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11.71 (1990)

Open

Let $A$ be a finite group with a normal subgroup $H$. A subgroup $U$ of $H$ is called an $A$-covering subgroup of $H$ if $\bigcup_{a\in A} U^a = H$. Is there a function $f : \mathbb{N} \to \mathbb{N}$ such that whenever $U < H < A$, where $A$ is a finite group, $H$ is a normal subgroup of $A$ of index $n$, and $U$ is an $A$-covering subgroup of $H$, the index $\lvert H : U \rvert \leqslant f(n)$? (We have shown that the answer is “yes” if $U$ is a maximal subgroup of $H$.)

Contributor: P. M. Neumann, C. E. Praeger

Progress

Comments of 2025: The conjecture is proved in the case where $H$ acts innately transitively on the coset space $[H : U]$ (M. Fusari, A. Previtali, P. Spiga, J. Group Theory, 27 (2024), 929–965); for $n = 3$ with $\lvert H : U \rvert \leqslant 10$ (L. Gogniat, P. Spiga, Preprint, 2025, https://arxiv.org/abs/2502.01287); and if $H = UL$ where $L$ is a minimal $A$-invariant subgroup of $H$ (M. Fusari, S. Harper, P. Spiga, Bull. Austral. Math. Soc., 2025, https://doi.org/10.1017/S00049725000176).

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