Kronecker classes, normal coverings and chief factors of groups

For a group $G$, a subgroup $U\le G$ and a group $\mathrm{I}\mathrm{n}\mathrm{n}(G)\le A\le \mathrm{A}\mathrm{u}\mathrm{t}(G)$, we say that $U$ is an $A$-covering group of $G$ if $G={\bigcup }_{a\in A}{U}^a$. A theorem of Jordan (1872) implies that if $G$ is a finite group, $A=\mathrm{I}\mathrm{n}\mathrm{n}(G)$ and $U$ is an $A$-covering group of $G$, then $U=G$. Motivated by a question concerning Kronecker classes of field extensions, Neumann and Praeger (1988) conjectured that, more generally, there is an integer function $f$ such that if $G$ is a finite group and $U$ is an $A$-covering subgroup of $G$, then $\mathrm{\mid }G:U\mathrm{\mid }\le f(\mathrm{\mid }A:\mathrm{I}\mathrm{n}\mathrm{n}(G)\mathrm{\mid })$. A key piece of evidence for this conjecture is a theorem of Praeger (1994), which asserts that there is a two-variable integer function $g$ such that if $G$ is a finite group and $U$ is an $A$-covering subgroup of $G$, then $\mathrm{\mid }G:U\mathrm{\mid }\le g(\mathrm{\mid }A:\mathrm{I}\mathrm{n}\mathrm{n}(G)\mathrm{\mid },c)$ where $c$ is the number of $A$-chief factors of~$G$. Unfortunately, the proof of this result contains an error. In this paper, using a different argument, we give a correct proof of this theorem.