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10.61 (1986)

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Suppose that $H$ is a proper subgroup of a group $G$, $a \in H$, $a^2 \neq 1$ and for every $g \in G \setminus H$ the subgroup $\langle a, a^g \rangle$ is a Frobenius group whose complement contains $a$. Does the set-theoretic union of the kernels of all Frobenius subgroups of $G$ with complement $\langle a \rangle$ constitute a subgroup? For definitions see 6.55; see also (A. I. Sozutov, Algebra and Logic, 34, no. 5 (1995), 295–305).

Contributor: A. I. Sozutov

Progress

Editors’ comment (2005): The answer is affirmative if the order of $a$ is even (A. M. Popov, A. I. Sozutov, Algebra and Logic, 44, no. 1 (2005), 40–45) or if the order of $a$ is not 3 or 5 and the group $\langle a, a^g \rangle$ is finite for any $g \notin H$ (A. M. Popov, Algebra and Logic, 43, no. 2 (2004), 123–127).

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