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Im not sure where to start. I first assume $|G|<{\infty}$ and $|G|=2k$ with $k\ge{1}$. Then take some element $g\in{G}$, but I can't see what I can show from the assumptions.

kam
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  • Do you know Cauchy's theorem? – B.Swan Jun 17 '19 at 16:00
  • No I do not know this theorem, is there an elementary way of proving the statement – kam Jun 17 '19 at 16:00
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    Yes, there is, via the equivalence relation $x \sim y \iff x=y $ or $y=x^{-1}$. Which classes have only $1$ element in them? What does it mean for other elements? – B.Swan Jun 17 '19 at 16:04

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We have : $$G:=A\sqcup(G\backslash A), $$ with $A:=\{x\in G,x^{-1}=x\}.$

Since $x\in G\backslash A$ iff $x^{-1}\in G\backslash A$, we have : $|G\backslash A|\equiv 0\pmod 2$, so $|A|\equiv 0\pmod 2$

Moreover, $e\in A$ and $|A|$ is even, so $|A|\ge 2$, and you can find $x\in G\backslash\{e\}$ with $x=x^{-1}$, i.e $x^2=e$. $\blacksquare$

Mishikumo2019
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