How many subgroups of $\mathbb{Z}_{p^3}\oplus\mathbb{Z}_{p^2}$ are there with order equal to $p^2$?
My attempt:
Let $H$ be a subgroup of $\mathbb{Z}_{p^3}\oplus\mathbb{Z}_{p^2}$ and $|H|=p^2$, then according to the structure theorem of finite Abelian group, $H\cong\mathbb{Z}_{p^2}$ or $H\cong\mathbb{Z}_p\oplus\mathbb{Z}_p$.
Case One: $H\cong\mathbb{Z}_{p^2}$, let $(a,b)$ be the generator of $H$ and $|(a,b)|=p^2$ (Here $|\cdot|$ represents the order of the element $\cdot$). We know that $$|(a,b)|=\mathrm{lcm}(|a|,|b|).$$
- $|a|=p^2$ and $b$ can be chosen arbitrarily since $|b|\leq p^2$, there are $\varphi(p^2)=p^2-p$ (Here $\varphi$ is the Euler phi-function) elements of order $p^2$ in $\mathbb{Z}_{p^2}$.
- $|a|=p$ and $|b|=p^2$. There are $p-1$ elements of order $p$ in $\mathbb{Z}_{p^3}$ (they are $[p^2],[2p^2],\cdots,[(p-1)p^2]$) and $\varphi(p^2)=p(p-1)$ elements of order $p^2$ in $\mathbb{Z}_{p^2}$, so the number of ways to choose $a,b$ is $p(p-1)^2$.
- $|a|=1$ and $|b|=p^2$. There is only choice for $a$, i.e. $a=[1]$. And there are $\varphi(p^2)=p(p-1)$ ways to choose $b$. So the number of ways to choose $a,b$ is $p(p-1)$.
Therefore, the number of ways to choose $a,b$ is $$(p^2-p)p^2+p(p-1)^2+p(p-1)=p^2(p^2-1).$$ But cyclic group $\mathbb{Z}_{p^2}$ has $\varphi(p^2)=p(p-1)$ generators, so there are $$\frac{p^2(p^2-1)}{p(p-1)}=p^2+p$$ different cyclic subgroups of order $p^2$ in $\mathbb{Z}_{p^3}\oplus\mathbb{Z}_{p^2}$.
My question is whether my attempt is correct? And how to consider the case in which $H\cong\mathbb{Z}_{p}\oplus\mathbb{Z}_p$?