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I am interested in why this proof is wrong since every proof I have seen uses much more knowledge than this and I can't figure out why it is incorrect.

Suppose $x^3 = x$ for every $x \in R$. Then $x = x^{-1}$ and similarly for every $y \in R$, $y = y^{-1}$. So then $xy \in R$ and $xy = x^{-1}y^{-1}$. Since $yx(x^{-1}y^{-1}$) = $e$, $(yx)^{-1} = (x^{-1}y^{-1})$. Then $xy = (yx)^{-1}$. Since $R$ is closed under multiplication, $yx \in R$ and so by the discussion above, $yx = (yx)^{-1}$. Therefore, $xy = yx$ for all $x$,$y \in R$ which means that $R$ is commutative.

stupidproofs123
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    Why do you assume that $x^{-1}$ exists for all $x\in R$? Well, Wuestenfux (+1) did essentially capture this. – Jyrki Lahtonen Jul 11 '19 at 08:55
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    Actually,is this a great proof that every group where $x^3=x$ is commutative! – David Jul 11 '19 at 08:57
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    I guess this is what happens when someone like me studies groups before rings... – stupidproofs123 Jul 11 '19 at 08:58
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    That's typical of everybody: walk before you crawl, fall down sometimes when you run. I think it would be a bad idea to learn about rings before learning anything about groups. – rschwieb Jul 11 '19 at 13:19

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Well, if $x^3=x$ in $R$, then by the shortening rule, $x^2=1$. But this only holds if $x$ is not a zero divisor.

Wuestenfux
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