Abstract
Let \phi be an automorphism of finite order of the nilpotent group G of class c and m and r positive integers with \phi^{m} = 1 . Consider the two (not usually homomorphic) maps \psi and \gamma of G given by \psi\colon g\longmapsto g\cdot g \phi \cdot g\phi^{2}\cdot\ldots\cdot g\phi^{m-1} \quad\text{and}\quad \gamma\colon g \mapsto g^{-1}\cdot g\phi\quad\text{for }g\in G. We prove that the subgroups X =\langle x\alpha\colon x\in\ker\:\psi, \alpha \in \mathrm {Aut}\: G, x^{r}\in\textstyle\bigcup_{s\geq 0}(G\gamma)^{s}\rangle, Y =\langle g\gamma\alpha\colon g\in G, \alpha\in\mathrm {Aut}G\:, (g\gamma)^{r}\in \mathrm {ker}\:\gamma\rangle, X^{\ast} =\langle x^{r}\alpha\colon x\in\mathrm {ker}\:\psi, \in \alpha\in\mathrm{Aut} \:G, x^{r}\in\textstyle\bigcup_{s\geq 0}(G\psi)^{s} \rangle, Y^{\ast}=\langle(g\gamma)^{r}\alpha\colon g\in G, \alpha\in\mathrm{Aut}\: G, (g\gamma)^{r}\in\mathrm {ker}\:\gamma\rangle=\langle((G\gamma)^{r}\cap\mathrm {ker}\:\gamma)\mathrm{Aut}\:G\rangle of G all have finite exponent bounded in terms of c , m and r only. This yields alternative proofs of the theorem of [4] and its related bounds.
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