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 $H$?


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.

  • $\begingroup$ I don't exactly understand the reduction to the case "$a=1$" inside the case $a\neq 0$. For given $(a,b)$ with $a\neq 1$, what is the new basis? $\endgroup$ – Billar Mar 8 '14 at 17:43
  • $\begingroup$ @Billar The new basis is also dependent on $p$, and there isn't an explicit formula for it in terms of $a$, $b$ and $p$, just an 'algorithmic' one. (It amounts, approximately, to computing $a^{-1}b\bmod p$, and there's no explicit formula for that) $\endgroup$ – Steven Stadnicki Mar 8 '14 at 17:45
  • $\begingroup$ @StevenStadnicki: It would still be nice to see. I am willing to accept a description in terms of $k$ and $l$. $\endgroup$ – Billar Mar 8 '14 at 17:47
  • $\begingroup$ @Billar: If $a \neq 0$, then the subgroup generated by $\{(a,b), (p,0), (0,p)\}$ has basis $(1, ka), (0,p)$ $\endgroup$ – spin Mar 8 '14 at 18:02
  • $\begingroup$ The basis I'm using in my answer of the group generated by $\{(a,b),(p,0),(0,p)\}$ where $a\neq 0$ is given by $\{(1,c),(0,p)\}$, where $c=kb-p\lfloor\frac{kb}p\rfloor$. Also note that $k$ may be taken to be $a^{p-2}$ by Fermat's little theorem. $\endgroup$ – Dejan Govc Mar 8 '14 at 18:07

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