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19.35 (2018)

Solved

Let $G$ be a finite group of order $n$. Is it true that for every factorization $n = a_1 \cdots a_k$ there exist subsets $A_1, \dots, A_k$ such that $|A_1| = a_1, \dots, |A_k| = a_k$ and $G = A_1 \cdots A_k$?

Contributor: M. H. Hooshmand

Progress

No, it is not true. A counterexample with $k = 3$ is given by the alternating group on 4 letters $G = A_4$ and $(a_1, a_2, a_3) = (2, 3, 2)$. (G. M. Bergman, Letter of 19 December 2019, https://math.berkeley.edu/~gbergman/papers/gp_factzn.pdf.)

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