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A werid isomorphism question.

  1. Oct 8, 2009 #1
    Let G be a group and let [tex]\phi[/tex] be an isomorphism from G to G. Let H be a subgroup.

    Hint: These subgroups should already be familiar to you.

    Let H={z in C:[tex]\phi[/tex](z)=z}

    This would be the subgroup of {-1,1}, this would be the group {-1,1} under multiplication.

    Let H={z in C: [tex]\phi[/tex](z)=-z}

    I'm not even sure where to start with this one.
     
  2. jcsd
  3. Oct 8, 2009 #2

    Office_Shredder

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    You don't seem to have posted an actual question. How did you get that H={-1,1} without having defined what phi or G is (and furthermore, what is C here?)

    Please post the complete question so we can understand what you're trying to do
     
  4. Oct 8, 2009 #3

    jambaugh

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    Firstly Dittos on "what's the actual question?"

    If [tex]\phi[/tex] is the identity automorphism then H is in fact G. In other cases H can be the trivial group {1}. Is G a specific group or do you know something more about G?

    From what you are given you can at best show that the subset H of such elements is in fact a subgroup. I think you can also show it is a normal subgroup. (Note all subgroups of an abelian group are normal subgroups.)
    You are using -z to indicate the inverse group element which indicates you are using addition to indicate the group product. Are you given that G is an abelian (commutative) group? Typically we represent group product as x+y if the group is abelian and xy if we do not assume it is abelian in which case the inverse is written as a -1 power.

    I will note this. Given the mapping:
    [tex]\phi: G\to G[/tex]
    such that
    [tex] \phi(g)=g^{-1}[/tex]
    is an isomorphism we observe that since:
    [tex]\phi(ab) = (ab)^{-1} = b^{-1}a^{-1}= \phi(b)\phi(a)[/tex]
    and since as an isomorphism:
    [tex]\phi(ab) = \phi(a)\phi(b)[/tex]
    we have that:
    [tex]b^{-1}a^{-1} = a^{-1}b^{-1}[/tex]
    which must mean that G is in fact abelian. (let a be x inverse, and b be y inverse and xy=yx).

    If your isomorphism doesn't map all elements to G to their inverses but only some then the subset on which it does must in fact be abelian but only if it is in fact a subgroup. But this need not be the case.

    Example: Let G be the group of permutations of 3 elements and let phi be the identity map.
    Since transposition of two elements square to the identity they are their own inverses and thus phi maps them to their own inverses. This is not the case for the 3-cycles so they are not in the set H. Thus the set H is the set of transpositions (and the identity) but this set does not close under group product (1,2)(2,3) = (1,2,3).

    If however you impose the condition that H is indeed a subgroup and not just a subset then you will have that H is abelian and I think also that H is a normal subgroup. I'm not 100% certain on the normal part though.
     
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