Subgroups of order $p^2$ present in a abelian group

Question

How many subgroups of order $p^2$ does the abelian group $\mathbb{Z_{p^3}} \times \mathbb{Z_{p^2}}$have ?

Answer 1

lemma: if there are $n$ elements of order $d$ in $G$, there are $\dfrac n {\phi(d)}$ cyclic subgroup of order $d$ where $\phi$ is Euler phi function.

By using this lemma and counting the elements of $G$ with order $p^2$, I found $p^2+p$ cyclic subgroup of $G$. The method is simple but calculation is boring.

Here is the calculation.

Let $G=Z_{p^2}\times Z_{p^3}$ and $a$ be an element of $G$ with order $p^2$ then $a=(x,y)$ where $lcd(|x||,|y|=p^2)$ then at least one of the order is $p^2$.

case 1: $|x|=p^2$

$\phi(p^2)=p^2-p$ if $|y|=p$ we have $p-1$ such $y$ as a result we get $(p-1)(p^2-p)$ elements in $G$.

if $|y|=p^2$ then we get $(p^2-p)$ such $y$ as a result $(p^2-p)^2$ elements in $G$.

$if |y|=1$ then we must have $y=e$ so $(p^2-p)$ elements in $G$.

In total, we have $p^2(p^2-p)$ elements in $G$.

case 2: if $|y|=p^2$ then it is enough to calculate when $|x|=p$ and $|x|=e$ .(we already calculated the case $|x|=|y|=p^2$).

Wits same manner we get $p(p^2-p)$ elements in $G$.

From case $1$ and case $2$ we get $(p^2+p)(p^2-p)$ elements of order $p^2$ in $G$.

Thus the number of the cyclic subgroup of $G$ with order $p^2$ is $\dfrac {(p^2+1)(p^2-1)}{\phi(p^2)}=p^2+p$

Edit: Notice that the group $\mathbb Z_p \times Z_p$ has $p^2-1$ elements of order $p$.

$\mathbb Z_{p^2}$ has $p-1$ elements of order $p$ and $Z_{p^3}$ has $p-1$ elements of order $p$.

and $G$ have $(p-1)p$+$1(p-1)=p^2-1$ elements of order $p$.

So $G$ has unique subgroup which is isomorphic to $\mathbb Z_p \times \mathbb Z_p $.

In total, $G$ has $p^2+p+1$ subgroup of order $p^2$.