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A group $G$ is called of generalized exponent $n$ if there exists elements $a_1,\dots,a_n \in G$ such that $x^{a_1}\cdots x^{a_n}=1$ for all $x\in G$, where $x^a=a^{-1}xa$. See the following question Generalized identities of (soluble) groups

Every nilpotent group $G$ of generalized exponent $m$ has finite exponent dividing $m^c$, where $c$ is the nilpotent class of $G$. If $m=p$ is prime, is it true that the nilpotent group $G$ has exponent dividing $p^2$. I have proved it for $p\in \{2,3,5,7\}$.

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  • $\begingroup$ $(\exists a_1,\dots,a_n \in G)(\forall x\in G)(x^{a_1}\cdots x^{a_n}=1)$ or $(\forall x\in G)(\exists a_1,\dots,a_n \in G)(x^{a_1}\cdots x^{a_n}=1)$? $\endgroup$ – user47958 Sep 17 '15 at 8:56
  • $\begingroup$ @user47958: The first one. This is a generalization of the notion of exponent. That is, every group of exponent dividing $n$ is of generalized exponent $n$. $\endgroup$ – Alireza Abdollahi Sep 17 '15 at 9:06
  • $\begingroup$ If the proof for $p=5,7$ is not merely by means of computations, perhaps it should be capable of generalizing to all primes. $\endgroup$ – M. Farrokhi D. G. Sep 17 '15 at 13:19
  • $\begingroup$ @M.FarrokhiD.G.: My proof depends on calculation and seemingly is ``$p$" sensitive. $\endgroup$ – Alireza Abdollahi Sep 17 '15 at 13:50

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