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Math Help - Kernel of a group action

  1. #1
    Super Member Bernhard's Avatar
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    Kernel of a group action

    Dummit and Foote Section 1.7 Group Actions - Exercise 4 is as follows:

    Let G be a group acting on a set A and fix some a \in A.
    Show that the following sets are subgroups of G:
    (a) the kernel of the action
    (b) {g \in G | ga = a} {stabilizer of a in G}

    ================================================

    I am working on part (a) and have not got far!

    We know that the kernel of the action of G on A is as follows:

    Kernel, K = {g \in G| ga = a for all a \in A}

    We need to show x y^{-1} \in K for all x, y \in K

    So assume k_1, k_2 \in K

    Then k_1a = a and k_2a = a

    Now need to show k_1 {k_2}^{-1} \in K

    That is ( k_1 {k_2}^{-1})a = a

    Now ( k_1 {k_2}^{-1})a = k_1( {k_2}^{-1}a)

    But ??? where to from here

    Can anyone help? Be grateful for some assistance!

    Peter
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  2. #2
    Senior Member slevvio's Avatar
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    Re: Kernel of a group action

    Hello,

    k_2 a = a \implies k_2^{-1} k_2 a = k_2 ^{-1} a \implies k_2^{-1} a = a ...

    is that cool?
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Kernel of a group action

    Thanks for the excellent response - so quickly!!

    Great mate!

    Peter
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  4. #4
    Senior Member slevvio's Avatar
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    Re: Kernel of a group action

    no problem!
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