Group $G$ of order $p^2$: $\;G\cong \mathbb Z_{p^2}$ or $G\cong \mathbb Z_p \times \mathbb Z_p$ [closed]

Hint:

Argue that $G$ must be abelian (why?)

Then use the Fundamental Theorem of Finitely Generated Abelian Groups to prove that any abelian groups of order $p^2$ must necessarily be isomorphic to one of the two groups $\mathbb Z_{p^2}$ or $\mathbb Z_p\times \mathbb Z_p$, which are non-isomorphic groups, since $\mathbb Z_m\times \mathbb Z_n \cong \mathbb Z_{mn} \iff \gcd(m,n) = 1$, and clearly, $\gcd(p, p) = p \neq 1$.

Hints:

What are the possible orders of elements in $\,G\,$ ? And if there is no element of order $\,p^2\,$ in $\,G\,$ , then can you find two elements

$$1\neq x,y\in G\;\;s.t.\;\;\langle x\rangle\cap\langle y\rangle = 1\ldots?$$

You can get by without the fundamental theorem of finitely Abelian groups in this case, once you know the group is Abelian. If there is only one subgroup of order $p,$ then consider the order of an element outside that subgroup. If there are two different subgroups $A$ and $B$ of order $p$. What is $A \cap B?$ What is $|AB|?$

It wasn't clear from what you wrote whether you already have proved for yourself that $G$ ( the whole group) a Abelian, or you know it for other reasons. Depending on what you have done so far on your course, this may or may not be straightforward. Have you proved that if $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is Abelian? Have you proved that a finite $p$-group has a non-trivial center? You didn't actually say that $p$ was a prime by the way, but I suppose that is clear from the context.