Problem
For each abelian group of order $p^2q^2$ determine the number of elements of order $pq$ and the number of elements of order $pq^2$ in $G$.
By the structure theorem we have that
$$(1) \space G \cong \mathbb Z_{p^2} \oplus \mathbb Z_{q^2}$$ $$(2) \space G \cong \mathbb Z_{p^2} \oplus \mathbb Z_q \oplus \mathbb Z_q$$ $$(3) \space G \cong \mathbb Z_p \oplus \mathbb Z_p \oplus \mathbb Z_{q^2}$$ $$(4) \space G \cong \mathbb Z_p \oplus \mathbb Z_p \oplus \mathbb Z_q \oplus \mathbb Z_q$$
For (4), it is easy to count the number of elements of order $p$ and $q$ in each component, the possibilities are $(p-1,0,0,q-1)$, $(p-1,0,q-1,0)$,$(0,p-1,q-1,0)$,$(0,p-1,0,q-1)$ or the case $(p-1,p-1,q-1,q-1)$. where in each coordinate I've put the number of elements of order $p$ and $q$ respectively. So, we have a total of $(p-1)(q-1)+(p-1)(q-1)+(p-1)(q-1)+(p-1)(q-1)+(p-1)^2(q-1)^2=4(p-1)(q-1)+(p-1)^2(q-1)^2$ elements of order $pq$.
I am not so sure how to count in the other cases, my very first doubt is how elements of order $p$ (order $q$) are in $\mathbb Z_{p^2}$ ($\mathbb Z_{q^2}$)?
I would appreciate suggestions to count the elements in the three remaining groups. Am I doing well so far?
Note that you can write, in each case, the group as $G = G_p \oplus G_q$ where $G_p$ is a group of order $p^2$ and $G_q$ a group of order $q^2$.
In each case, the number of elements of order $pq$ in $G$ is $n_pn_q$ where $n_p$ is the number of elements of order $p$ in $G_p$ and $n_q$ the number of elements of order $q$ in $G_q$.
Likewise, in each case, the number of elements of order $pq^2$ in $G$ is $n_pn_q'$ where $n_p$ is the number of elements of order $p$ in $G_p$ and $n_q'$ the number of elements of order $q^2$ in $G_q$.
The number of elements of order $q^2$ in $Z_{q^2}$ is $\varphi(q^2) = q^2 -q$, where $\varphi$ is Euler totient. The remaining elements, except for the $0$ element, have order $q$, so there are $q-1$.
For $Z_q \oplus Z_q$ the number of elemnts of order $q$ is $q^2-1$ and/as there are (of course) none of order $q^2$.
The remaining calculation should not pose a problem.