This question addresses how to find the cyclic subgroups of order $p^r$ in $\mathbb Z _ {p^r} \times \mathbb Z_{p^r}$. I wonder if it is possible to find all subgroups with two generators via the following argument.
Attempt: Any element $z$, which is coprime to $p^r$ will generate all of $\mathbb Z _ {p ^r}$, which will probably lead to a subgroup of order greater than $p^r$. I believe that the order of $p^l$ is just $p^{r - l}$ so that
$$|\langle (p^l,0),(0,p^{r-l})\rangle| = p^r$$
for $0 < l < r$. (We have $p^{r-l}$ choices for the first coordinate and $p^l$ choices for the second.) I believe that this method can find $r-1$ groups with two generators. Originally, I wanted to believe that a property similar to linear independence would hold, where if we had a 3rd generator, we could necessarily express it in terms of the first two. On second thought, such a result would be too good to be true. By Lagrange's theorem, any subgroup must have an order dividing $p^r$, so all subgroups will be of the order $p^l$ for $0 \leq l \leq r$. The identity $$|H_1 \cup H_2| = |H_1| + |H_2| - |H_1 \cap H_2|$$ implies that if a subgroup $H_0$ can be written as the union as the union of two smaller subgroups $H_1$, $H_2$, then $$p^{l_0} = p^{l_1} + p^{l_2} - p^{l_{12}}$$ where $|H_i| = p^{l_i}$, $H_1 \cap H_2$ is a subgroup, and $|H_1 \cap H_2| = p^{l_{12}}$. I'm not entirely sure if this equation has very many solutions, so I conclude that any other subgroup made from more than one generator cannot be the simple union of smaller subgroups and must instead be formed from the interactions (sums and differences in particular) between its generators.
Question: Did this method find all subgroups with two nontrivial generators? Can it generalize to more than two generators?
