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Thread: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A

  1. #1
    MHF Contributor Amer's Avatar
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    Prove that G acts transitively on normal subgroup orbits on nonempty finite set A

    The Question is let $\displaystyle G $ acts transitively on a nonempty finite set A

    let H be normal subgroup of G,Let Orbits of H on A be $\displaystyle O_1,O_2,...,O_r$

    prove that G acts transitively on $\displaystyle O_1,O_2,...,O_r$

    My work I want to prove that for any two oribts $\displaystyle O_i,O_j$ there exist $\displaystyle g\in G $ such that $\displaystyle O_i = gO_j$

    I will call $\displaystyle O_i$ elements with "a" and $\displaystyle O_j$ with "b"

    pick $\displaystyle a_1\in O_i \;\;\;\;, b_1\in O_j $
    there exist g in G such that $\displaystyle a_1 = g. b_1 $ since G acts transitively on A
    let $\displaystyle b_s \in O_j \;\;\; b_s = h.b_1$
    want to show that $\displaystyle g.b_s $ is an element of $\displaystyle O_i$
    $\displaystyle g.b_s = g.h.b_1 $
    $\displaystyle g.b_s = g.h.g^{-1} .a_1 \;\;\; a_1=g.b_1 \Rightarrow g_{-1}.a_1 = b_1 $
    $\displaystyle g.b_s = (ghg^{-1}).a_1 $
    but H is normal subgroup so b_s in O_i

    b_s is arbitrary so for any b in O_j g.b is an element of O_i
    $\displaystyle O_i = gO_j $ ends

    is it correct ?? is there any shorter way ? any ideas
    Thanks
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  2. #2
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    Re: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A

    is it already given that a normal subgroup H of G induces a well-defined action on the orbits of H? (it's true, but what i'm asking is: have you already proven this earlier?)

    aside from that, your proof of the transitivity of this action of G on the orbits of H looks fine to me.
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  3. #3
    MHF Contributor Amer's Avatar
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    Re: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A

    Quote Originally Posted by Deveno View Post
    is it already given that a normal subgroup H of G induces a well-defined action on the orbits of H? (it's true, but what i'm asking is: have you already proven this earlier?)

    aside from that, your proof of the transitivity of this action of G on the orbits of H looks fine to me.
    you are wright it is not the whole question i proved that before, any orbit of H say O, gO still orbit of H

    let a,b in orbit O, we have a = h.b for some h in H
    if I can show that g.a , g.b is in the same orbit I am done

    $\displaystyle ghg^{-1} \in H \;\;\; say \;\;\; ghg^{-1}=h' \Rightarrow h = g^{-1}h'g $

    $\displaystyle a = g^{-1}h'g.b$

    $\displaystyle g.a = h'.g.b $
    so g.a , g.b in same orbit in H
    is it Okay ??
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  4. #4
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    Re: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A

    that's right, and it shows why H has to be normal for this to work.
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  5. #5
    MHF Contributor Amer's Avatar
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    Re: Prove that G acts transitively on normal subgroup orbits on nonempty finite set A

    in the same question i asked to prove that all orbits has the same cardinality
    My work,I will show that the stabalizer of each representative element of the orbits has the same cardinality, let
    $\displaystyle O_i , O_j$ orbits of H on A let a,b be the element that present each orbit

    $\displaystyle G_a = gG_bg^{-1}\;\;,\text{such that} G_a,G_b \;\; \text{stabalizer of a,b in G}$

    Stab of a in H $\displaystyle H_a = G_a \cup H^c $ so $\displaystyle H_a = H_b$

    and $\displaystyle \mid O_a\mid = \mid H:H_a\mid$ which end the proof
    I did not post another thread because it is one question
    is it correct ?
    Thanks
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