I need to prove that for any integers $m$, $n$ $\geq 2$, there exists a non-cyclic group of order $n^{m}$.
I have a result that says that if $G$ is a finite group, and $G = H_{1}\times H_{2}$ then $G$ is cyclic iff both $H_{1}$ and $H_{2}$ are cyclic and $\gcd\left( |H_{1}|,|H_{2}|\right) = 1$.
If I could extend this result for any $G = H_{1} \times H_{2} \times \cdots \times H_{k}$ $\forall k \in \mathbb{N}$, then, could I exhibit $ \displaystyle \prod_{n=1}^{m} Z_{n} $ as such a non-cyclic group of order $n^{m}$? Even though each copy of $Z_{n}$ is cyclic, its direct product won't be, because each copy will have the same order, right?
If this isn't the best way to prove this, what is the best way? Thank you.
Yes, this is correct. In order for a group $G$ to be cyclic, we need an element of order $|G|$.
In the case of $\prod_{k=1}^{m} \mathbb{Z}_{n}$, $m$ copies of the cyclic group $\mathbb{Z}_n$, every element has order that divides $n$. So no element has order $n^m \gt n$ since $n,m \ge 2$.