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Normal subgroups, isomorphisms, and cyclic groups

  1. Oct 30, 2005 #1

    hgj

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    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.
     
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  3. Oct 31, 2005 #2

    AKG

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    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.
     
  4. Oct 31, 2005 #3

    Hurkyl

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    Hrm. Just to make sure I haven't made a silly mistake, H and K are both abelian groups, right?
     
  5. Oct 31, 2005 #4

    AKG

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    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.
     
  6. Oct 31, 2005 #5

    Hurkyl

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    It helps for showing something else though!
     
  7. Nov 1, 2005 #6

    AKG

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    What else needs to be shown?
     
  8. Nov 1, 2005 #7

    Hurkyl

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    Well, if H and K are both abelian groups, then so is HxK!
     
  9. Nov 1, 2005 #8

    AKG

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    Why do we need to know that?
     
  10. Nov 1, 2005 #9

    Hurkyl

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    Well, any group isomorphic to an abelian group must be abelian, right?
     
  11. Nov 1, 2005 #10

    AKG

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    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.
     
  12. Nov 1, 2005 #11

    Hurkyl

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    Well, G isn't abelian, and thus cannot be isomorphic to HxK.
     
  13. Nov 2, 2005 #12

    AKG

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    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.
     
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