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Group action on a subgroup

  1. Nov 28, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm trying to prove that, if H is a subgroup of an arbitrary group G, then H^g, the action of a given element in G on H, is isomorphic to H.


    2. Relevant equations



    3. The attempt at a solution

    Let \sigma denote a given action of G on H. We are considering the map \sigma(g, *) : H -> H^g (where g is fixed and * denotes a variable element of H). I think that the kernel of this map is the set of all elements x in H such that \sigma(g, x) = g (this is the part I'm not sure of). Since x lies in H and H is a subgroup of G, it follows that x must be equal to e, the identity element of G. Thus, ker{\sigma(g, x)} is trivial, and, as a result,
    [tex] H^g \cong G/ \mbox{ker}(\sigma(g, *)) = G, [/tex]
    by the first isomorphism theorem.
    Does this seem correct? If so, can you please justify the statement that
    [tex] \mbox{ker} \sigma(g, *) = \{ x| \sigma(g, x) = g \ \forall x \in H \} [/tex]?
     
    Last edited: Nov 28, 2007
  2. jcsd
  3. Nov 28, 2007 #2

    morphism

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    This doesn't make sense - an action is not a group. To have any chance of solving this problem correctly, you should write down what H^g actually is.
     
  4. Nov 28, 2007 #3
    H^g would be a group in certain cases (i.e. conjugation), but not in others. So the proof above certainly does not apply in general.
     
    Last edited: Nov 29, 2007
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