINITE GROUP TO BE CYCLIC
J. H. E. COHN
Abstract. A finite group, with the property that for every prime power $q=p^k$,
there are at most $p^{k+1}-1$ elements in the group
whose $q$th power is the identity, is a cyclic group
It seems to be well known that a finite group, which need not be given to be abelian, must be cyclic if for every positive integer $n$, the number of elements for which $x^{n}$ is the identity does not exceed $n$. The object of this note is to prove the stronger
THEOREM. Let $G$ be a finite group, with identity $e$, such that for every prime power $q=p^{k}$ there exist at most $p^{k+1}-1$ elements satisfying $x^{q}=e$. Then $G$ is cyclic.
Proof. Suppose that $p \mid o(G)$, with say $p^{m} \| o(G)$. Let $H$ denote any Sylow $p$-subgroup. Then choosing $q=p^{m-1}$ it follows that $H$ is cyclic, since it has at least one element whose order does not divide $q$.
Furthermore, $H$ is the only Sylow $p$-subgroup, and so is a normal subgroup of $G$. For in the contrary case, there would exist at least $p+1$ distinct cyclic Sylow $p$-subgroups, no two of which could contain an element of order $p^{m}$ in common. Thus they would contain together at least $(p+1) p^{m-1}(p-1)+p^{m-1}=p^{m+1}$ distinct elements, all of which would satisfy $x^{q}=e$ with $q=p^{m}$, contrary to hypothesis.
Thus $G$ must be the direct product of its Sylow subgroups, each one of which is cyclic, and accordingly $G$ must be cyclic.
Corollary. Every noncyclic finite group has for at least one prime power $p^{k}$, at least $p^{k-1}\left(p^{2}-1\right)$ elements of order $p^{k}$ exactly.
Proof. Suppose on the contrary that for every prime power $p^{k}$ there were less than $p^{k-1}\left(p^{2}-1\right)$ elements of order $p^{k}$. Now the number of such elements is divisible by $p^{k-1}(p-1)$, for each one generates a cyclic subgroup with this number of generators. Thus there can be at most $p^{k}(p-1)$ elements of order $p^{k}$ for every prime power. But then there would be at most $p^{k+1}-p+1 \leqq p^{k+1}-1$ elements satisfying $x^{q}=e$ for every prime power $q=p^{k}$, and the group would be cyclic by the theorem. | 
