Why Are There an Odd Number of Elements in a Finite Group Where g^3 Equals 1?

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SUMMARY

The discussion centers on the proof that a finite group \( G \) contains an odd number of elements \( g \) such that \( g^3 = 1 \). The solution provided by user castor28 demonstrates that the elements satisfying this condition form a subgroup of \( G \), and by applying Lagrange's theorem, it is concluded that the order of this subgroup must be odd. Therefore, the total number of elements \( g \) for which \( g^3 = 1 \) is confirmed to be odd.

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Euge
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Here is this week's POTW:

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Show that if $G$ is a finite group, then there are an odd number of elements $g\in G$ for which $g^3 = 1$.

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This week's problem was solved correctly by castor28. You can read his solution below.
The elements of order $3$ can be grouped in pairs $\{g,g^{-1}\}$. Together with the identity, that makes an odd number of elements satisfying $g^3=1$.
 

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