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作者  george (...) 站內  ALGEBRA
標題  Re: [問題] 一個關於order的問題
時間  2009/11/10 Tue 20:30:24

※ 引述《kyod ( )》之銘言:
> G is an abelian group and a,bεG  s.t │a│= m,│b│= n
> show G contains an element whose order is  [m,n]
> (pf)
> Case1:gcd(m,n)=1

  ∵(m,n)[m,n]=mn => [m,n]=mn

> Claim: mn is the order of ab , ie │ab│= mn
          mn   mn  mn     m n   n m             mn
  pf. (ab)  = a   b   = (a )  (b )  = e  => (ab)  = e => order(ab) | mn (1)

>                                                  mn
> check: mn is the least positive integer s.t  (ab)  = e

  Let k = order(ab)

                    k        k k
          Then  (ab) = e => a b = e
>                            k    -k
>                        => a  = b
>                            k    -k
>                       ∵  a  = b  ε <a>
>                            k    -k
>                           a  = b  ε <b>
>                            k    -k
>                       ∴  a  = b  ε <a>∩<b> = {e}
                             k
                         => a  = e => m|k=order(ab)

          Similarly, n|k=order(ab).

          Since (m,n)=1 , mn|order(ab)  (2)

          By (1)(2) mn = order(ab)

>  ∴ [m,n] = mn is the order of ab with (a,b)=1
>                                                #


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→ kyod :哈哈~~有些地方有小小筆誤...Orz                            09/11/10

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