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Math Help - Problem with proof

  1. #1
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    Problem with proof

    This proof is from Dummit & Foote and parts from it are spread throughout the book. I'm having problems proving thing or two the way the authors meant it to be proved.

    Let H and K be subgroups (any subgroups) of group G and HK=\{hk | h \in H, k \in K\}
    Prove that:

    a/ The number of elements of HK is |HK|=|H||K| / |H \cap K|
    b/ Each element of HK can be written in |H \cap K| different ways as product of h and k
    c/ In particular if |H \cap K|=1 then each element of HK can be written in unique way as product of h and k


    a/ Goes by the book. HK is really union of all cosets of K in G, which are of type hK for some h \in H. Distinct cosets are always disjoint and have equal number of elements equal to the elements of the subgroup - in this case K. So we seek how many such distinct cosets are there and multiply their number by |K|.
    So when in principle are two cosets h_1K and h_2K equal in G? If h_1K=h_2K, then h_{2}^{-1}h_{1}\in  K. But we are interested only in the case when our h_1 and h_2 are elements from H. This gives another property - for h_{1} and h_{2} in H, h_{2}^{-1}h_{1}\in H too. Therefore h_{2}^{-1}h_{1}\in  (H \cap K) and therefore h_1(H \cap K)=h_2(H \cap K) (only applies when our h_1 and h_2 are from H)
    So the number of distinct cosets of K in HK is equal to the number of distinct cosets of (H \cap K) in H. The number of cosets of H \cap K in H is |H|/|(H \cap K)| by Lgarange's theorem and this gives the end result of |K||H|/|(H \cap K)| elements of HK.
    c/ It's easy to derive H by simple combinatorial argument. All different sequences hk for which the first element is from H and the second from K are |H||K|. All distinct elements of HK are |H||K|/|(H \cap K)| by a). If |(H \cap K)| is bigger than 1 it means that the number of distinct elements is strictly smaller than the number of sequences hk which means that some of them denote the same element.
    b/ But how to prove that each element of HK can be written by exactly |(H \cap K)| different ways? In other words if |(H \cap K)| is 2 how to prove that each element has exactly two sequences of type hk and there are no elements that have three such sequences to represent them and there are no elements that have one such sequence to balance the number.
    D&F left this as an excercise mentioning that it follows easily from a) and it implies c). I cannot do it.
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    Re: Problem with proof

    Come to think of it, my proof of c) needs more elaboration. If |(H \cap K)| is bigger than 1 it means that it's impossible each element of HK to have only one representation of type hk. But it does not prove that it's the case when |(H \cap K)|=1.
    It's intuitive, but I doubt how well I can formalize it. By definition of HK, each sequence hk represents element of HK so there is "map" between the set of representations and the set of distinct elements. This map is surjective by def. of HK and the two sets have equal cardinality, hence the map is bijective. Injectivity follows...
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    Re: Problem with proof

    I did some nummeric examples with Z/12Z and subgroups. Altough the subgroups are normal, it seems that the trick is in the cosets that are equal/overlap. Each coset, represented as h_1K gives one way of representing each element this coset as product of type hk. If the same coset is represented as h_2K this gives another way to represent the same elements as products of type hk.
    Only need to formalize this. Like to prove that the number of disjoint cosets in HK divides the number of of all elements in H, which is equal to all coset representations as hK.
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    Re: Problem with proof

    I'm so stupid.
    The proof of a) basically says everything needed. All elements of one distinct coset of H \cap K give rise to different "representations" hK of one and same distinct coset in HK. So there are as many different representations of one coset in HK as is the order of H \cap K.
    Only need to state this more formally.
    Last edited by mrproper; August 24th 2014 at 08:59 AM.
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  5. #5
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    Re: Problem with proof

    If $h_1k_1 = h_2k_2$, then:

    $h_2^{-1}h_1 = k_2k_1^{-1}$, and this element is in $H \cap K$.

    It then follows (as you pointed out) that $h_1(H \cap K) = h_2(H \cap K)$. If $HK$ is a group, you have shown:

    $[HK:K] = [H:H\cap K]$

    In fact, in the special case where $K$ is normal in $G$ (and thus normal in $HK$), we can say more: $H \cap K$ is normal in $H$, and the two factor groups are isomorphic.

    The isomorphism is, as you might suspect: $(hk)K \mapsto h(H\cap K)$ (what you have proved is that this is a bijection, even when $K$ is not normal, as $(hk)K = h(kK) = hK$, since $k \in K$).

    Note that if $|H\cap K| = 1$, the only element of this subgroup is $e$, in which case:

    $h_2^{-1}h_1 = e \implies h_1 = h_2$ and similarly for the $k$'s.
    Last edited by topsquark; August 26th 2014 at 12:42 PM.
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  6. #6
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    Re: Problem with proof

    Here's an addendum to Deveno's post:

    Thanks from mrproper
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    Re: Problem with proof

    JohnG, putting things like relation and proving it's a function is a beautiful way to write things. All the time I spent on this problem I felt like I lack of "notation" to describe the bijection.

    Let me try to fill the proof:

    Let S=\{hK: h \in H\} be the set of left cosets of K in G and T=\{h(H\cap K) : h \in H\} be the set of left cosets of (H\cap K) in H. Define relation R as {(hK, h(H\cap K)): h\in H)}. R has domain S and codomain T.
    1/ R is a funnction
    We have to prove that h_1K=h_2K implies h_1(H \cap K) = h_2(H \cap K). We did just that in the previous posts.

    2/ R is surjective

    By design of relation the range of R is T

    3/ R is injective. Assume h_1(H \cap K) = h_2(H\cap K). h_1(H \cap K) = h_2(H\cap K) \iff h_2^{-1}h_1 \in K \iff h_1K=h_2K .

    How did I do ?
    Last edited by mrproper; August 26th 2014 at 06:05 AM.
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  8. #8
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    Re: Problem with proof

    Exactly correct. Good job.
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    Re: Problem with proof

    Correct problem and proof shared here. Thanks!


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    http//www.acadsoc.com
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