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Math Help - Orders of Groups

  1. #1
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    Orders of Groups

    Hi i have completed the answer to this question. Just need your verification on whether it's completely correct or not:

    Question:
    If G is a group and xEG we define the order ord(x) by:
    ord(x) = min{ r \geq 1: x^r = 1}

    If \theta: G --> H is an injective group homomorphism show that, for each xEG, ord( \theta(x)) = ord(x)

    My answer: Please verify
    If \theta(x) = { x^r: r \epsilon Z} then ord( \theta(x)) = ord(x).

    For any integer r, we have x^r = e (or 1) if and only if ord(x) divides r.

    In general the order of any subgroup of G divides the order of G. If H is a subgroup of G then "ord (G) / ord(H) = [G:H]" where [G:H] is an index of H in G, an integer.
    So order for any xEG divides order of the group. So ord( \theta(x)) = ord(x)


    any suggestions or changes please? thnx
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  2. #2
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    If ord(x)=a then \left[ {\phi (x)} \right]^a  = \left[ {\phi (x^a )} \right] = \phi (e) = e'.
    Now suppose that ord\left[ {\phi (x)} \right] = b < a.
    Then
    \left[ {\phi (x)} \right]^b  = e' = \left[ {\phi (x)} \right]^a
    \phi (x^b ) = \phi (x^a )
    x^b  = x^a (injective)
    x^{a - b}  = e

    Do you see the contradiction?
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  3. #3
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    yes i think i see the contradiction.

    i believe to do this question we have to show 2 things:

    i) (\theta(x))^a = e'
    ii) 0 < b < a \implies (\theta(x))^b \neq e'.

    therefore the contradiction seems to arise from the fact that if x^{a} = x^{b}, then \left[ {\phi (x)} \right]^b = e' which is not what statement (ii) says.
    am i correct in this assumption? any ideas on how to deal with this?
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