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Homework Help: Only homomorphism from rationals to integers

  1. Jun 6, 2012 #1
    Kindly see the attached image.i cant understand the step where he write f(x)=f(1+1+.....+1)=xf(1).But the homomorphism is from <Q,.> to <Z,+>

    Attached Files:

  2. jcsd
  3. Jun 6, 2012 #2
    Sorry, but this is unreadable
  4. Jun 6, 2012 #3
    Sorry for bad image quality.
    Show that only homomorphism from the group <Q*,.> to <Z,+> is the zero homomorphism where Q*=Q-{0}
    Let 0≠f be a homomorphism from <Q*,.> to <Z,+>
    Let f(1)=n[itex]\in[/itex]Z. Suppose f(1)=0
    Then f(x)=f(1+1+...............+1) if x>0,x[itex]\in[/itex]Z
    =x f(1)=0
    This is the step I couldnt understand as the homomorphism is from <Q*,.> to <Z,+> not from <Q*,+> to <Z,+> .
    Thus f(-x)=f(-1χx)=f(-1)+f(x)=f(-1)+0=-f(-x) [itex]\Rightarrow[/itex] f(-x)=0

    Further he proved that f(r/s)=0
  5. Jun 7, 2012 #4
    Indeed something is very fishy about this proof.
    Furthermore I think he is trying to prove something which is wrong.
    Unless I miss something,
    define for any x of Q being a multiple of a power of 2 f(x)= said power
    for instance, f(-1/2)=f(15/2)=-1
    and fx)=0 otherwise, than it looks to me that this is a counter example, but I could miss something...

  6. Jun 8, 2012 #5
    Thanks for your reply.Perhaps it is serving as counterexample to what the question was given.In some other book the same question was there with <Q,+> instead of <Q,.>
    .Maybe the solution provided by the author is wrong
  7. Jun 8, 2012 #6
    Well i,n fact Q+ sounds like a much better candidate to do the job, so my counter example would be even simpler, but I probably miss something.
    Anyway, the proof you linked to is completely bogus as far as I can tell, and, just think about exp(x) ln(x); if we were to talk about R+ -> R (or the other way around, pick one) you would have a perfect example of the same idea, so I think the proof should be a lot more involved than that and.

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