Thread: [SOLVED] Subgroups of the Group Z^2

1. [SOLVED] Subgroups of the Group Z^2

I am not really sure if this is the correct section for this question, but the question I am given is this:

Describe all the subgroups of the group Z^2 which consists from all 2 dimensional vectors with integer components.

I am not looking for an answer at all.

I just don't really understand the question, know where to start, or even a method to do this. Any help that you are able to give would be most appreciated.

2. Originally Posted by sto4432
I am not really sure if this is the correct section for this question, but the question I am given is this:

Describe all the subgroups of the group Z^2 which consists from all 2 dimensional vectors with integer components.

I am not looking for an answer at all.

I just don't really understand the question, know where to start, or even a method to do this. Any help that you are able to give would be most appreciated.
Hi sto4432.

Subgroups of $\displaystyle \mathbb Z$ are of the form $\displaystyle n\mathbb Z=\{nk:k\in\mathbb Z\}$ for some non-negative integer $\displaystyle n$. Obviously, if $\displaystyle m\mathbb Z$ and $\displaystyle n\mathbb Z$ are subgroups of $\displaystyle \mathbb Z,$ then $\displaystyle m\mathbb Z\times n\mathbb Z$ is a subgroup of $\displaystyle \mathbb Z\times\mathbb Z.$

Conversely, suppose $\displaystyle \mathcal Z$ is a subgroup of $\displaystyle \mathbb Z\times\mathbb Z.$ $\displaystyle \mathcal Z$ can be $\displaystyle \{(0,0)\},$ $\displaystyle \{0\}\times n\mathbb Z,$ or $\displaystyle m\mathbb Z\times\{0\}.$ Otherwise, there is some $\displaystyle (a,b)\in\mathcal Z$ with $\displaystyle a,b\ne0.$ Then $\displaystyle (-a,-b)\in\mathcal Z.$ Hence there is a positive integer $\displaystyle a$ such that $\displaystyle (a,b)\in\mathcal Z$ for some integer $\displaystyle b,$ and a positive integer $\displaystyle b$ such that $\displaystyle (a,b)\in\mathcal Z$ for some integer $\displaystyle a.$

[Edit: The rest of the proof I originally posted is wrong. It led to the following conclusion (highlighted in red) which, as ThePerfectHacker points out, is clearly incorrect. I am editing to remove my false proof.]

$\displaystyle \color{red}\therefore\ \mathcal Z\ =\ m\mathbb Z\times n\mathbb Z$

The subgroups of $\displaystyle \color{red}\mathbb Z\times\mathbb Z$ are therefore the direct products of subgroups of $\displaystyle \color{red}\mathbb Z.$

3. Originally Posted by TheAbstractionist
$\displaystyle \therefore\ \mathcal Z\ =\ m\mathbb Z\times n\mathbb Z$

The subgroups of $\displaystyle \mathbb Z\times\mathbb Z$ are therefore the direct products of subgroups of $\displaystyle \mathbb Z.$
What about $\displaystyle \left< (2,2) \right>$?

4. Originally Posted by ThePerfectHacker
What about $\displaystyle \left< (2,2) \right>$?
Hi ThePerfectHacker.

Thank you for pointing this out. Looking through the post I made last night, I have just noticed a flaw in my proof. I have edited my post to remove the erroneous proof (I hope this will not affect the flow of the thread).

5. Originally Posted by sto4432
Describe all the subgroups of the group Z^2 which consists from all 2 dimensional vectors with integer components.
Hi sto4432.

Here is my second stab at the problem. Let $\displaystyle \mathcal Z$ be a subgroup of $\displaystyle \mathbb Z\times\mathbb Z.$ If $\displaystyle \mathcal Z$ is not $\displaystyle Z_1\times\{0\}$ or $\displaystyle \{0\}\times Z_2$ for any subgroup $\displaystyle Z_1,Z_2$ of $\displaystyle \mathbb Z,$ then there is a least positive integer $\displaystyle m$ such that $\displaystyle (m,b)\in\mathcal Z$ for any integer $\displaystyle b$ and a least positive integer $\displaystyle n$ such that $\displaystyle (a,n)\in\mathcal Z$ for any integer $\displaystyle a.$

Thus $\displaystyle (a_0,n),(m,b_0)\in\mathcal Z$ for some integers $\displaystyle a_0,b_0.$

If $\displaystyle (a,b)\in\mathcal Z,$ write $\displaystyle b=qn+r$ with $\displaystyle 0\le r<n.$ Then $\displaystyle (a-qa_0,r)=(a,b)-q(a_0,n)\in\mathcal Z$ and so, by minimality of $\displaystyle n,$ we need $\displaystyle r$ to be 0. So $\displaystyle n$ divides $\displaystyle b.$ By a similar argument, we have that $\displaystyle m$ divides $\displaystyle a.$ Thus every element of $\displaystyle \mathcal Z$ is of the form $\displaystyle (pm,qn)$ for some integers $\displaystyle p,q.$

Let $\displaystyle b_0=p_0m$ and $\displaystyle a_0=q_0n$ so $\displaystyle (m,q_0n),(p_0m,n)\in\mathbb Z.$ Hence $\displaystyle ([p_0q_0-1]m,0)=q_0(p_0m,n)-(m,q_0n)\in\mathcal Z.$

Now consider two cases: (1) there is no nonzero integer $\displaystyle a$ such that $\displaystyle (a,0)\in\mathcal Z,$ (2) there is a nonzero integer $\displaystyle a$ such that $\displaystyle (a,0)\in\mathcal Z.$

Case (1):
In this case $\displaystyle p_0q_0-1=0$ and so $\displaystyle p_0,q_0=\pm1.$ It follows that $\displaystyle \mathcal Z$ is cyclic generated by $\displaystyle (m,n)$ or $\displaystyle (m,-n).$

It remains to consider Case (2). For this case, we may assume that there is a nonzero integer $\displaystyle b$ such that $\displaystyle (0,b)\in\mathcal Z.$ Otherwise, an argument similar to the above will lead to the conclusion that $\displaystyle \mathcal Z=\left<(\pm m,n)\right>.$ Indeed, Case (1) together with the possibilities $\displaystyle \mathcal Z=Z_1\times\{0\}$ and $\displaystyle \mathcal Z=\{0\}\times Z_2$ for subgroups $\displaystyle Z_1,Z_2$ of $\displaystyle \mathbb Z$ covers all the cyclic subgroups of $\displaystyle \mathbb Z\times\mathbb Z.$ So the condition that $\displaystyle \mathcal Z$ is not cyclic is equivalent to the condition that there exists a nonzero integer $\displaystyle a$ such that $\displaystyle (a,0)\in\mathcal Z$ and a nonzero integer $\displaystyle b$ such that $\displaystyle (0,b)\in\mathcal Z.$

Let $\displaystyle a_0,b_0$ be the least positive integers such that $\displaystyle (a_0,0),(0,b_0)\in\mathcal Z.$ Then if $\displaystyle \mathcal Y=\{(r,s)\in\mathcal Z:1\le r\le a_0,1\le s\le b_0\},$ $\displaystyle \mathcal Z=\left<a_0\mathbb Z\times b_0\mathbb Z,\,\mathcal Y\right>.$ I think that’s the best way to describe these noncyclic subgroups of $\displaystyle \mathbb Z\times\mathbb Z.$

6. Thanks very much for your help and quick responses. So all of the subgroups will be of the form above then?

7. Originally Posted by sto4432
Thanks very much for your help and quick responses. So all of the subgroups will be of the form above then?
Hi sto4432.

The problem was trickier than I thought at first. I initially jumped to the happy conclusion that all subgroups were of the form of the direct product $\displaystyle m\mathbb Z\times n\mathbb Z$ but ThePerfectHacker was quick to point out that I missed some cyclic subgroups. In the end, I found that each subgroup is either cyclic, a direct product, or a combination of both. If I have missed any more, I’m sure ThePerfectHacker or someone else will point it out to me again.

As an example of a subgroup that is a combination of a direct product and a cyclic subgroup, try $\displaystyle a_0=2,$ $\displaystyle b_0=4$ and $\displaystyle \mathcal Y=\{(1,2)\}.$ Then the subgroup $\displaystyle \mathcal Z$ is $\displaystyle \{(2m+k,4n+2k):m,n\in\mathbb Z,\,k\in\{0,1\}\}=2\mathbb Z\times4\mathbb Z\,\cup\,\{(2m+1,4n+2):m,n\in\mathbb Z)\}.$ You can verify directly that this is a subgroup (note for example that the inverse of $\displaystyle (2m+1,4n+2)$ is $\displaystyle (2[-m-1]+1,4[-n-1]+2))$ – and it’s a subgroup that is neither cyclic nor a direct product of subgroups of $\displaystyle \mathbb Z,$ but a combination of both.