How to prove that G is a cyclic Group?

Question

Suppose $G$ is a finite Abelian group and, $\forall\ n\in \mathbb{N} $, there exist at most $n$ elements in $G$ which satisfy $x^n=1$. Prove $G$ is cyclic.

Thanks for your help.

Answer 1

Lets prove it by contradiction.

Let $G$ is not cyclic and finite abelian, we can find an element $g\in G$ which has maximal order among elements of $G$, say order $k$ and $k < |G|$ . Now pick an element $h\in G$ \ $<g>$ and let $|h|=m$ then obviously $m \le k$ as $k$ is maximal among orders of elements of $|G|$.

Now consider $l=\text{lcm} (k,m) $. As $G$ is abelian there exist an element of order $l$ and obviously $l \ge k$, so it implies by maximality of $k$, that $l=k$. But $l=k \implies m|k\ $ that is $k=tm$ for some positive integer $t$. Now note that all elements $\{1,g,g^2,\dots g^{k-1}\} \cup \{h\}$ satisfies $x^k=1$, thus more than $n$ elements satisfies $x^k=1$.

Hence Proved