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Prove that G has an element of order 2.

  1. Sep 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Let G be a group of finite cardinality. By considering the size of the set {x ∈ G: x is of order ≥ 3} show that |G| is even iff there is an element of G with order 2.

    2. Relevant equations

    Perhaps Lagrange's Theorem

    3. The attempt at a solution

    It's obvious that if g ∈ G has order 2 then, since the subgroup generated by g has order dividing |G|, |G| must be even.

    Now we assume |G| is even.
    Considering the size of the set given, I have concluded that if the set has even cardinality, then there must be at least one g with order 2, and specifically an odd number of these elements. Otherwise G = [{x ∈ G: order of x ≥ 3} + 1] which is odd, and contradicts our assumption. But I'm not sure why the set has to have odd cardinality.

    Footnote: the question uses notation o(x) to denote what I think is the order of x. It seems to make sense for earlier questions.
     
  2. jcsd
  3. Sep 21, 2011 #2
    Does this make sense?

    Every element of {x ∈ G: order of x ≥ 3} is not equal to its inverse. So for every element of the set, we can find a distinct element of the set which is its inverse. As inverses are unique, these two elements will be inverses ONLY to each other. In this way the set can be divided into pairs of inverses, so is of even cardinality?

    If that works, the rest of the proof is straightforward.
     
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