The coprime commutators $\gamma_j^*$ and $\delta_j^*$ were recently
introduced as a tool to study properties of finite groups that can be expressed
in terms of commutators of elements of coprime orders. They are defined as
follows. Let $G$ be a finite group. Every element of $G$ is both a
$\gamma_1^*$-commutator and a $\delta_0^*$-commutator. Now let $j\geq 2$ and
let $X$ be the set of all elements of $G$ that are powers of
$\gamma_{j-1}^*$-commutators. An element $g$ is a $\gamma_j^*$-commutator if
there exist $a\in X$ and $b\in G$ such that $g=[a,b]$ and $(|a|,|b|)=1$. For
$j\geq 1$ let $Y$ be the set of all elements of $G$ that are powers of
$\delta_{j-1}^*$-commutators. The element $g$ is a $\delta_j^*$-commutator if
there exist $a,b\in Y$ such that $g=[a,b]$ and $(|a|,|b|)=1$. The subgroups of
$G$ generated by all $\gamma_j^*$-commutators and all $\delta_j^*$-commutators
are denoted by $\gamma_j^*(G)$ and $\delta_j^*(G)$, respectively. For every
$j\geq2$ the subgroup $\gamma_j^*(G)$ is precisely the last term of the lower
central series of $G$ (which throughout the paper is denoted by
$\gamma_\infty(G)$) while for every $j\geq1$ the subgroup $\delta_j^*(G)$ is
precisely the last term of the lower central series of $\delta_{j-1}^*(G)$,
that is, $\delta_j^*(G)=\gamma_\infty(\delta_{j-1}^*(G))$.
In the present paper we prove that if $G$ possesses $m$ cyclic subgroups
whose union contains all $\gamma_j^*$-commutators of $G$, then $\gamma_j^*(G)$
contains a subgroup $\Delta$, of $m$-bounded order, which is normal in $G$ and
has the property that $\gamma_{j}^{*}(G)/\Delta$ is cyclic. If $j\geq2$ and $G$
possesses $m$ cyclic subgroups whose union contains all
$\delta_j^*$-commutators of $G$, then the order of $\delta_j^*(G)$ is
$m$-bounded.Comment: Final version, referee's suggestions adde