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I am aware that it is straightforward to prove that a group of even order has an odd number of elements of order 2 just by noting that elements of order greater than 2 come in pairs necessarily (i.e. $|a|>2 \implies a\neq a^{-1}$).

However, I am curious to know whether there is a way to prove this quickly just by using properties of cosets, maybe Lagrange's theorem?

Thanks for any ideas!

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up vote 1 down vote accepted

Lagrange's theorem is necessary here because a group containing any elements of order $2$ ("involutions") must have even order. I don't think there is alternative proof using cosets, however, because the set of involutions doesn't necessarily form a subgroup, even when including the identity (nor does the set of elements with order $>2$). The easiest example of this is $D_4$, which has order $8$, yet is generated by (even just two of) its five involutions. The $|a|>2\Rightarrow a\not= a^{-1}$ element is already quite simple, though. I would just stick with that.

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