If $a^{3}=a$ for all $a\in R$, show that the ring $R$ is commutative.
I tried to follow Ehsan M. Kermani's answer, but I thought wouldn't the following be sufficient?
Let $a,b\in R$. Then $a^{3}=a$ and $b^{3}=b$. Then $ab=a^{3}b^{3}=(ba)^{3}=ba$.
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7How did you go from $a^3 b^3$ to $(ba)^3$? I'm pretty sure that step is non-trivial. – Oct 29 '18 at 21:40
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1Based on link. Because for any $a$, $a^2\in Z(R)$ we got that $ab=(ab)^3=ababab=a(ba)^2b=(ba)^2ab=baba^2b=ba^3b^2=bab^2=b^3a=ba$. – maciek97x Oct 29 '18 at 22:33
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Why does $a(ba)^2b = (ba)^2ab$? – fleablood Oct 29 '18 at 22:40
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By definition $Z(R)={a\in R |\forall_{b\in R} ab=ba}$. – maciek97x Oct 29 '18 at 22:45
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1Possible duplicate of If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative. – lisyarus Nov 01 '18 at 15:53