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Proof about Cyclic Groups

  1. Oct 16, 2014 #1
    Hello everyone,

    I am trying to understand the proof given in this link:

    https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic

    I understand everything up until the part where they conclude that ##r## must be ##0##. Their justification for this is, that ##m## is the smallest integer, and so this forces ##r=0##. Is it not possible that ##m=3## could be the smallest integer? Couldn't ##r## then be ##2##?
     
    Last edited: Oct 16, 2014
  2. jcsd
  3. Oct 16, 2014 #2

    HallsofIvy

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    No, you are missing the point. m was, by hypothesis, the smallest positive integer such that [itex]a^m\in H[/itex]. We also have that [itex]a^r\in H[/itex]. Once we have that [itex]0\le r< m[/itex], we cannot have r any positive integer because m is the smallest such integer.
     
    Last edited: Oct 16, 2014
  4. Oct 16, 2014 #3

    RUber

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    It seems like the critical information is in this line "Since ## a^m∈H## so is ##(a^m)^{−1}## and all powers of the inverse by closure ."
     
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