Prove that $G=\mathbb{Z}_{n_1} \times ... \times \mathbb{Z}_{n_k}$ is cyclic $\iff n_1,...,n_k$ are pairwise coprime

Question

Prove that $G=\mathbb{Z}_{n_1} \times ... \times \mathbb{Z}_{n_k}$ is cyclic $\iff n_1,...,n_k$ are pairwise coprime.

In order to prove this fact, I was asked to prove:

1. $lg=e_G$ (the identity element of G), $\forall g\in G$ where $l=LCM(n_1,...,n_k)$

2. $\exists g\in G$ such that $o(g)=l$, where $o(g)$ denotes the order of $g$.

I have done both of these smaller proofs and am curious about how they fit into the larger picture. Hints are appreciated, thank you!

Answer 1

Using those is easy, notice that the least common multiple of $n_1,n_2, \dots n_k$ is equal to the product if and only if they are pairwise coprime.

So if $n_1,n_2\dots n_k$ are not pairwise coprime we have that $l<|G|$, and we have by the first result that the order of every element in $G$ is smaller than $|G|$, so the group cannot be cyclic.

On the other hand if $n_1,n_2,\dots n_k$ are pairwise coprime then $l=|G|$, so by result two we have that there is an element $g\in G$ with $o(g)=|G|$, and so the group is cyclic.