Amount of elements of order $p^2$

Question

Let $p \neq 5$ be a prime number. How many elements of order $p^2$ are there in $\mathbb{Z_p} \times \mathbb{Z}_{p^5} \times \mathbb{Z}_{25}$?

I have no idea how to even approach this... Any hints?

Answer 1

Hint: The number of elements of order $p^2$ is equal to the number of solution of $p^2x = 0$ for $x \in \mathbb{Z}_p \times \mathbb{Z}_{p^5} \times \mathbb{Z}_{25}$ minus the number of solution of $px = 0$ for $x \in \mathbb{Z}_p \times \mathbb{Z}_{p^5} \times \mathbb{Z}_{25}$.

Answer 2

The order of $(x,y,z)\in\Bbb{Z}_p\times\Bbb{Z}_{p^5}\times\Bbb{Z}_{25}$ is the least common multiple of the orders of $x$, $y$ and $z$. For this to be $p^2$, the orders of $x$, $y$ and $z$ must divide $p^2$, and one of them must equal $p^2$. The order of every $x\in\Bbb{Z}_p$ divides $p^2$, so there are $p$ choices for $x$. The only element $z\in\Bbb{Z}_{25}$ of order dividing $p^2$ is $z=0$, because $p\neq5$, so there is $1$ choice for $z$. For the least common multiple of the orders to be $p^2$, the order of $y\in\Bbb{Z}_{p^5}$ must be $p^2$. How many choices are there for $y$?