JournalsrsmupVol. 136pp. 61–68

# Automorphisms of finite order of nilpotent groups IV

• ### B.A.F. Wehrfritz

Queen Mary University of London, UK ## 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  and its related bounds.