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Proof problem, help!

  1. Mar 20, 2005 #1
    I would really appreciate if anyone could help me with this problem.

    F is a group homomorphism from G= (E, *) to H= (F,#).
    If , for all x e E , x*x e ker(f).
    Show that for all x e E, f(x-1)=f(x)

    Now I don't know how I should start the proof. Also, I would like to know if I can assume that Ker(f) is equal to the identity element.
     
  2. jcsd
  3. Mar 20, 2005 #2

    AKG

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    If [itex]x*x \in Ker(f)[/itex] then this means precisely that [itex]f(x*x) = 1_F[/itex] where [itex]1_F[/itex] is the identity element of H. You want to show something like:

    [tex]f(x*y) = f(x)#f(y)[/tex]

    and

    [tex]f(x^{-1}) = (f(x))^{-1}[/tex]

    In fact, you probably already have these proved somewhere. If you do, then use them and the rest should be very simple. Actually, I don't remember the definitions, but I have a feeling that the first thing (that f(x*y) = f(x)#f(y)) might just be a property of homomorphisms. The second property follows from the first at least in the case where x*x is in Ker(f), and probably in general too.
     
  4. Mar 21, 2005 #3

    matt grime

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    You've got F as a homomorphism, and the underlying set of H (at least that's what I presume H=(F,#) means. Usually there is no reason to distinguish the group from its elements like that. Also, if you're not going to tex it then it's common to indicate inverses by putting the minus 1 in an exponent, since x-1 doesn't really make sense for a group. You may not assume ker(f) is the identity element, since that is saying f is an injection which is not given.

    From the definition y in ker f iff f(y)=e conclude something about f(xx), and use AKG's stuff above which is the (usual) definitoin of what a homomorphism is.

    And remember showing that y is the inverse of x is showing exactly that xy=yx=e.
     
  5. Mar 21, 2005 #4

    AKG

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    Matt, I believe he/she meant f (lower case) to be the homomorphism, and F to be the set underlying group H.
     
  6. Mar 21, 2005 #5
    Is that what the guy is talking about, F(x-1) is actually F(x^(-1))? Just written differently? No wonder the problem look mysterious! This is a completely simple problem now that it is written so it can be read! If F(x*x) goes to 1, then x is its own inverse, since by group property there can be only a unique inverse.

    You gotta be careful about how you write the inverse if you use a calculator, you know!
     
    Last edited: Mar 21, 2005
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