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Math Help - Factor exponent

  1. #1
    Senior Member sfspitfire23's Avatar
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    Factor exponent

    K \lhd G has index m , show that g^m \in K for all g \in G


    Thanks guys
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    Quote Originally Posted by sfspitfire23 View Post
    K \lhd G has index m , show that g^m \in K for all g \in G


    Thanks guys

    What is the order of the group G/K? Then....

    Tonio
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  3. #3
    Senior Member sfspitfire23's Avatar
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    order will be g^m=1 where m is the copies of g
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    Quote Originally Posted by sfspitfire23 View Post
    order will be g^m=1 where m is the copies of g

    No. The order is m and thus any element in the quotient group raised to the m-th power is the group's unit and thus...

    Tonio
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  5. #5
    Senior Member sfspitfire23's Avatar
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    So, If Kg was raised to the mth power, it would be come the inverse, bringing Kg back to K
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    Quote Originally Posted by sfspitfire23 View Post
    So, If Kg was raised to the mth power, it would be come the inverse, bringing Kg back to K

    No. You'd get (Kg)^m=Kg^m=K, SO....

    Tonio
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  7. #7
    Senior Member sfspitfire23's Avatar
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    So, kg^mk^{-1}=k \in K because K\lhd G then g^m\in K
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    Quote Originally Posted by sfspitfire23 View Post
    So, kg^mk^{-1}=k \in K because K\lhd G then g^m\in K

    Uuuh?? Ok, let us go back to the basics, shall we? First, we must know that Ka=Kb\Longleftrightarrow ba^{-1}\in K, so we get that Kx=K \Longleftrightarrow x\in K.
    Second...what I already wrote in my past messages: as G/H is a finite group of order m, any of its elements raised to the m-th power is the group's unit, and this means that (Kg)^m=K\,\,\forall\,g\in G. But  (Kg)^m=Kg^m , according to the definition of the operation that makes G/H a group, so this togehter with the first lines in this message gives...

    Tonio
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