Find all subgroups of $\mathbb{Z}\times\mathbb{Z}$.

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.

Let $e_1 = (1, 0), e_2 = (0, 1)$ be the canonical basis of $\mathbb{Z}\times\mathbb{Z}$. Let $p_i\colon \mathbb{Z}\times\mathbb{Z} \rightarrow \mathbb{Z}$ be the $i$-th projection for $i = 1, 2$. Let $N \neq 0$ be a subgroup of $\mathbb{Z}\times\mathbb{Z}$. If $p_1(N) = 0$, then $N \subset \mathbb{Z}e_2$. Hence $N = \mathbb{Z}be_2$ for some integer $b > 0$. If $p_1(N) \neq 0$, then $p_1(N) = \mathbb{Z}a$ for some integer $a > 0$. Hence there exists $y_1 \in N$ such that $p_1(y_1) = a$. Let $x = x_1e_1 + x_2e_2 \in N$. Then $x_1$ is divisible by $a$. Hence $x - ny_1 \in \mathbb{Z}e_2$ for some integer $n$. Since $x - ny_1 \in N \cap \mathbb{Z}e_2$, $N = \mathbb{Z}y_1 \oplus (N \cap \mathbb{Z}e_2)$. If $N \cap \mathbb{Z}e_2 = 0$, then $N = \mathbb{Z}y_1$. If $N \cap \mathbb{Z}e_2 \neq 0$, 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 $y_1 = ae_1 + be_2, a > 0, c > 0$

You should find it quite easy to write out all the finite subgroups of $\mathbb{Z}$. (If you're still struggling, here's a hint: a finite subgroup will have a largest element...) This should give you a clue as to how many finite subgroups there are of $\mathbb{Z}\times\mathbb{Z}$.

I accidentally made an incorrect comment, and the question still hasn't been answered entirely, so it seems fair that I should come back and answer the other half...

The question is: have you found all the infinite subgroups of $\mathbb{Z}\times\mathbb{Z}$? And the answer is no. Where are the rest?

The subgroups $m\mathbb{Z} \times n\mathbb{Z}$ are generated by two elements: (m,0) and (0,n). This generates a nice, rectangular-looking grid (unless either m or n is zero, in which case we get a horizontal/vertical line, or 0 itself). What else is there? Well, clearly there's the line generated by (1,1), or indeed (1,2), (1,3), (5,7), or anything else. There are also all kinds of parallelogram-shaped grids, e.g. the one generated by (3,0) and (1,1). Do we get anything more exotic? These look a little reminiscent of 0-, 1- and 2-dimensional subspaces of a vector space...

In fact, the answer is that we don't get anything more exotic - all of our subgroups look like parallelogram-shaped grids, with two generators, except for the degenerate cases (the lines and 0). Why not? Well, we can't appeal to linear algebra directly (vector spaces and finitely generated abelian groups are similar, but subtly different things), but we can certainly borrow a trick from it.

Let me give you an example, and hopefully you can fill in the details. Suppose your group is generated by 3 (or more) things: $g_1, g_2, g_3$. Then it's obviously also generated by $g_1, g_2, (g_3+g_2)$, or by $g_1, g_2, (g_3-g_1)$, or similar things. In other words, one at a time, we can add/subtract integer multiples of one generator to/from another. So let's look at the group generated by (5,0), (1,1) and (3,-4):

(5,0), (1,1), (3,-4) - these are my current generators. I'm going to add the second to the third, 4 times:

(5,0), (1,1), (7,0) - and these still generate the same group. Now I'll subtract the first from the third:

(5,0), (1,1), (2,0) - and now I'll subtract the third from the first, twice:

(1,0), (1,1), (2,0) - and now I'll subtract the first from the third, twice:

(1,0), (1,1), (0,0) - and now I only really have 2 generators, because (0,0) doesn't generate anything. So it's a 2-dimensional subgroup after all.

Can you fill in the details for the general case?

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 $y_1 = ae_1 + be_2, a > 0, c > 0$