$\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$

Question

I try to prove if de groups $\mathbb{Z}_{p^{2}}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$ are isomorphic. I was using the fact that $ \mathbb{Z}_ {mn} $ is isomorphic to $ \mathbb{Z}_m \times \mathbb{Z}_n $ if and only if $ (m, n) = 1 $ taking $ m = n = p $,in this case as $ p $ is a prime greater than 1, then $ (p, p) = p \neq 1 $ and so I don't have the isomorphism, but I don't know if this is true.

Answer 1

I claim that $\mathbb{Z}/p^2$ is not isomorphic to $\mathbb{Z}/p\times \mathbb{Z}/p$. By definition, the former is the cyclic group of order $p^2$, which implies that it has a generating element of order $p^2$. Any element of $\mathbb{Z}/p\times \mathbb{Z}/p$ is of form $(a,b)$, with $a,b\in \mathbb{Z}/p$. But observe that $p(a,b)=(pa, pb)=(0,0)$, so the order of any such $(a,b)$ is at most $p$. Since isomorphisms preserve order, no such isomorphism is possible.

Answer 2

It is. The statement $\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{mn}$ is equivalent to $(m,n)=1$.