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Thread: 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{$\displaystyle r \geq 1: x^r = 1$}

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

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


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

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

    therefore the contradiction seems to arise from the fact that if $\displaystyle x^{a} = x^{b}$, then $\displaystyle \left[ {\phi (x)} \right]^b = e'$ 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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