(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Prove that G has an element of order 2.

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