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Isomorphism between HK and H x K

  1. Mar 8, 2017 #1
    1. The problem statement, all variables and given/known data
    H and K are normal subgroups of G such that the intersection of H and K is the identity. Also, G = HK = {hk | h in H and k in K}. Find an isomorphism between G and H x K

    2. Relevant equations


    3. The attempt at a solution

    I was thinking that an isomorphism could be ##\mu : G \rightarrow H \times K## where ##\mu(g) = \mu(hk) = (h, k)##. I think that this will work, and I just want some verification.
     
  2. jcsd
  3. Mar 9, 2017 #2

    Dick

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    It's easy to show if ##G## is commutative, right? Can you show it without assuming that?
     
    Last edited: Mar 9, 2017
  4. Mar 9, 2017 #3
    Yes. Since both h and k are normal and since both h and k are in G, they commute with each other.
     
  5. Mar 9, 2017 #4

    Dick

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    If you are quoting a theorem that says that, then it's wrong. Normality only tells you e.g. ##hkh^{-1}=k'## where ##k'## is in ##K##. You need to use another premise.
     
  6. Mar 9, 2017 #5
    I am not sure. I thought that normality said that hg = gh, but since all the k are in hk = kh
     
  7. Mar 9, 2017 #6

    Dick

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    No, that's not what normality means. Might be time to look up the definition...
     
  8. Mar 9, 2017 #7
    In my textbook: "A subgroup H of a group G is normal of its left and right cosets coincide, that is, if gH = Hg for all g in G."

    So we know that gH =Hg for all g and gK = Kg for all g.
     
  9. Mar 9, 2017 #8

    fresh_42

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    Yes, but this doesn't mean elementwise, therefore it's only ##gh=h'g## and ##gk=k'g##.
     
  10. Mar 9, 2017 #9
    I am not sure how to proceed, because it seems that the statement of normality involves cosets, but I am trying to find a statement involving just elements, hk = kh
     
  11. Mar 9, 2017 #10
    Could I do soomething like ##hk = h(h^{-1} k h) = kh##?

    edit: nevermind
     
  12. Mar 9, 2017 #11

    fresh_42

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    You don't need to and won't find one. Whether you write it as cosets ##gH=Hg## or elementwise ##gh=hg'## doesn't matter.
    The point is: Why is ##HK## a group? Or: How do you manage to sort elements of the kind ##hkh'k'## such that you can separate them into ##H \times K##, i.e. pairs ##(h'',k'')\,##?
     
  13. Mar 9, 2017 #12
    Well, since e is in both H and K e is in HK.
    We need to show that (hk)(h'k') is also in HK. To do this we would need to have commutativity so that (hk)(h'k') = h(kh')k' = h(h'k)k' = (hh')(kk'). But I haven't shown commutativity yet
     
  14. Mar 9, 2017 #13

    fresh_42

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    You don't need commutativity. Only associativity and ##kh'=h''k##. It is important to end up in the groups ##H## and ##K##, not that the elements are invariant. After that you have to show that ##\mu ## is surjective (obvious) and injective (what do you need here to use?) and a group homomorphism (which is basically the same as to show that ##HK## is a group.)
     
  15. Mar 9, 2017 #14
    Take any two elements of HK, hk and h’k’. Then, we want to show that (hk)(h’k’) is in HK. We note that K is normal, which implies that for kh’, there is some k’’ in K such that kh’=h’k’’ since for all a in G, aK=Ka. Thus,
    (hk)(h’k)’=h(kh’)k’=h(h’k’’)k’=hh’k’’k’. Then since hh’ is in H and k’’k’ is in K, we have that hkh’k’ is in HK.

    So will my definition of ##\mu## suffice if I show that it is injective and a homomorphism?

    Also, don't I need to eventually show that h and k commute? I am still not sure how to show that
     
  16. Mar 9, 2017 #15

    Dick

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    Yes, I think you do need to show ##hk=kh##. I'll give you a big hint. Use that the intersection of ##H## and ##K## is the identity.
     
  17. Mar 9, 2017 #16
    Let ##g = hkh^{-1} k^{-1}##. Since ##hkh^{-1} \in K##, ##g \in K##. Since ##kh^{-1} k^{-1} \in H##, we have that ##g \in H##. So g is in the intersection of H and K. But that is just the identity, so ##hk h^{-1} k^{-1} = e## and ##hk = kh##.

    Now do I use this to show that ##\mu (g) = \mu(hk) = (h, k)## is a homomorphism?
     
  18. Mar 9, 2017 #17

    fresh_42

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    Yes, right.
    It's quite obvious, and if you had written it down, it would have been basically of the same length as your question. Only injectivity needs an argument, not really a proof, but a reason.
     
  19. Mar 9, 2017 #18

    Dick

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    Well, what would showing ##\mu## is a homomorphism require you to prove??
     
  20. Mar 9, 2017 #19
    I know that I would use commutativity to show to show that it's a homomorphism, and injectivity is easy to show.
    I'm wondering though whether mu is well-defined. It seems like a strange definition because the image does not involve g but rather h and k. Is the function well-defined precisely because each g is uniquely represented as a product hk for some h and some k? Also, if I were to use the function, given a g in G how would I know how to decompose it into the product hk in order to use the function?
     
  21. Mar 9, 2017 #20

    fresh_42

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    The general case is ##H/(H\cap K) \cong HK/K## for subgroups ##H,K## of ##G## from which ##K## is normal.
    In the given case this translates to ##H \cong H/\{1\} =H/(H \cap K) \cong HK/K \cong (H \times K)/K \cong G/K \cong H## and is the reason, why ##H \cap K = \{1\}## is important to answer those questions.
    What is ##G## here? The product of subgroups ##HK## or the product ##H\times K##? Which one do you mean?
    In the latter case, it's obvious, and in the former, that's what you have shown: that you can sort the elements. So again, what is ##G## in your question?
     
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