Subgroups between $p\mathbb{Z}\oplus p\mathbb{Z}$ and $\mathbb{Z}\oplus \mathbb{Z}$ I'm looking for a nice description of all proper subgroups of $G=\mathbb{Z}\oplus \mathbb{Z}$ that contain $K=p\mathbb{Z}\oplus p\mathbb{Z}$ properly ($p$ prime).
I know how to get all such subgroups. I look at the quotient $G/K$. It's a vector space over $\mathbb{F}_p$ of dimension $2$. I take a nonzero vector in $\mathbb{F}_p^2$, pull it back to $G=\mathbb{Z}\oplus\mathbb{Z}$ and see what it spans together with $K=p\mathbb{Z}\oplus p\mathbb{Z}$.
In other words, I choose integers $a,b$ in the range $0,\dotsc,p-1$, not both zero, and see what abelian group $(a,b)$ generates together with $(p,0)$ and $(0,p)$.
(I know that different choices and $a,b$ may give the same subgroup. That's ok).
The part that I don't like about this description is the part "see what subgroup $(a,b)$ generates together with $(0,p)$ and $(p,0)$". This part involves "Gauss elimination" over the integers, and it's not so clear what the basis for the resulting subgroup is.
Here's a concise form of my question:

Let $p$ be prime. Let $a,b$ be integers in the range $0,\dotsc,p-1$, not both zero. Let $H$ be the subgroup of $\mathbb{Z}\oplus \mathbb{Z}$ generated by $\{(a,b),(p,0),(0,p)\}$. Is there a "nice" basis for $H$?

 A: I'm not sure if this is the "nicest" possible description or not, but you might proceed as follows.
Case 1. $a\neq 0$
In this case, $a$ is coprime to $p$, so there exist integers $k,l$ such that $ka+lp = 1$. This means that $k(a,b)+l(p,0)=(1,kb)$. This means that we can assume that $a=1$.
Note that the subgroup generated by $\{(1,b),(p,0),(0,p)\}$ is already generated by $\{(1,b),(0,p)\}$, since $(p,0)=p(1,b)-b(0,p)$. Also note that if $b_1\neq b_2$ (where $b_i\in\{0,1,\ldots,p-1\}$), the two groups generated by $\{(1,b_1),(0,p)\}$ and by $\{(1,b_2),(0,p)\}$ are distinct. (If not, $(0,1)$ would have to be an element of the group, which would imply that $(1,0)$ is also an element.)
Case 2. $a=0$
In this case, $b$ is coprime to $p$, so there are integers $k,l$ such that $kb+lp=1$. This means that $k(a,b)+l(0,p)=(0,1)$. This implies that the group is generated by $\{(0,1),(p,0)\}$, so it is equal to $p\mathbb Z\oplus\mathbb Z$.
In particular, this shows that there are precisely $p+1$ groups of the form you are interested in.
