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Math Help - Group Action, Double costs

  1. #1
    MHF Contributor Amer's Avatar
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    Group Action, Double costs

    Let H,K subgroups of G for each x in G define the double cost HK of x

    HxK=\{hxk\mid h\in H ,k\in K\}

    a) Prove that HxK is the union of the left costs x_1K,x_2K,...,x_nK where \{x_1K,x_2K,...,x_nK\} is the orbit containing xK of H acting by left Multiplication on the set of left costs of K

    My work
    I want to prove that HxK = \bigcup_i^n x_iK it is clear that
     1xK = xK so   \bigcup_i^n x_iK \subseteq HxK

    let hxK \in HxK hx in G so hxK in the union

    d) Prove that  \mid HxK \mid = \mid H \mid .\mid H : H\bigcap xKx^{-1}\mid

    my question is about d) I put a) because I think it will help in solving d)
    I was thinking about the order of orbit equal to the index group of the stabilizer

     \mid HxK\mid = \mid \bigcup x_iK\mid = \mid O \mid = \mid H: Stab_H(O)\mid

     Stab_H(O)= \{ h \in H \mid hxK = xK \}

    Stab_H(O) = \{h \in H \mid x^{-1}hx \in K \Rightarrow h\in xKx^{-1}\}

    Stab_H(O) = H \bigcap xKx^{-1}
    so we will have

    \mid HxK\mid = \mid H: H \bigcap xKx^{-1}\mid which is not like what i want to prove

    THanks
    Last edited by Amer; November 19th 2011 at 09:19 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Group Action, Double costs

    Quote Originally Posted by Amer View Post
    Let H,K subgroups of G for each x in G define the double cost HK of x

    HxK=\{hxk\mid h\in H ,k\in K\}

    a) Prove that HxK is the union of the left costs x_1K,x_2K,...,x_nK where \{x_1K,x_2K,...,x_nK\} is the orbit containing xK of H acting by left Multiplication on the set of left costs of K

    My work
    I want to prove that HxK = \bigcup_i^n x_iK it is clear that
     1xK = xK so   \bigcup_i^n x_iK \subseteq HxK

    let hxK \in HxK hx in G so hxK in the union

    d) Prove that  \mid HxK \mid = \mid H \mid .\mid H : H\bigcap xKx^{-1}\mid

    my question is about d) I put a) because I think it will help in solving d)
    I was thinking about the order of orbit equal to the index group of the stabilizer

     \mid HxK\mid = \mid \bigcup x_iK\mid = \mid O \mid = \mid H: Stab_H(O)\mid

     Stab_H(O)= \{ h \in H \mid hxK = xK \}

    Stab_H(O) = \{h \in H \mid x^{-1}hx \in K \Rightarrow h\in xKx^{-1}\}

    Stab_H(O) = H \bigcap xKx^{-1}
    so we will have

    \mid HxK\mid = \mid H: H \bigcap xKx^{-1}\mid which is not like what i want to prove

    THanks
    I think you are going about the last part wrong. You know that HxK should be equal to \displaystyle \bigsqcup_{gK\in\mathcal{O}_{xK}}|gK| where \mathcal{O}_{xK} is orbit of xK under the left H-action on G/K. But, each |gK| is equal to |K| and so |HxK|=|K||\mathcal{O}_{xK}|. But, using the orbit stabilizer theorem, as you indicated, you can show that |\mathcal{O}_{xK}|=[H:H\cap xKx^{-1}] and so |HxK|=|K|[H:H\cap xKx^{-1}].

    For more information you can see my blog post here.


    EDIT: I noticed that I did the "opposite side" than you wanted, but by symmetry they're the same.
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  3. #3
    MHF Contributor Amer's Avatar
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    Re: Group Action, Double costs

    Quote Originally Posted by Drexel28 View Post
    I think you are going about the last part wrong. You know that HxK should be equal to \displaystyle \bigsqcup_{gK\in\mathcal{O}_{xK}}|gK| where \mathcal{O}_{xK} is orbit of xK under the left H-action on G/K. But, each |gK| is equal to |K| and so |HxK|=|K||\mathcal{O}_{xK}|. But, using the orbit stabilizer theorem, as you indicated, you can show that |\mathcal{O}_{xK}|=[H:H\cap xKx^{-1}] and so |HxK|=|K|[H:H\cap xKx^{-1}].

    For more information you can see my blog post here.


    EDIT: I noticed that I did the "opposite side" than you wanted, but by symmetry they're the same.
    Thanks very much for the direction I did not noticed that but there still a small problem the question show that

    \mid HxK \mid = \mid H \mid . \mid H : H \bigcap xKx^{-1} \mid
    not
    \mid HxK \mid = \mid K \mid . \mid H : H \bigcap xKx^{-1} \mid

    you can check the book David Dammit same source which you gave in your Blog
    chapter 4
    is it a typo ?

    nice blog I like it
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Group Action, Double costs

    Quote Originally Posted by Amer View Post
    Thanks very much for the direction I did not noticed that but there still a small problem the question show that

    \mid HxK \mid = \mid H \mid . \mid H : H \bigcap xKx^{-1} \mid
    not
    \mid HxK \mid = \mid K \mid . \mid H : H \bigcap xKx^{-1} \mid

    you can check the book David Dammit same source which you gave in your Blog
    chapter 4
    is it a typo ?

    nice blog I like it
    What I wrote is correct, namely |HxK|=|K|[H:H\cap xKx^{-1}]. Of course, since |HxK|=|KxH| (by the obvious bijection) we also have that |HxK|=|H|[K:K\cap xHx^{-1}].


    And, thanks about the blog!
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  5. #5
    MHF Contributor Amer's Avatar
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    Re: Group Action, Double costs

    when i said typo i mean in the book I made an example

    let G=Z_{24} , H = <3> = \{0,3,6,9,...,21\} , K = <4> = \{0,4,8,...,20\}

    \mid H\mid = 8 , \mid K \mid = 6

    H acting on xK by left additive
    G/K = \{K, 1+K,2+K,3+k\}

    the orbit of H acting on the left cost will be   \{K, 1+K,2+K,3+k\}

    \mid HxK \mid = \mid K \mid . \mid H:Stab_H(xK)\mid = 6.4 = 24

    but if it is \mid HxK \mid = \mid H \mid . \mid H :Stab_H(xK) \mid = 8.4 = 32 ?? which is not correct since the order should equal to the G order
    so there is a typo in book
    am I right ?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Re: Group Action, Double costs

    Quote Originally Posted by Amer View Post
    when i said typo i mean in the book I made an example

    let G=Z_{24} , H = <3> = \{0,3,6,9,...,21\} , K = <4> = \{0,4,8,...,20\}

    \mid H\mid = 8 , \mid K \mid = 6

    H acting on xK by left additive
    G/K = \{K, 1+K,2+K,3+k\}

    the orbit of H acting on the left cost will be   \{K, 1+K,2+K,3+k\}

    \mid HxK \mid = \mid K \mid . \mid H:Stab_H(xK)\mid = 6.4 = 24

    but if it is \mid HxK \mid = \mid H \mid . \mid H :Stab_H(xK) \mid = 8.4 = 32 ?? which is not correct since the order should equal to the G order
    so there is a typo in book
    am I right ?
    Yes, you are correct. There is a typo. It is not |HxK|=|H|[H:H\cap xKx^{-1}] but [HxK]=|K|[H:H\cap xKx^{-1}.
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  7. #7
    MHF Contributor Amer's Avatar
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    Re: Group Action, Double costs

    Quote Originally Posted by Drexel28 View Post
    Yes, you are correct. There is a typo. It is not |HxK|=|H|[H:H\cap xKx^{-1}] but [HxK]=|K|[H:H\cap xKx^{-1}.
    Thanks for the fast respond I appreciate that
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