Let $G$ be a finite Abelian group: $G=\mathbb{Z}_{n_1}\times\cdots\times\mathbb{Z}_{n_k}$ and $n_k \mid n_{k-1} \mid \cdots \mid n_1$.
Show that $G$ contains an element of order $m$ iff $m\mid n_1$.
Let $G$ be a finite Abelian group: $G=\mathbb{Z}_{n_1}\times\cdots\times\mathbb{Z}_{n_k}$ and $n_k \mid n_{k-1} \mid \cdots \mid n_1$.
Show that $G$ contains an element of order $m$ iff $m\mid n_1$.
Exercise 1. Show that, for any finite groups $G$ and $H$,
$\mathrm{ord}_{G\times H}(a,b)=\mathrm{lcm}\big(\mathrm{ord}_G(a),\mathrm{ord}_H(b)\big).$
Proceed further to show that $\mathrm{ord}_{G_1\times\cdots\times G_k}(a_1,\cdots,a_k)=\mathrm{lcm}\big(\mathrm{ord}_{G_1}(a_1),\cdots,\mathrm{ord}_{G_k}(a_k)\big)$.
Exercise 2. Show that the $\mathrm{lcm}$ of any number of divisors of $n_1$ will divide $n_1$.
Now suppose you have a tuple in $\Bbb Z_{n_1}\times\cdots\times\Bbb Z_{n_k}$. The order (in $\Bbb Z_{n_j}$) of the $j$th component will divide $n_j$ by Lagrange's theorem, and by transitivity of the $\mid$ relation will also divide $n_1$. By exercise one, the order of the tuple itself in the product will be the lcm of these component orders, and by exercise two will divide $n_1$. Conversely, for every divisor $d|n_1$, $\Bbb Z_{n_1}$ will have an element of order $d$, which can easily be transplanted into the product group (put it inside the first coordinate and make all the other coordinates zero).