Def. If $X\subseteq G$, we say that $X$ is $k$-large in $G$ if the intersection of any $k$ left translates of $X$ is non-empty. (See the notion of largeness which was introduced in "Largeur et nilpotence".)
Let $G$ be a soluble group with derived length $n$, $p$ be a prime number and $\phi\in\text{Aut}(G)$ be of order $p$. Assume that
- The set $\{x\in G: \prod_{k=0}^{p-1}x^{\phi^k}=1\}$ is $2^{n+1}$-large,
- $G/G^{(n-1)}$ is nilpotent and its nilpotency class is $2^n$,
- For all $x\in G$, $\prod_{k=0}^{p-1}x^{\phi^k}\in G^{(n-1)}$.
Then is $G$ nilpotent (of nilpotency class $2^{n+1}$)?
($G^{(r)}$ stands for the $r$th derivation of $G$)
