TITLE: Find all subgroups of $\mathbb{Z}\times\mathbb{Z}$.
QUESTION [16 upvotes]: Find all subgroups of $\mathbb{Z}\times\mathbb{Z}$.

I can find all the infinite subgroups of the form $n\mathbb{Z}\times m\mathbb{Z}$, where $n,m$ run over $\mathbb{Z}$. But I don't know how to write out all the finite subgroups in a uniform expression.
Any suggestions? Thanks in advance.

REPLY [2 votes]: By reason of comments underneath Makoto Koto's answer and spacing, I reworked the answer.
Let ■ $\{(1, 0),(0, 1)\}$ be the canonical basis of $\mathbb{Z}\times\mathbb{Z}$.
■ $p_i\colon \mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Z}$ be the $i$-th projection for $i = 1, 2$. Hence $p_1(x, y) = x, p_2(x, y) = y$
■ $N \neq \emptyset$ be a subgroup of $\mathbb{Z}\times\mathbb{Z}$.
If $p_1(N) = 0$, then $N \subset \mathbb{Z}(0, 1)$.
Hence $N = \mathbb{Z}b(0,1)$ for some integer $b > 0$.
If $p_1(N) \neq 0$, then $p_1(N) = \mathbb{Z}(1,0)$ for some integer $a > 0$.
Hence there exists $y_1 \in N$ such that $p_1(y_1) = a$.
Let $x = x_1(1,0) + x_2(0,1) \in N$.  Then $x_1$ is divisible by $a$.
Hence $x - ny_1 \in \mathbb{Z}(0,1)$ for some integer $n$.
Since $y_1 \in N$ and $n$ is an integer, hence $ny_1 \in N$. Notice it's possible $n$ $\not\in N$.
Since $x - ny_1 \in \color{purple}{N \cap \mathbb{Z}(0,1)}$, $N = \mathbb{Z}y_1 +(\color{purple}{N \cap \mathbb{Z}(0,1)})$.
Since $\mathbb{Z}y_1 \cap (\color{purple}{N \cap \mathbb{Z}(0,1)}) = 0, N = \mathbb{Z}y_1 \oplus (\color{purple}{N \cap \mathbb{Z}(0,1)} \,). 
$
If $N \cap \mathbb{Z}(0,1) = \emptyset$, then $N = \mathbb{Z}y_1$.
If $N \cap \mathbb{Z}e_2 \neq \emptyset$, then $N \cap \mathbb{Z}e_2 = \mathbb{Z}ce_2$ for some integer $c > 0$. Hence $N = \mathbb{Z}y_1 \oplus \mathbb{Z}ce_2$
Therefore the non-zero subgroups of $\mathbb{Z}\times\mathbb{Z}$ are classified as follows.
(1) $N = \mathbb{Z}be_2$ for some integer $b > 0$.
(2) $N = \mathbb{Z}y_1$, where $y_1 = ae_1 + be_2, a > 0$
(3) $N = \mathbb{Z}y_1 \oplus \mathbb{Z}ce_2$, where $y_1 = ae_1 + be_2, a > 0, c > 0$