A finite abelian group $A$ is cyclic iff for each $n \in \mathbb{N}$, $\#\{a \in A : na = 0\}\le n$ 
Let $A$ be a finite abelian group. Prove that $A$ is cyclic iff for each $n \in \mathbb{N}$
$$\#\{a \in A : na = 0\}\le n.$$

Any help or hint will be appreciated.
 A: Here is a proof that appears in Weil's gem Number theory for beginners and in several other books.
Let $m$ be the order of $A$ and 
let $\psi(n)$ be the number of elements of order $n$ in $A$.
Then, by Lagrange's theorem, $m=\sum_{d\mid m} \psi(d)$ because every element has an order that is a divisor of $m$.
If $\psi(d)>0$, then there is an element of order $d$ and the subgroup generated by it has order $d$ and so the hypothesis on $A$ implies that $\psi(d) = \phi(d)$.
Therefore, for all $d \mid m$, we have $\phi(d) \ge \psi(d)$.
Now $m = \sum_{d\mid m} \phi(d) \ge \sum_{d\mid m} \psi(d) = m$. (*)
This means that $\psi(d)=\phi(d)$ for all $d \mid m$.
In particular, $\psi(m)>0$, which means that $A$ has an element of order $m$ and so is cyclic.
(*) The first equality comes from counting the fractions $1/m,\dots,m/m$ when they are reduced.
A: As pointed out in the comment section by Arthur, your statement is partially false. Let us show your statement for finite abelian groups.
Let $G$ be a finite abelian group, then there exists $n\in\mathbb{N}$ and $(d_i)_{i\in\{1,\ldots,n\}}\in{\mathbb{N}_{\geqslant 2}}^n$ such that: $$\forall i\in\{2,\ldots,n\},d_{i-1}\vert d_i\textrm{ and }G\cong\bigoplus_{i=1}^n\mathbb{Z}/(d_i).$$


*

*Assume that for all $m\in\mathbb{N}$, $|\{g\in G\textrm{ s.t. }mg=0\}|\leqslant m$. One has to show that $n\in\{0,1\}$. Assume by contradiction that $n\geqslant 2$, let us take $m:=d_2$ and notice that for all $(x,y)\in\mathbb{Z}/(d_1)\oplus\mathbb{Z}/(d_2)$, one has: $$m(x,y,0,\ldots,0)=0.$$
Which is a contradiction, since we have found $d_1d_2>m$ elements in $\{g\in G\textrm{ s.t. }mg=0\}$.

*Assume that $G$ is cyclic, since $G$ is finite, $G\cong\mathbb{Z}/(|G|)$ and I let you show that the given condition holds.
