normal subgroups, isomorphisms, and cyclic groups

hgj

I'm really stuck on these two questions, please help!
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[tex]^-^1[/tex] 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[tex]^n[/tex],b[tex]^n[/tex])=(g[tex]^i[/tex],h[tex]^j[/tex]) for any i,j. This is where I'm stuck.

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)
(a^m,b^m) = (aⁿ, bⁿ⁺¹)
(a^m-n,b^m-n-1) = (e_G, e_H)

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.

Hurkyl

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

AKG

They are abelian subgroups, but this doesn't help in showing that they're normal. Consider A_n 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.

Hurkyl

It helps for showing something else though!

AKG

What else needs to be shown?

Hurkyl

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

AKG

Why do we need to know that?

Hurkyl

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

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.

Hurkyl

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

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.