Every factorization $|G|=ab$ realized by subsets: must $G=AB$ with $|A|=a,\ |B|=b$? (Kourovka 20.37, Hooshmand)
Statement
Let $G$ be a finite group and write $|G|=ab$ for positive integers $a,b$. Must there exist subsets $A,B\subseteq G$ with $|A|=a$ and $|B|=b$ such that $G=AB$, i.e. every $g\in G$ can be written $g=\alpha\beta$ with $\alpha\in A,\ \beta\in B$? Decide this for all finite groups $G$ and all factorizations $|G|=ab$, or exhibit a triple $(G,a,b)$ for which no such subset pair exists.
Acceptance. FULLY RESOLVES: either (a) a machine-checkable counterexample — a specific finite group $G$ (Cayley table or generators) and factorization $|G|=ab$ with an exhaustive certificate that no pair $A,B$ of sizes $a,b$ has $AB=G$ (a finite, decidable search for small $G$); or (b) a proof for all finite groups, reduced via Bildanov-Goryachenko-Vasil'ev to the residual simple groups and closed uniformly. ADVANCES: verify the property for further nonabelian simple groups without prime-index subgroups (for each factorization $|G|=ab$, an explicit $(A,B)$ with $AB=G$), e.g. new $\mathrm{PSL}_2(q)$ or sporadic cases beyond orders 168, 360. Provide the sets and a script that recomputes $AB$.
Background
Problem 20.37 of the Kourovka Notebook (M. H. Hooshmand, 20th issue 2022; formerly 19.35): the two-factor case of Hooshmand's 'logarithmic factorization' question. The general $k\ge 3$ version ($|G|=n_1\cdots n_k$, seek subsets $A_i$ of size $n_i$ with $G=A_1\cdots A_k$) is FALSE — G. Bergman observed that $G=A_4$ with $(n_1,n_2,n_3)=(2,3,2)$ admits no such factorization — so the OPEN content is precisely the two-subset case $k=2$. Best known: Bildanov, Goryachenko & Vasil'ev (Sib. Electron. Math. Rep. 17 (2020), 683-689) proved any minimal counterexample to the two-subset case must be a nonabelian simple group with no subgroup of prime index; Hooshmand (Comm. Algebra 49 (2021)) established basic reductions; Kabenyuk (arXiv:2401.09306, 2024) verified the property for the simple groups of order 168 and 360. Whenever a subgroup of order $a$ or $b$ exists the coset structure gives $A,B$ immediately, so the crux is the residual simple groups lacking such subgroups. An attacker needs: (i) a covering/subset-design search (greedy + SAT/ILP over the Cayley table) that for a given $(G,a,b)$ either constructs $(A,B)$ or certifies none exists; (ii) for the general theorem, the CFSG list of simple groups and their maximal-subgroup index data.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Unsolved Problems in Group Theory: The Kourovka Notebook (No. 21) | arxiv |
| REF-02 | Factorizations of simple groups of order 168 and 360 (Kabenyuk, 2024) | arxiv |
| REF-03 | A note on factorizations of finite groups | arxiv |
Investigations · 0
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