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Prove Beta is an isomorphism of groups

  1. Nov 22, 2004 #1
    i cant grasp these concepts, 1-to-1 and onto have always annoyed me.

    here's 1 question, (i dont know how to post symbols so Beta ..)
    (C is Complex numbers)
    Let Beta:<C,+> -> <C,+> by Beta(a+bi)=a-bi (that is, the image is a +(-b)i).
    Prove Beta is an isomorphism of groups.


    i have alot of these problems but maybe a one may help me understand the rest
     
  2. jcsd
  3. Nov 22, 2004 #2
    1-1: f(a+bi)=y
    f(c+di)=y ->
    f(a+bi)=f(c+di)->a-bi=c-di
    ->a=c & -bi=-di
    ->-b=-d->b=d as inverse is unique in integers under addition
    so if f(x)=f(n)->x=n

    onto: if y is an element of f(x) then y=a-bi for some a,b element of integers. But, c+di is an element of <C,+> for all c, d elements of integers. a, b are integers so a+bi is an element of <C,+> so the function is onto.

    f(a+bi + c+di) = a-bi + c-di = f(a+bi) + f(c+di)...
     
  4. Nov 22, 2004 #3
    thanks nnnnnnnn
     
  5. Nov 23, 2004 #4

    matt grime

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    Science Advisor
    Homework Helper

    don't forget that since f has an obvious inverse map , namely itself, then it must be a bijection, so it suffices to prove it is a homomorphism of additive groups.
     
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