How to prove that $\Bbb Z_{p^2}$ is not isomorphic to $\Bbb Z_{p} \oplus\Bbb Z_{p}$ as a $\Bbb Z$-module [duplicate]

Solution 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.

Solution 2:

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