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Let $n$ be a positive integer or infinity (denoted $\infty$), $k$
a positive integer. We denote by $\Omega_k(n)$ the class of groups
$G$ such that, for every subset $X$ of $G$ of cardinality $n+1$,
there exist distinct elements $x,y\in X$ and integers $t_0, t_1,
\ldots, t_k$ such that $[x_0^{t_0}, x_1^{t_1}, \ldots,
x_k^{t_k}]=1$, where $x_i\in \{x,y\}, i=0,1,\ldots, k,x_0\neq
x_1$. If the integers $t_0, t_1, \ldots, t_k$ are the same for any
subset $X$ of $G$, we say that $G$ is in the class
$\overline{\Omega}_k(n)$. The class ${\cal U}_k(n)$ is defined
exactly as $\Omega_k(n)$ with the additional conditions
$x_i^{t_i}\neq 1$. Let $t_2,t_3,\ldots, t_k$ be fixed integers. We
denote by $\overline{{\cal W}}^*_k$ the class of all groups $G$
such that for any infinite subsets $X$ and $Y$ there exist $x\in
X, y\in Y$ such that $[x_0, x_1, x_2^{t_2}..., x_k^{t_k}]=1$,
where $x_i\in \{x,y\}, x_0\neq x_1, i=2,3,\ldots, k$. Here we
prove that
\begin{itemize}
\item[1] If $G\in {\cal U}_k(2)$ is a finitely generated soluble group. then $G$ is nilpotent.
\item[2] If $G\in\Omega_k(\infty)$ is a finitely generated soluble group. then $G$ is nilpotent-by-finite.
\item[3] If $G\in\overline{\Omega}_k(n)$, $n$ a positive integer, is a finitely generated residually finite group. then $G$ is nilpotent-by-finite.
\item[4] If $G$ is an infinite $\overline{{\cal W}}_k^*$-group in which every non-trivial finitely generated subgroup has a non-trivial finite quotient. then $G$ is nilpotent-by-finite.
\end{itemize}
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