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Cyclic subgroup proof

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex] a \in <b>[/itex]
    Then [itex]<a> = <b>[/itex] iff a and b have the same order (let the order be n - the group is assumed to be finite for the problem).


    Proof:
    Suppose a and b have the same order (going this direction I'm trying to show that <a> is contained in <b> and <b> is contained in <a> ergo <a> = <b>).

    Since [itex]a \in <b>[/itex] it is obvious that [itex]<a> \subseteq <b>[/itex].

    Since a is in <b>, [itex]a = b^{m}[/itex] for some m.
    So [itex]<a> = <b^{m}> \supseteq <b>[/itex]
    Hence <a> = <b>

    Suppose <a> = <b> and <b> has order n.
    [itex]a = b^{m}[/itex]'
    [itex]a^{n} = (b^{m})^{n} = (b^{n})^{m} = e^{m} = e[/itex]

    Hence a has order n.

    This all look good?
     
  2. jcsd
  3. Nov 3, 2011 #2
    For the first part, it doesn't look like you used the fact that a and b have the same order... That might be a problem. Also, it's not true that the subgroup generated by b is contained in the subgroup generated by b^m. It should go the other way.
     
  4. Nov 3, 2011 #3
    Yeah, I see that now...
    Well, its obvious that <a> is contained in <b>

    Not sure how I can use the fact that a and b have the same order to prove the other direction, any hints?
     
  5. Nov 3, 2011 #4
    You know that the order of any subgroup of a group must be a divisor of the order of the group.

    Oh, that is confusing to read.

    Let me try again. You know that some group has order n. You should also know that the order of any subgroup of that group is going to divide n.
     
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