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Painstaking Proof

  1. Sep 3, 2006 #1
    I'm working on the proof of the following statement: Prove that any finite group G of even order contains an element of order 2.

    What frustrates me about this is that I know from a previous class that this statement immediately follows from Cauchy's Theorem. So, in a way, I don't wanna prove it using group axioms.

    But I hafta, so whining aside, I will carry on. But I seem to be stuck.

    So far, I am looking at the subset of G that contains elements that are not of order 2. That is:

    [tex] H = \{ g \in G \ | \ g \neq g^{-1} \} [/tex]

    I think that this set must have an even number of elements since for each h in H, the inverse of h must also be in H, and, they can't be the same.

    Does this make sense?

    Then, H complement must include at least the identity of G. So, since G is the union of H and H complement, it seems that there must be at least one more element, a, in H complement, so that the total number of elements in G is indeed even.

    Thus, a fulfills the role of the element of order 2.

    Mathematical casualness aside, does this sound good? Thanks...

    EDIT: Ok, maybe "painstaking" was a bit overdramatic...
    Last edited: Sep 3, 2006
  2. jcsd
  3. Sep 3, 2006 #2


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    That's perfectly fine, why the lack of confidence?
  4. Sep 3, 2006 #3
    Oh, it's a way of life.:eek:
  5. Sep 3, 2006 #4


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    snap out of it. be critical of your work but accept the fact that not being able to find an error sometimes means there are no errors and you are right on the money :bugeye:
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