If $G$ is a non-cyclic group of order $n^2$, then $G$ is isomorphic to $\mathbb{Z_n} \oplus \mathbb{Z_n}$ I've independently come up with a question (I know it's been asked before, but I can't find the question online) involving the external direct product, non-cyclic groups and isomorphisms.  So, is the following statement true?  
Claim:  "If $G$ is a non-cyclic group of order $n^2$, then $G$ is isomorphic to $\mathbb{Z_n} \oplus \mathbb{Z_n}$."
Where $\mathbb{Z_n} \oplus \mathbb{Z_n}$ is the external direct product of $\mathbb{Z_n}$ and $\mathbb{Z_n}$.
I've been thinking about this for a few hours, but really can't figure out a good place to start (or a counter-example).  
 A: First, $\mathbb{Z}_n \oplus \mathbb{Z}_n$ is abelian, while there are many non-cyclic groups that are non-abelian (take $S_3$ for example), so the answer to your question as written is immediately no.
However, what if we only consider abelian non-cyclic groups? Then $\mathbb{Z}_2 \oplus \mathbb{Z}_6$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$ are two counterexamples you might consider. [After OP's edit: a counterexample is $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2$, which has order $16$ but is not $\mathbb{Z}_4 \oplus \mathbb{Z}_4$.]
[Note: this is a relevant result that you might already know: if $m$ and $n$ are coprime, then $\mathbb{Z}_m \oplus \mathbb{Z}_n \cong \mathbb{Z}_{mn}$.]
What you might then ask is if every abelian group can be written as the direct product of cyclic groups, and this is true, but not obvious: Classification of finitely generated abelian groups.
A: Asking what every non-cyclic group of a given order looks like is just as hard as asking what every group of a given order looks like. This is a very hard problem. For example see this article.
Now if you wanted to know what every abelian group of a given order looks like that is possible. 
A: The question should be formulated as follows.


*

*Let $p$ be a prime.  Consider an abelian group $A$ of order $p^2.$  Prove that either $A$ is cyclic or $A={\mathbb Z}/p{\mathbb Z}\oplus {\mathbb Z}/p{\mathbb Z}.$

*Let $n$ have three or more divisors (i.e. either there are two different primes dividing $n$, or there is one prime $p$ such that $p^2|n$).  Then there are more than two abelian groups of order $n^2$, up to isomorphism, that are non cyclic.
I would suggest the following to start with: think of Lagrange's theorem (for the first part).
A: The non cyclic-group $G=Z_2\times Z_2$ is of order $4$, but is not isomorphic to $Z_4\oplus Z_4$.
