If $p$ is a prime number, how many elements of order $p$ are there in $\mathbb Z_{p^2}\oplus \mathbb Z_{p^2}$? 
If $p$ is a prime number, how many elements of order $p$ are there in $\mathbb Z_{p^2}\oplus \mathbb Z_{p^2}$? 

From what I know, the group is not cyclic and has order $p^4$.
 A: Suppose $(m, n) \in \mathbb{Z}_{p^2}\oplus\mathbb{Z}_{p^2}$ has order $p$, then $(0, 0) = (m, n)^p = (pm, pn)$ so $pm =0$ and $pn = 0$. Therefore the order of $m$, which I will denote $|m|$, divides $p$, so $|m| = 1$ or $|m| = p$; likewise, $|n| = 1$ or $|n| = p$.
The group $\mathbb{Z}_{p^2}$ has elements $0, 1, \dots, p^2 - 1$ with group operation addition modulo $p^2$. How many of them have order $1$ or $p$? If $m \in \mathbb{Z}_{p^2}$ has order $p$, then $p^2 \mid pm$ so $m$ is a multiple of $p$; there are $p$ such elements: $0, p, 2p, \dots, (p-1)p$.
As there are $p$ possibilities for $m$ and $p$ possibilities for $n$, there are $p^2$ elements $(m, n) \in \mathbb{Z}_{p^2}\oplus\mathbb{Z}_{p^2}$ with $(m, n)^p = (0, 0)$. However, we need to rule out those elements with order less than $p$. If $(m, n)^p = (0, 0)$, then the order of $(m, n)$ divides $p$, but as $p$ is prime, the only other possible order is one, and the only element with order one is the identity of the group: $(0, 0)$. 
Therefore $\mathbb{Z}_{p^2}\oplus\mathbb{Z}_{p^2}$ has $p^2 - 1$ elements of order $p$.
