The OP is here, but this post was deleted by the author. So I put it here. If $G=\mathbb{Z}_{p^3}\times \mathbb{Z}_{p^2}$, how many subgroups in $G$ are with the order $p^2$?
Here is my solution:
Since $G$ is Abelian, its subgroups are either cyclic or non-cyclic. For cyclic subgroups case, first we count the number of elements whose order is $p^2$, since each them can generate a cyclic subgroup with order $p^2$. For an arbitrary element $g\in G$, we write it as $g=(x,y), x\in \mathbb{Z}_{p^3}, y\in \mathbb{Z}_{p^2}$. If we solely look at the first component $x$, there are $p^3-p^2$ generators for $\mathbb{Z}_{p^3}=\langle1\rangle$, and each generator has an order of $p^3$, but their order are greater than what we need. We discard them, so the remaining $p^3-(p^3-p^2)=p^2$ elements are with order $1, p,$ or $p^2$.
$\mathbb{Z}_{p^3}$ has a unique subgroup $\langle p\rangle$ with order $p^2$, and $\langle p\rangle \simeq \mathbb{Z}_{p^2}$, hence, there are $p^2-p$ generators for $\langle p\rangle$. Since the first component $x$ has $p^2-p$ choices with order $p^2$, and the order of the second component $y$ is always a factor of $p^2$. Therefore, the second component $y$ has $p^2$ choices. There are $(p^2-p)p^2$ combinations for this Case.$A$.
Similarly, if we solely look at the second component $y$, there are $p^2-p$ generators for $\mathbb{Z}_{p^2}$, and each generator has an order of $p^2$. For the first component $x$, we have shown in the beginning, there are $p^3-(p^3-p^2)=p^2$ elements are with order $1, p,$ or $p^2$. So there are $p^2(p^2-p)$ combinations for this Case.$B$.
BUT, there are repeated countings in Case.$A$ and Case.$B$. Those repeated countings are the elements with order for both the first component $|x|=p^2$ and the second component $|y|=p^2$, which are $|A\cap B|=(p^2-p)(p^2-p)$ cases.
So there are totaly
$$\begin{align}|A\cup B|&=|A|+|B|-|A\cap B|\\ \\ &=(p^2-p)p^2+p^2(p^2-p)-(p^2-p)(p^2-p)\\ \\ &=(p^2-p)(p^2+p)\end{align}$$
distinct elements with order $p^2$. Each element generates a cyclic subgroup with order $p^2$, hence $(p^2-p)(p^2+p)$ cyclic subgroups with order $p^2$. BUT, they are counted repeatedly. Since for each cyclic subgroup with order $p^2$, it contains $p^2-p$ generators inside its own group, so we need to divide this factor.
$$\frac{(p^2-p)(p^2+p)}{p^2-p}=p^2+p$$
There are totally $p^2+p$ distinct cyclic subgroups with order $p^2$.
For non-cyclic subgroup case, there is only one subgroup with order $p^2$, which is $\langle p^2\rangle\times\langle p\rangle$.
In conclusion, there are totally $p^2+p+1$ subgroups with order $p^2$.