In a finite group $G$, a commutator $[x,y]$ is called a coprime commutator if $(|x|,|y|)=1$. In an article of Shumyatsky, he defines a certain notion involving these coprime commutators which he in turn uses to investigate some structural properties of solvable groups.
Recall that the derived series of a group is the following: $G^{(0)}\supseteq G^{(1)}\supseteq G^{(2)}\supseteq \cdots\;$ where $G^{(0)}=G$ and $G^{(n)}=[G^{(n-1)},G^{(n-1)}]$. Alternatively, let $\delta_0=x_1$ and $\delta_k=[\delta_{k-1}(x_1,x_2,\ldots,x_{2^{k-1}}),\delta_{k-1}(x_{2^{k-1}+1},\ldots,x_{2^k})]$. Then $G^{(k)}=\delta_k(G)$ is precisely the verbal subgroup of $G$ induced by the word $\delta_k$.
Shumyatsky defines the following: Every element of $G$ is a $\delta_0^{*}$-commutator. Let $k\geq 1$. Let $Y$ be the set of all elements of $G$ consisting of all powers of $\delta_{k-1}^*$-commutators. An element is a $\delta_k^{*}$ commutator if it is of the form $[x,y]$ where $x,y\in Y$ and $(|x|,|y|)=1$. Define $\delta_k^{*}(G)$ as the subgroup of $G$ generated by $\delta_k^{*}$-commutators. He goes on to prove the following:
A finite group is solvable of fitting height atmost $k$ if and only if $\delta_{k}^*(G)=1$.
My question is on the definition of $\delta_k^{*}(G)$. We could have defined it as follows: $\delta_0^{*}(G)=G$. $\delta_1^*(G)$ is the subgroup generated by all coprime commutators in $G^{(1)}=[G,G]$ and in general $\delta_k^{*}(G)$ is the subgroup generated by the coprime commutators of $G^{(k)}$. Is this the same as what he defines as $\delta_k^*(G)$. I am confused with the set $Y$ being taken as the power of $\delta_{k-1}^*$-commutators and what is actually going on there. Any kind of help will be highly appreciated. Thank you.