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Proof by contradicton

  1. Jul 26, 2010 #1
    lal means the order of a


    Let G be a group and a[tex]\in[/tex]G. Then lal=la-1l


    Case 1, if lal=1

    Case 2, if lal=n

    Case 3, if lal=infinity

    i understand case 1 and 3, so i'll be post the proof when need,

    but, case 2

    Here's the proof

    Suppose lal=n

    Then an=1 and ar[tex]\neq[/tex]1, 1[tex]\leq[/tex]r<n-----------(1)

    To show that (a-1)n=1 and (a-1)r[tex]\neq[/tex]1, 1[tex]\leq[/tex]r<n.

    Clearly, (a-1)n=(an)-1=(1)-1=1

    Suppose (a-1)r=1, for some 1[tex]\leq[/tex]r<n

    => (ar)-1, 1[tex]\leq[/tex]r<n

    => ar=1, for some 1[tex]\leq[/tex]r<n

    but this contradict (1)

    So, (a-1)r[tex]\neq[/tex]1, 1[tex]\leq[/tex]r<n

    Hence, lal=n=la-1l

    I don't understand why, it suppose "(a-1)r=1, for some 1[tex]\leq[/tex]r<n"

    then say it contradict with (1), i cannot see how they contradict.

    Help, T_T
  2. jcsd
  3. Jul 26, 2010 #2


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    You need to finish this line by having an equation, not an expression. I suspect you meant to put =1 in there.

    You showed that if the order of a-1 is r with r<n, then ar=1. But the order of a was n, so that's the contradiction
  4. Jul 26, 2010 #3
    (a-1)r=a-r=an-r, so if 0<r<n and a-r=e, order of a is not n.
  5. Jul 26, 2010 #4
    OOOOOOOOOOOOOOOOOOOOO, i get it, now i'm trying to catch what losiu99 try to convey.
    in the mean time

    do you mind checking this? please :D

  6. Jul 26, 2010 #5
    Sorry, my post was a bit off topic, I thought you didn't understand the part on reaching ar=1 starting from (a-1)r.
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