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If $R$ is a ring such that $x^5=x$ for all $x\in R$, is $R$ commutative?

If the answer to the above question is yes, then what is the least positive integer $k \ge 6$, such that there exists a noncommutative ring $R$ with $x^k=x$ for all $x\in R$?

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    $\begingroup$ Sounds like a hard homework (: $\endgroup$ – curious Jan 6 '11 at 5:27
  • $\begingroup$ No. I came across with proofs that rings with $x^k=x$ for $k=2,3,4$ are commutative; hence the question. $\endgroup$ – TCL Jan 6 '11 at 5:34
  • $\begingroup$ @Jasper. Not necessary, from the link by curious. $\endgroup$ – TCL Jan 6 '11 at 5:39
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The answer is that if $R$ is a ring such that for all $x \in R$ there is an integer $n(x) > 1$ such that $x^{n(x)} = x$ then $R$ is commutative. See Herstein's "Non Commutative Rings".

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