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Proof on isomorphisms

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that for any isomorphism [tex]\phi[/tex] : G--> H |[tex]\phi[/tex](x)| = |x| for all x in G. is the result true if [tex]\phi[/tex] is only assumed to be a homomorphism?

    Using the solution to the above proof or otherwise, show that any 2 isomorphic groups have the same number of elements of order n, for every positive integer n.

    2. Relevant equations

    The definition of a homomorphism states that given two arbitrary groups G and H. A function f : G---> H is called a homomorphism if f(ab) = f(a)f(b) for all a, b in G.

    3. The attempt at a solution

    What I started doing was that since we know that [tex]\phi[/tex] is an isomorphism, we know that it is bijective and a homomorphism. Therefore,
    [tex]\phi[/tex](ab) = [tex]\phi[/tex](a)[tex]\phi[/tex](b) (because it is a homomorphism)

    I'm not too sure what I can do from here...
     
    Last edited: Oct 14, 2009
  2. jcsd
  3. Oct 14, 2009 #2

    Office_Shredder

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    Do this in two parts

    If the order of x is k, show that [tex] \phi(x)^k = e[/tex] e the identity. Then show that if the order of [itex] \phi(x) [/itex] is k, that xk is also the identity element
     
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