x^3 = e for any element x in finitely-generated group G. How to prove that G is finite?
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11$\begingroup$ See mathshistory.st-andrews.ac.uk/HistTopics/Burnside_problem or en.wikipedia.org/wiki/Burnside_problem -- this follows because the Burnside group $B(m,3)$ is finite. $\endgroup$– Max HornDec 12 at 12:47
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6$\begingroup$ See Marshall Hall's book Group Theory, where this and other similar statements are proved. $\endgroup$– kabenyukDec 12 at 12:47
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1 Answer
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For a (partial) proof:
First, check in some way that for any 4 elements in such a group we have $[x,[y,[z,t]]]=1$. I.e., in the free group, this commutator $[x,[y,[z,t]]]$ is a product of cubes. Thus, a single formula (equality in the free group $F_4$) does the job (I'll add it if I can compute it or if somebody knows it).
As a consequence, every group of exponent 3 is 3-step nilpotent. The finiteness follows by dévissage.
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1$\begingroup$ Could you expand on what you mean by "dévissage" in this case? $\endgroup$– PedroDec 12 at 17:00
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$\begingroup$ @Pedro If $G$ is an $n$-step solvable finitely generated torsion group, its abelianization is finite, so $[G,G]$ has finite index hence is finitely generated, and hence is an $(n-1)$-step solvable finitely generated torsion group. So by induction on $n$ (starting with $n=1$), every $n$-step solvable finitely generated torsion group is finite. $\endgroup$– YCorDec 12 at 21:02