21.18 (2026)
OpenSuppose that $G$ is a finite group, and $A_1, A_2, A_3$ are subsets of $G$ such that the multiplication map $A_1 \times A_2 \times A_3 \to G$ is bijective. Must the subgroup $\langle A_2 \rangle$ generated by $A_2$ have order divisible by the cardinality $|A_2|$? This is Question 8 in (G. M. Bergman, J. Iranian Math. Soc., 1 (2020), 157–161).
It is known (ibid.) that the corresponding statement is true for the subgroups $\langle A_1 \rangle$ and $\langle A_3 \rangle$. Moreover, $|A_2|$ will at least divide the order of the least subgroup containing $A_2$ and closed under conjugation by members of $A_1$, and similarly of the least subgroup containing $A_2$ and closed under conjugation by members of $A_3$.
Progress
*No, it need not (M. I. Kabenyuk, Preprint, 2021, https://arxiv.org/abs/2102.08605).
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