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Let $R$ be a finite group which has even number of elements. Prove that at for least two elements $r\in R$ the following equation is valid: $r^{2}=1$.

Note: $1$ is the element of unity in $R$

So I have been sitting with that problem for some hours now and do not know how to approach the problem. Studying Algebra as an extra curriculum activity, so any help that takes me back on the right track is greatly appreciated.

Goodylan
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    Hint: consider the pairs ${g, g^{-1}}$ for $g\in R$. Do these pairs partition $R$? Can all these sets have size $2$? – Mathmo123 Sep 07 '20 at 07:02
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    It seems to me that they both should have the equal size of two but.... when we have the element 1 in there and the group should have only even number of elements then... It seems that all the partitions can't have the size 2... Or am I missing or misreading something? – Goodylan Sep 07 '20 at 08:02
  • I can find at least one element in there and it is 1 itself (because every group has the element of unity... Is that plain or should I prove that as well? – Goodylan Sep 07 '20 at 08:08

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The number of elements $g$ such that $g\ne g^{-1}$ is even, since each element has a unique inverse. That means there's an even number of elements with $g^2=e$. But, of course, $e^2=e$.

Alternatively, use Cauchy's theorem, to get an element of order $2$.

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    Depending on how you prove Cauchy’s theorem, using Cauchy’s theorem here is overkill! In fact the first argument generalizes to a very clean counting proof of Cauchy’s theorem: https://qchu.wordpress.com/2013/07/09/the-p-group-fixed-point-theorem/ (I am not the downvoter, though.) – Qiaochu Yuan Sep 07 '20 at 16:21
  • @QiaochuYuan Yes. Cauchy's theorem would be the sledgehammer. Btw, i like the proof of Cauchy's theorem using Sylow subgroups. –  Sep 07 '20 at 16:23