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Thread: 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

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

    a) Prove that HxK is the union of the left costs $\displaystyle x_1K,x_2K,...,x_nK$ where $\displaystyle \{x_1K,x_2K,...,x_nK\}$ 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 $\displaystyle HxK = \bigcup_i^n x_iK $ it is clear that
    $\displaystyle 1xK = xK $ so $\displaystyle \bigcup_i^n x_iK \subseteq HxK$

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

    d) Prove that $\displaystyle \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

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

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

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

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

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

    THanks
    Last edited by Amer; Nov 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

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

    a) Prove that HxK is the union of the left costs $\displaystyle x_1K,x_2K,...,x_nK$ where $\displaystyle \{x_1K,x_2K,...,x_nK\}$ 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 $\displaystyle HxK = \bigcup_i^n x_iK $ it is clear that
    $\displaystyle 1xK = xK $ so $\displaystyle \bigcup_i^n x_iK \subseteq HxK$

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

    d) Prove that $\displaystyle \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

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

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

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

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

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

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

    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

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

    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

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

    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 $\displaystyle |HxK|=|K|[H:H\cap xKx^{-1}]$. Of course, since $\displaystyle |HxK|=|KxH|$ (by the obvious bijection) we also have that $\displaystyle |HxK|=|H|[K:K\cap xHx^{-1}]$.


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

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

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

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

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

    but if it is $\displaystyle \mid HxK \mid = \mid H \mid . \mid H :Stab_H(xK) \mid = 8.4 = 32 ??$ 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 $\displaystyle G=Z_{24}$ , $\displaystyle H = <3> = \{0,3,6,9,...,21\}$ , $\displaystyle K = <4> = \{0,4,8,...,20\}$

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

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

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

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

    but if it is $\displaystyle \mid HxK \mid = \mid H \mid . \mid H :Stab_H(xK) \mid = 8.4 = 32 ??$ 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 $\displaystyle |HxK|=|H|[H:H\cap xKx^{-1}]$ but $\displaystyle [HxK]=|K|[H:H\cap xKx^{-1}$.
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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 $\displaystyle |HxK|=|H|[H:H\cap xKx^{-1}]$ but $\displaystyle [HxK]=|K|[H:H\cap xKx^{-1}$.
    Thanks for the fast respond I appreciate that
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