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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.

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    Technically you have found all subgroups *up to isomorphism*, but I wonder if you have found all the subgroups themselves. For example, $$(1,1)\Bbb Z=\{(n,n):n\in\Bbb Z\}\le\Bbb Z\times\Bbb Z$$ cannot be written as the internal direct sum $$n(\Bbb Z\times 0)\oplus m(0\times\Bbb Z)=\{(nx,my):x,y\in \Bbb Z\}$$ for any $n,m\in\Bbb Z$, although is is isomomorphic to $n(\Bbb Z\times0)\oplus0(0\times\Bbb Z)$. (Indeed the subgroups you identify are all isomorphic for nonzero $n,m$ but are not the same subgroups.)2012-09-11
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    @anon What do you mean by "isomorphism" in this context? What are its origin and target?2012-09-11
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    I mean, for example, $\{(2x,0):x\in\Bbb Z\}$ is [isomorphic](http://en.wikipedia.org/wiki/Isomorphism) to $\{(x,x):x\in\Bbb Z\}$, via the map $(2a,0)\mapsto(a,a)$, though these are not the same subgroup of $\Bbb Z\times\Bbb Z$.2012-09-11
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    @anon But these two subgroups seem both in $n\mathbb{Z}\times m\mathbb{Z}$... Confused...2012-09-11
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    The second one is *inside* of $n\Bbb Z\times m\Bbb Z$ if and only if $n=m=1$, but it is not *equal* to any subgroup of $\Bbb Z\times\Bbb Z$ of the form you have (meaning of the form $\{(mx,ny):x,y\in\Bbb Z\}$).2012-09-11
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    @anon But I didn't say $n$ can't equal to $m$, and both of them can't equal to $1$.2012-09-11
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    So what's your point? Let me reiterate, so as not to derail with these sidesteps: not a single one of the subgroups you've pointed out is equal to (for example) the subgroup $(1,1)\Bbb Z=\{(n,n):n\in\Bbb Z\}$ $\le\Bbb Z\times\Bbb Z$, so you have not found all of the subgroups of $\Bbb Z\times\Bbb Z$. (The subgroups you have pointed out are of the form $\{(nx,my):x,y\in\Bbb Z\}$; there is no choice of $n,m$ that makes this equal to $(1,1)\Bbb Z$.)2012-09-11
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    @anon Yeah, you are right. I'm sorry that I confused $(1,1)\mathbb{Z}$ with $\mathbb{Z}\times\mathbb{Z}$. Then does there exist a way to find _all_ subgroups?2012-09-11
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    Every subgroup should be of the form $(m,n)\Bbb Z$ or $m\Bbb Z\times n\Bbb Z$. A proof escapes me at the moment, but it might involve basic integral linear algebra.2012-09-11
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    @anon Could you tell me what is "_integral_ linear algebra"?2012-09-11
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    I mean matrix stuff over $\Bbb Z$.2012-09-11
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    I'm afraid it gets worse than that, @anon: the subgroup generated by (1,1) and (1,-1), for instance.2012-09-11
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    @Billy Very good point. Thanks.2012-09-11

5 Answers 5

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

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    This is a clean proof.2012-09-11
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    Can't this be simplified to $ N = \mathbb{Z}y_1 + \mathbb{Z}y_2$ with $y_1, y_2 \in \mathbb{Z}\times\mathbb{Z}$?2012-09-11
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    @lhf The bases are explicitly given.2012-09-11
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    I wonder what is the "i-th projection". What are $p_1$ and $p_2$? Thanks.2012-09-11
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    @JackWitt $p_1(x, y) = x, p_2(x, y) = y$2012-09-11
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    I don't understand why this statement is true--"Since $x - ny_1 \in N \cap \mathbb{Z}e_2$, $N = \mathbb{Z}y_1 \oplus (N \cap \mathbb{Z}e_2)$."2012-09-11
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    @JackWitt Since $x - ny_1 \in N \cap \mathbb{Z}e_2$, $N = \mathbb{Z}y_1 + (N \cap \mathbb{Z}e_2)$. Since $\mathbb{Z}y_1 \cap (N \cap \mathbb{Z}e_2) = 0$, $N = \mathbb{Z}y_1 \oplus (N \cap \mathbb{Z}e_2)$.2012-09-11
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    Excuse me, I wonder why $x-ny_1$ is in $N$, or why $ny_1$ is in $N$. Thanks.2012-09-17
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    @JackWitt I guess you mean $y_1$ instead of $y$. In that case, because $x$ and $y_1$ belong to $N$ and $N$ is a subgroup.2012-09-17
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    But why is $ny_1$ in $N$? $n$ is not necessarily in $N$...2012-09-17
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    @JackWitt $n$ is an integer.2012-09-17
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    @JackWitt Since $y_1 \in N$ and $n$ is an integer, $ny_1 \in N$.2012-09-18
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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 $\mathbb{Z}\times\mathbb{Z}$.

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    There are infinitely many finite subgroups, I know. But can I write them in one expression?2012-09-11
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    Are there? (Hint: no.)2012-09-11
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    Perhaps that was a rather mean hint. So let me be more explicit. Z has no finite subgroups except {0}, because if any subgroup contains the element n (n =/= 0), then it also contains n+n, n+n+n, n+n+n+n, ..., all of which are non-zero in Z. Perhaps you're confusing subgroups with quotient groups?2012-09-11
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    Like ${(-n,-n),(0,0),(n,n)}$, where $n$ runs over positive integers. Doesn't it generate infinite many?2012-09-11
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    That set doesn't form a group - if it contains (n,n), then it should contain (n,n) + (n,n) too, for example. In general, if g and h are elements of a group G, then g*h must be too - otherwise it's not a group.2012-09-11
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    Yes, I'm silly... So for $\mathbb{Z}\times\mathbb{Z}$, the only finite subgroup should be $(0,0)$? Do I write the infinite subgroups correctly?2012-09-11
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    Yes, all correct.2012-09-11
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    Oops - I'm afraid I misread your original post. You haven't found all subgroups of Z*Z, but someone else is helping you by now. Sorry! :)2012-09-11
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    Not that I'm complaining or envying. I have no idea why this answer got 6 up votes.2012-09-11
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    @MakotoKato: nor have I. Especially as it turned out to be unhelpful for the main part of the question (which was a hasty accident on my part). But I have now corrected this in a later answer, so perhaps the upvotes are simply all coming to this post rather than being split between the two.2012-09-11
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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 $y_1 = ae_1 + be_2, a > 0, c > 0$

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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?

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    You know you could edit your original answer with this information?2012-09-11
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    I know that. I don't know for certain that it notifies Jack that I've edited my answer (i.e. that he'll ever see it)...2012-09-11
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    Thanks, @Billy. Is $\mathbb{Z}\times\mathbb{Z}$ finitely generated? I guess I know how things go here by your well-explained example, but I still can't do the general case...2012-09-11
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    Yes - it's generated by (1,0) and (0,1), for instance. (You *can* pick an infinite set of generators, but the point is that all but two of them are redundant.) Suppose I give you three elements (a,b), (c,d) and (e,f) which are supposed to generate a subgroup of $\mathbb{Z}\times\mathbb{Z}$. Can you show that one of them is redundant, by using the same adding/subtracting trick? (Hint: suppose a and c are non-zero. Prove that you can kick it into the form (0,b'), (c',d'), (e',f'). What happens if b' = 0? If b' =/= 0 and d' (or f') =/= 0? If b' =/= 0 and d' = f' = 0? Try some examples.)2012-09-11
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If you can use linear algebra, then consider $V$ the subspace of $\mathbb R^2$ generated by a subgroup $H$ of $\mathbb Z \times \mathbb Z $.

  • If $\dim V=0$, then $H=0$.

  • If $\dim V=1$, take $u\in H$ with smallest positive length. Then $H=\mathbb Z u$.

  • If $\dim V=2$, take $u\in H$ with smallest positive length and take $v \in H\setminus\mathbb Z u$ with smallest positive length. Then $H=\mathbb Z u + \mathbb Z v$.

Thus, all subgroups of $\mathbb Z \times \mathbb Z $ are of the form $\mathbb Z u + \mathbb Z v$, where one or both $u$ and $v$ may be zero.

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    In your second step, I wonder what "length" means here.2012-09-11
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    "length" means Euclidean length of a vector.2012-09-11
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    The only finite subgroup is $\{(0,0)\}$.2012-09-16