## A divisibility criterion for orders of elements in a finite abelian group

Let $G$ be a finite abelian group of type $(n_1,n_2,\ldots,n_k)$. (That is, $n_1,\ldots,n_k$ are the invariant factors of $G$.) Prove that $G$ contains an element of order $m$ if and only if $m|n_1$. Deduce that $G$ is of exponent $n_1$.

$(\Rightarrow)$ Suppose $G$ contains an element $x$ of order $m$. Now $x = (a_1,a_2,\ldots,a_k)$ for some $a_i \in Z_{n_i}$, and we have $m = \mathsf{lcm}(|a_1|,|a_2|,\ldots,|a_k|)$. By the divisibility criterion for invariant factors, $n_1$ is a common multiple of the $n_i$, hence of the $|a_i|$. Thus $m|n_1$.

$(\Leftarrow)$ Suppose $m|n_1$. Now $G$ has a subgroup isomorphic to $Z_{n_1}$, which (being cyclic) has an element of order $m$. Thus $G$ has an element of order $m$.

We conclude that if $x \in G$ has order $m$, then since $n_1 = tm$ for some $t$, $n_1x = tmx = 0$. Hence the exponent of $G$ is finite and divides $n_1$. Moreover, the exponent of $G$ is at least $n_1$ since $G$ has an element of order $n_1$. Thus $G$ has exponent $n_1$.