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Properties of finite groups

  1. May 4, 2005 #1
    Assume G is a finite group and [tex]H = \left\{ {g \in G|g^n = e} \right\}[/tex] for any [tex]n>0[/tex]. e is identity.
    I have been able to show that if G is cyclic, then H has at most n elements.
    However, I can't go the other way. That is, assuming H has at most n elements, I haven't been able to say anything about whether G is cyclic, abelian or neither.
    Any suggestions?
     
    Last edited: May 4, 2005
  2. jcsd
  3. May 4, 2005 #2

    Hurkyl

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    Can you add an element to G without adding an element to H?
     
  4. May 4, 2005 #3

    matt grime

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    Look at D_3=S_3 the nonabelian group of order 6. How many elements of order 3 does it have?
     
  5. May 4, 2005 #4
    Since G is a finite group, then every element in G must equal identity for some n. That means that for some n the element must be added to H. So I think the answer is no. But I can't make any connection to my problem.

    If I haven't made any mistakes, it has 2 elements of order 3 and 3 elements of order 2. But that means that H has 3 elements for n=2 (which does not agree with the assumption that H has at most n elements), doesn't it?

    I'm quite confused now...
     
  6. May 4, 2005 #5

    Hurkyl

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    You seem to be treating n as both a constant and a variable -- that might be the source of confusion.
     
  7. May 4, 2005 #6
    I don't mean to treat n as a variable, only as an unknown constant. Where do I treat it as a variable?
     
  8. May 4, 2005 #7

    Hurkyl

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    "Since G is a finite group, then every element in G must equal identity for some n. That means that for some n the element must be added to H."
     
  9. May 4, 2005 #8

    matt grime

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    S_3 is a group that has fewer than 3 elements of order 3. How does that not prove useful unless you're being odd about n: at least quantify it: there exists an n, or "for all n".

    It apparently seems you wish it to be "for all n", which you didn't bother to specify.
     
  10. May 4, 2005 #9
    But a certain value of n defines a certain H, so I have to consider the different values n can take.

    What I mean is for any n.
     
  11. May 4, 2005 #10

    Hurkyl

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    So, what you want to prove (or disprove) is that if:

    For all n > 0, Hn has no more than n elements

    Then G is cyclic?
     
  12. May 4, 2005 #11
    Yes, and if it's not cyclic: Is it "at least" abelian?

    I'm sorry about the poor specification.
     
    Last edited: May 4, 2005
  13. May 4, 2005 #12

    mathwonk

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    try the direct sum of the cyclic groups of orders 2,3,5,7,11,13,17,19,23,29. and look for elements of order 2.
     
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