Show that $Z_{p^2} \oplus Z_{p^2}$ has exactly one subgroup isomorphic to $Z_p \oplus Z_p$ Show that $Z_{p^2} \oplus Z_{p^2}$ has exactly one subgroup isomorphic to $Z_p \oplus Z_p$
Attempt: $Z_p \oplus Z_p$ has $p^2-1$ elements of order $p$ . Hence, all non trivial elements of $Z_p \oplus Z_p$ are of order $p$. Number of cyclic sub groups of order $p$ = $p+1$
$Z_{p^2} \oplus Z_{p^2}$ has $\{p (p^3-3p+1)\}$ elements of order $p^2$ and $(p^2-1)$ elements of order $p$ .Number of cyclic sub groups of order $p$ = $p+1$
Now , example of a generator of order $p$ in $Z_{p^2} \oplus Z_{p^2}$ is $(1,p)$ and example of a generator of order $p$ in $Z_p \oplus Z_p$ is $(1,0)$
How do i proceed next in these question where isomorphism is sought to be displayed? Which means a mapping must be specified as well. Help will be appreciated.

Unless I am mistaken, the idea of the question is as follows: You know that $G:= \mathbb{Z}_{p^2}\oplus \mathbb{Z}_{p^2}$ has one subgroup $H$ isomorphic to $\mathbb{Z}_p\oplus \mathbb{Z}_p$. The question is asking you to prove that this $H$ is the unique such subgroup.
So suppose $K$ is any other subgroup isomorphic to $\mathbb{Z}_p\oplus\mathbb{Z}_p$, then $K$ has $p^2-1$ elements of order $p$. But $G$ has only $p^2-1$ elements of order $p$, all of which are in $H$ - hence $K\subset H$ and vice-versa.
This is all you need to show.
