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| Main Author: | |
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| Format: | Preprint |
| Published: |
2021
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| Subjects: | |
| Online Access: | https://arxiv.org/abs/2102.08605 |
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Table of Contents:
- A finite group $G$ is called $k$-factorizable if for every ordered factorization $|G|=a_1\cdots a_k$ into integers each greater than $1$ there exist subsets $A_1,\dots,A_k\subseteq G$ such that $|A_i|=a_i$ for each $i$ and $G=A_1\cdots A_k$. The main results are as follows. 1. For every integer $k\geq3$ there exists a finite group $G$ such that $G$ is not $k$-factorizable. 2. Let $G$ be a finite group of order $4m$. If a Sylow $2$-subgroup of $G$ is elementary abelian, all involutions of $G$ are conjugate, and the centralizer of every involution has a normal Sylow $2$-subgroup, then $G$ has no factorization of the form $G=ABC$ with $|A|=|C|=2$ and $|B|=m$. 3. Only $8$ groups of order at most $100$ fail to be $k$-factorizable for some $k$.