The number of subgroups of $Z_{p^2}$ $\oplus$ $Z_{p^2}$ which are isomorphic to $Z_{p^2}$.

Question

How many subgroups of $Z_{p^2}$ $\oplus$ $Z_{p^2}$ are isomorphic to $Z_{p^2}$, where $p$ is a prime?

Let $G$ = $Z_{p^2}$ $\oplus$ $Z_{p^2}$.

I found that there are $p^2−1$ elements of order $p$ in $G$ and the number of subgroups of order $p$ in $G$ is $p+1$, but how do I show which are isomorphic to $Z_{p^2}$?

Answer 1

$Z_{p^2} \oplus Z_{p^2}$ has $p^4-p^2$ elements of order $p^2$, each of which generates a subgroup isomorphic to $Z_{p^2}$, and $Z_{p^2}$ has $p^2 - p$ elements of order $p^2$. Therefore, we can partition the $p^4-p^2$ elements into sets of $p^2-p$ elements, each set giving a distinct subgroup.

The number of sets in this partition, and therefore the number of subgroups isomorphic to $Z_{p^2}$, is $$\frac{p^4-p^2}{p^2-p} = p^2+p$$

Answer 2

$\pi:g\mapsto \langle g\rangle$ defines a map from $G-\text{ker}(\psi)$ onto the family $\mathcal F$ of all cyclic $G$ subgroups of size $p^2$ where $G$ is an abelian group of exponent $p^2$ and $\psi:g\mapsto g^p$. Each $\pi$ fiber has size $\Phi(p^2)=p^2-p$ and therefore $|\mathcal F|=\frac{|G-\text{ker}(\psi)|}{p^2-p}$. When $G=\Bbb Z_{p^2}^+\times\Bbb Z_{p^2}^+$ then $\text{Im}(\psi)=H\times H$ where $H$ is the prime $\Bbb Z_{p^2}$ subgroup and we have $$|\mathcal F|=\frac{|G-\text{ker}(\psi)|}{p^2-p}=\frac{p^4-p^2}{p^2-p}=p^2+p$$.