# Normal subgroups of Z X Z

1. Apr 6, 2013

### dumbQuestion

EDIT:( Sorry I meant for the title of this to say conjugacy classes of subgroups of Z X Z) I have a question.

I am trying to figure out, what are all of the normal subgroups of Z X Z? Well, I know that Z has the following subgroups: the trivial group {0}, itself Z, and then aZ, one for each a in Z. (the group of all multiples of an integer a).

Well, this makes me assume that Z X Z has the following subgroups:

0, Z X Z, Z X aZ (one for each positive integer a), bZ X Z (one for each positive integer b) and cZ X dZ (one for each set of positive integers c,d)

First off, I was wondering is this logic correct? Is there a relevant theorem which says that if you have a cross of two groups, that the subgroups of the new cross product is the cross of its subgroups?

Also, am I right in assuming that all of the groups of the form Z X aZ, cZ X dZ and bZ X Z are in the same conjugacy class?

2. Apr 6, 2013

### Bacle2

If you're using the direct product, then ,yes, if A is a subgroup of G and B is a subgroup of G' , then the
product AB is a subgroup of GG'. Same is true for normal subgroups,and their product is normal on the direct product GG'. Also, AFAIK, for n,m positive,nZ iso.mZ iff n=m.

Last edited: Apr 6, 2013
3. Apr 6, 2013

### jbunniii

I assume your group is $\mathbb{Z} \times \mathbb{Z}$, the direct product of the additive group of integers with itself, with the usual operation of componentwise addition: $(a,b) + (c,d) = (a+b,c+d)$.

If that is the case, your question doesn't make much sense to me. Of course $\mathbb{Z} \times \mathbb{Z}$ is abelian, so all of its subgroups are normal, and the conjugacy classes are all trivial.

Also, there are subgroups of $\mathbb{Z} \times \mathbb{Z}$ which are not of the form $H \times K$ where $H,K$ are subgroups of $\mathbb{Z}$. For example, the cyclic subgroup generated by the element $(1,1)$ is not of that form.

4. Apr 6, 2013

### dumbQuestion

Hey guys, thanks for the responses. The reason I am confused is because, the torus T^2 has fundamental group isomorphic to Z X Z. Now, it's a result that if a space X' covers the torus, then X' is homeomorphic to either R^2, S1 X R, or T^2. But I know that the isomorphisms of covering spaces of a space X should be in one to one correspondence with the conjugacy classes of subgroups of X. Now I know that R^2 is the universal cover of the torus, and so it corresponds with the trivial class. T^2 obviously corresponds with the conjugacy class of T^2 itself. And so there is this third class, which has a correspondence with S1 X R.

5. Apr 7, 2013

### Bacle2

But the product of covering spaces is a covering space for the product; since S^1 and R both cover S^1, any of the products S^1 x S^1 ~ T^2 ; S^1 x R and Rx R all cover S^1 . Is that what you meant?

6. Apr 7, 2013

### dumbQuestion

oooooh yeah. I'd forgotten about the product of covering spaces being a covering space for the product. Well, at least it makes sense how to understand think of it in one direction. But I'm still confused about something: am I wrong in thinking that the covering spaces of the torus can be broken in to three distinct groups: ones that are homeomorphic to S1 X S1 (which is the torus), ones homeomorphic to S1 X R, and ones homeomorphic to R X R? The ladder three categories should be distinct because S1 X S1 has fund. group ZXZ, S1 X R is homotopic to S1 (because R is contractible), while R X R has trivial fundamental group - so neither three of these representative are homeomorphic as none are homotopic. So we have 3 distinct "isomorphism groups" of covering spaces of the torus. Right? And I thought the correspondance between the isomorphism groups of covering spaces X' of X and the conjugacy classes of subgroups of pi_1(X) was a one-to-one correspondance?

7. Apr 7, 2013

### Bacle2

Right, for each n , there is a cover associated with nZ ; the map z^n : S^1-->S^1 represents the cover associated with nZ . Notice the induced mapgives you multiplication by n . Maybe playing around with these will give you all covers over T^2.