Find the number of subgroups of $\mathbb Z_{p^3} \times \mathbb Z_{p^2}$ [duplicate]

The given group is $\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$. We will use the formula $$ |(m,n)|=\mathrm{lcm}(|m|,|n|),$$ for any $ m \in \mathbb{Z}_{p^3}$ and $n \in \mathbb{Z}_{p^2}$, where $|.|$ denote the order.

Since $m,n$ are the elements of $\mathbb{Z}_{p^3}$ and $\mathbb{Z}_{p^2}$ respectively, we have $ |m| $ divides $ p^3$ and $|n|$ divides $p^2$.

(i) Show that there is only one subgroup of order of $ \mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$ isomorphic to $ \mathbb{Z}_p \oplus \mathbb{Z}_p$.

In this step we calculate the number of elements $(m,n)$ of order $p$. You easily check that there are $p^2-1$ elements having order $p^2$. All these $p^2-1$ elements along with the identity element $(0,0)$ form a unique subgroup of order $p^2$ that is isomorphic to $\mathbb{Z}_p \oplus \mathbb{Z}_p$.

(ii) Show that there are $p^2+p$ subgroups of $ \mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$ isomorphic to $ \mathbb{Z}_{p^2}$.

In this step we will count the number of cyclic subgroups of order $p^2$. In other words we have to find all elements $(m,n)$ of the group $ \mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$ having order $p^2$. To do this we need to find

$(a)$ all elements $(m,n) \in \mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$, such that $|m|=p^2 \in \mathbb{Z}_{p^3}$,

$(b)$ all elements $(m,n) \in \mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$, such that $|n|=p^2 \in \mathbb{Z}_{p^2}$

From part $(a)$ and part $(b)$ try to show that there are total $p^4-p^2$ elements of order $p^2$ and each cyclic group order $p^2$ has $p^2-p$ elements of order $p^2$.

Hence the number of cyclic subgroup of order $p^2$ is equal to $\frac{p^4-p^2}{p^2-p} =p^2+p$.

Thus there is a total of $p^2+p+1$ subgroups of the group $ \mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$ of order $p^2$.

To answer the complete question it is helpful to know that, in a finite abelian group, the number of subgroups of order $n$ is equal to the number of index $n$ for any $n$. This is basically because finite abelian groups $G$ are isomorphic to their dual groups ${\rm Hom}(G,{\mathbb C}^\times)$, and subgroups of the group correspond to quotients of the dual.

So the final answer is $2(1 + (p+1) + (p^2+p+1)) = 2p^2+4p+6$.