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Math Help - normal subgroup and orbits

  1. #1
    Senior Member abhishekkgp's Avatar
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    normal subgroup and orbits

    Let the group G act on the finite set A and let H \trianglelefteq G. Let O_1, O_2,\ldots,O_r be the distinct orbits of H on A.

    now read the following:
    g \cdot a_i = a_j, where g \in G, a_i \in O_i, a_j \in O_j
    let b_i = h_1 \cdot a_i \text{ for some } h_1 \in H \text{ and } a_i \neq b_i.
    now g \cdot b_i = g \cdot (h_1 \cdot a_i)=(g h_1) \cdot a_i = (g h_1 g^{-1} g) \cdot a_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j \text{ where }g h_1 g^{-1} = h_2; h_2 \in H.
    now h_2 \cdot a_j = c_j \text { for some } c_j \in O_j. also it can be shown that a_j \neq c_j very easily.

    from the above argument can i say that |O_i|=|O_j|
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    Quote Originally Posted by abhishekkgp View Post
    Let the group G act on the finite set A and let H \trianglelefteq G. Let O_1, O_2,\ldots,O_r be the distinct orbits of H on A.

    now read the following:
    g \cdot a_i = a_j, where g \in G, a_i \in O_i, a_j \in O_j
    let b_i = h_1 \cdot a_i \text{ for some } h_1 \in H \text{ and } a_i \neq b_i.
    now g \cdot b_i = g \cdot (h_1 \cdot a_i)=(g h_1) \cdot a_i = (g h_1 g^{-1} g) \cdot a_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j \text{ where }g h_1 g^{-1} = h_2; h_2 \in H.
    now h_2 \cdot a_j = c_j \text { for some } c_j \in O_j. also it can be shown that a_j \neq c_j very easily.

    from the above argument can i say that |O_i|=|O_j|


    I'm not sure: why won't you directly show a bijection between O_i\,,\,O_j ?

    Tonio
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    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by tonio View Post
    I'm not sure: why won't you directly show a bijection between O_i\,,\,O_j ?

    Tonio
    Thats what i tried to do. is there another way to show a bijection between the two??
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    Quote Originally Posted by abhishekkgp View Post
    Thats what i tried to do. is there another way to show a bijection between the two??

    "Another" way? Which one has been shown? If you think that your calculations in your first post show

    this then do define directly and unmistakenly the bijection, otherwise search for one...

    Tonio
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    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by tonio View Post
    "Another" way? Which one has been shown? If you think that your calculations in your first post show

    this then do define directly and unmistakenly the bijection, otherwise search for one...

    Tonio
    In the question i forgot to mention that G is transitive on A(sorry for that).
    with this in mind there exists a g \in G \text{ such that } g \cdot a_i = a_j \text{ where } a_i \in O_i, a_j \in O_j
    for such a g there exists a bijection:
    g \colon O_i \rightarrow O_j
    g \cdot m_i \mapsto m_j \text{ where } m_i \in O_i, m_j \in O_j
    to prove this consider  a_i, b_i \in O_i \text{ such that } a_i \neq b_i
    now  b_i = h_1 \cdot a_i \text{ for some } h_1 \in H
    from my first post  g \cdot b_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j
    to show that the mapping is injective i must show that g \cdot a_i \neq g \cdot b_i, \text{ that is, } a_j \neq h_2 \cdot a_j
    i will use contradiction to do so. Assume  a_j=h_2 \cdot a_j \Rightarrow a_j=(g h_1 g^{-1}) \cdot a_j \Rightarrow a_j= (g h_1) \cdot a_i \Rightarrow a_j=g \cdot b_i \Rightarrow g^{-1} \cdot a_j = b_i \Rightarrow a_i =b_i which is a contradiction. so the map is injective.

    has the injection been shown correctly? If this is correct then i can attempt to show the surjection too. Please comment.
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    Quote Originally Posted by abhishekkgp View Post
    In the question i forgot to mention that G is transitive on A(sorry for that).
    with this in mind there exists a g \in G \text{ such that } g \cdot a_i = a_j \text{ where } a_i \in O_i, a_j \in O_j
    for such a g there exists a bijection:
    g \colon O_i \rightarrow O_j
    g \cdot m_i \mapsto m_j \text{ where } m_i \in O_i, m_j \in O_j


    Uuh?? You define a function between two sets naming it the same as an element in G? And besides this,

    how is the function defined? Because we know that g\cdot a_i=a_j , but what happens with some

    a_i\neq x\in O_i?? How's defined the map you're talking about on x? Can you be sure that g\cdot x\in O_j

    to begin with at all? So, how do you define the map g ?


    to prove this consider  a_i, b_i \in O_i \text{ such that } a_i \neq b_i
    now  b_i = h_1 \cdot a_i \text{ for some } h_1 \in H

    from my first post  g \cdot b_i = (g h_1 g^{-1})g \cdot a_i = h_2 g \cdot a_i = h_2 \cdot a_j
    to show that the mapping is injective i must show that g \cdot a_i \neq g \cdot b_i, \text{ that is, } a_j \neq h_2 \cdot a_j
    i will use contradiction to do so. Assume  a_j=h_2 \cdot a_j \Rightarrow a_j=(g h_1 g^{-1}) \cdot a_j \Rightarrow a_j= (g h_1) \cdot a_i \Rightarrow a_j=g \cdot b_i \Rightarrow g^{-1} \cdot a_j = b_i \Rightarrow a_i =b_i which is a contradiction. so the map is injective.

    has the injection been shown correctly? If this is correct then i can attempt to show the surjection too. Please comment.
    .

    This looks fine, if we address the above questions I wrote.

    About surjectivity; Since |O_i|=|O_j|<\infty , a function between them is

    injective iff it is surjective.

    Tonio
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    Senior Member abhishekkgp's Avatar
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    I have committed a mistake in defining the bijection, what i should have written is the following:
     g \colon O_i \rightarrow O_j
     x_i \mapsto g \cdot x_i

    you have asked whether it can be said for sure that  g \cdot x \in O_j \text{ when } x \in O_i, x \neq a_i .
    I show that it is sure to hold. proof: since  x \in O_i \text{ there exists } h \in H \text{ such that } h \cdot a_i = x
    now  g \cdot x = g \cdot ( h \cdot a_i)
     \Rightarrow g \cdot x = (g h g^{-1})g \cdot a_i
    but we know that  g \cdot a_i =a_j \text{ and } g h g^{-1} \in H
    \Rightarrow g \cdot x = h^\prime \cdot a_j
    now since  h^\prime \cdot a_j \in O_j \text{ we have } g \cdot x \in O_j

    Thank you for your help. is the definition of the bijection alright now? it was a blunder on my part
    Last edited by abhishekkgp; February 14th 2011 at 06:06 AM. Reason: typo
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