# Normal subgroups, isomorphisms, and cyclic groups

1. Oct 30, 2005

### hgj

1. Let G={invertible upper-triangular 2x2 matrices}
H={invertbile diagonal matrices}
K={upper-triangular matrices with diagonal entries 1}
We are supposed to determine if G is isomorphic to the product of H and K. I have concluded that this is true, but I'm having trouble proving it.
I need to show three things: (1)H,K are normal subgroups of G; (2)the intersection of H and K is the identity; (3)HK=G
I can do (3), I can see that (2) is true (though I haven't written it up formally yet), and I can show that K is a normal subgroup of G.
I'm having trouble showing H is a normal subgroup of G. I tried finding a homomorphism between G and K and showing the kernal of that homomorphism is H (this is how I showed K is normal), but I can't find a function that works. I've also tried showing that there are elments g in G and h in H such that ghg$$^-^1$$ is in H, but to no avail. So I'm stuck.
2. Prove that the product of two infinite cyclic groups is not infinite cyclic.
So far, this is what I have:
Let H and G be two infinite cyclic groups. Let H be generated by h and G by g. Also, the product of G and H, GxH={(g,h) such that g is in G and h in H}. So I need to show there is no element (a,b) in GxH suth that (a,b) generates GxH. That is, show there is no (a,b) in GxH such that (a$$^n$$,b$$^n$$)=(g$$^i$$,h$$^j$$) for any i,j. This is where I'm stuck.

2. Oct 31, 2005

### AKG

No, what you have to do is show that for ALL h in H and for ALL g in G, ghg-1 is in H. This can be shown by a simple computation.

Suppose (a,b) generates GxH. Then for all g in G, h in H, there is an m such that (a,b)m = (g, h). In particular, there must be some m such that this holds when g = an and h = bn+1 for some n, so:

(a,b)m = (an, bn+1)
(am,bm) = (an, bn+1)
(am-n,bm-n-1) = (eG, eH)

So either one of a, b is the identity, or both m-n and m-n-1 is zero. Clearly, the second option is impossible. But if one of a, b is identity, you can easily show that (a,b) won't generate the group.

3. Oct 31, 2005

### Hurkyl

Staff Emeritus
Hrm. Just to make sure I haven't made a silly mistake, H and K are both abelian groups, right?

4. Oct 31, 2005

### AKG

They are abelian subgroups, but this doesn't help in showing that they're normal. Consider An for n > 5. The group is simple, thus has no non-trivial normal subgroups, but any subgroup generated by a single element is of course abelian.

5. Oct 31, 2005

### Hurkyl

Staff Emeritus
It helps for showing something else though!

6. Nov 1, 2005

### AKG

What else needs to be shown?

7. Nov 1, 2005

### Hurkyl

Staff Emeritus
Well, if H and K are both abelian groups, then so is HxK!

8. Nov 1, 2005

### AKG

Why do we need to know that?

9. Nov 1, 2005

### Hurkyl

Staff Emeritus
Well, any group isomorphic to an abelian group must be abelian, right?

10. Nov 1, 2005

### AKG

Yes, I still don't see why it's relevant. There's a theorem which states that if H and K are normal subgroups of G, H and K intersect only at identity, and HK = G, then G is isomorphic to H x K. Why does he need to know anything about them being abelian? My book actually states a slightly different theorem, replacing the condition that H and K be normal with the condition that every element of H commute with every element of K. However it can be proved that if H and K are normal, then every element of H commutes with every element of K. At no point, however, do you need to consider whether or not H or K are abelian.

11. Nov 1, 2005

### Hurkyl

Staff Emeritus
Well, G isn't abelian, and thus cannot be isomorphic to HxK.

12. Nov 2, 2005

### AKG

I see the problem. I erred when I said that:

No, what you have to do is show that for ALL h in H and for ALL g in G, ghg-1 is in H. This can be shown by a simple computation.

H is not even normal. I assumed it was because I thought the original poster had concluded that it was true, and figured that it was. If H were normal, then a simple computation would have sufficed to prove it. In fact, H is not (which is why the theorem doesn't apply), and a simple computation is enough to show this. Of course, showing H is not normal is unnecessary (and insufficient) to show that G is not isomorphic to H x K, what Hurkyl said is perfect.