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Question about invariant w.r.t. a group action

  1. Sep 23, 2014 #1
    Hello,

    I have a group [itex](G,\cdot)[/itex] that has a subgroup [itex]H \leq G[/itex], and I consider the action of H on G defined as follows:
    [itex]\varphi(h,g)=h\cdot g[/itex]
    In other words, the action is simply given by the group operation.

    Now I am interested in finding a (non-trivial) invariant function w.r.t. the action of H, which means finding a function [itex]\chi:G\rightarrow G'[/itex] such that [itex]\chi(h\cdot g)=\chi(g)[/itex] for all [itex]h\in H[/itex] and [itex]g\in G[/itex].

    I realized that I can easily impose sufficient conditions on [itex]\chi[/itex] to ensure that it is an invariant function w.r.t. H.
    Such conditions are:
    1. [itex]\chi[/itex] has the form [itex]\chi(g) = \gamma(g)^{-1}\cdot g[/itex]
    2. [itex]\gamma[/itex] is an automorphism of the group [itex]G[/itex]
    3. [itex]\gamma[/itex] fixes the subgroup [itex]H[/itex], [itex]\quad[/itex]i.e. [itex]\gamma(h)=h[/itex] for all [itex]h\in H[/itex]
    The proof is very easy (just apply [itex]\chi[/itex] as defined in 1. to [itex](h\cdot g)[/itex] and use 2. and 3.)

    My question is: Are the above conditions already well-known, perhaps in a more general form?
    Is my construction just a specific application of some more general theorem in group theory?
     
  2. jcsd
  3. Oct 6, 2014 #2

    lavinia

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    Your three conditions seem to not work but maybe I don't understand what you are saying.

    In any case,,any group homomorphism whose kernel contains the normalizer of H would seem to work.
     
  4. Oct 6, 2014 #3
    Hi Lavinia,

    thanks for replying. I was wondering why you said that the three conditions I listed do not work. The proof went like this:

    [itex]\chi(hg)= \gamma(hg)^{-1}hg [/itex] [itex]\quad\quad[/itex] (definition of [itex]\chi[/itex])
    [itex] = \gamma(g)^{-1} \gamma(h)^{-1} hg [/itex] [itex]\quad\quad (\gamma[/itex] is an automorphism)
    [itex] = \gamma(g^{-1}) \gamma(h^{-1}) hg [/itex] [itex]\quad\quad(\gamma[/itex] is an automorphism)
    [itex] = \gamma(g^{-1}) h^{-1}hg [/itex] [itex]\quad\quad(\gamma[/itex] fixes H)
    [itex] = \gamma(g^{-1}) g [/itex]
    [itex] = \chi(g) [/itex]

    Did I miss something? I think I have used all the three assumptions mentioned in my previous post.

    I feel that what I have done here is essentially the consequence of some well-known construction/theorem in group theory, but I don't know which one. I didn't get completely your argument about the normalizer.
     
  5. Oct 6, 2014 #4

    lavinia

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    Your calculation is correct. I was confused.

    A homomorphism whose kernel contains H will satisfy your required condition since all of the elements of H are sent to the identity. The normalizer is automatically in the kernel.
     
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