Finite abelian group A cyclic iff for every $n\in \mathbb{N}$, $|\{a\in A \mid a^n = 1\}| \leq n$

Question

I'm trying to prove that a finite group A is cyclic if and only if $\forall n\in\mathbb{N}$ , there are at most n elements in A with $a^n=1$.

Heres what I've come up with so far:

Suppose that A is cyclic. We know that $H=\{a\in A \mid a^n =1\}$ is a cyclic subgroup of A, thus is generated by an element of order n or less. Therefore H has n or fewer elements.

Now for the converse, what I feel inclined do is assume that A is not cyclic and then get that there must be more than n elements with $a^n=1$.

Any thoughts on how I might finish this one off are appreciated.

Answer 1

By the FTFAG, $A$ is not cyclic only if it contains a subgroup isomorphic to $\Bbb Z_p\times \Bbb Z_p$, which implies there are $p^2-1\gt p$ elements of order $p$.