On factorizations of finite groups

Mikhail I. Kabenyuk

Published 2021-02-17Version 1

Let $G$ be a finite group and let $\{A_1,\ldots,A_k\}$ be a collection of subsets of $G$ such that $G=A_1\ldots A_k$ is the product of all the $A_i$ and ${\rm card}(G)={\rm card}(A_1)\ldots{\rm card}(A_k)$. We shall write $G=A_1\cdot\ldots\cdot A_k$, and call this a $k$-fold factorization of the form $({\rm card}(A_1),\ldots,{\rm card}(A_k))$. We prove that for any integer $k\geq3$ there exist a finite group $G$ of order $n$ and a factorization of $n=a_1\ldots a_k$ into $k$ factors other than one such that $G$ has no $k$-fold factorization of the form $(a_1,\ldots,a_k)$.