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Orders of Isomorphisms

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data

    let phi : G --->G' be an isomorphism of groups. let x element of G and let x'=phi(x)

    Prove that the orders of x and x' are equal

    3. The attempt at a solution

    I dont even know what the order of a isomorphism means. As far as i know, an isomorphism is just a bijective map from G to G'. How does this have order?
    Last edited: Oct 21, 2008
  2. jcsd
  3. Oct 21, 2008 #2


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    The question isn't asking you to find the order of an isomorphism. It's asking you to look at the orders of x and x'.

    Also, generally speaking, an isomorphism can have an order: there are groups whose elements are maps.
  4. Oct 21, 2008 #3


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    An isomophism is NOT just a "bijective map from G to G'". It is a bijective map that preserves the operation: phi(x*y)= phi(x).phi(y) where * is the operation in G and . is the operation in G'.

    As morphism told you, the question does not ask anything about "order of an isomophism"- it asks about the orders of x and phi(x), a member of G and a member of G'.
  5. Oct 21, 2008 #4
    Ok im still lost on this problem.

    I know we want to show that xn=eg and phi(x)n = eg' for some integer n.

    but i dont know how to do this.
  6. Oct 21, 2008 #5


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    No, you want to show that IF xn= eG, then (phi(x)n= eG'. try applying phi to both sides of the first equation.
    Last edited: Oct 22, 2008
  7. Oct 21, 2008 #6
    ok so phi(xn)= phi(eg)
    ==> phi(xn)= eg'

    because phi is an isom

    phi(xn)= phi(x)n

    and phi(x)n=eg'

    thus both x and x' have order n


    That good?
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