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> **Definition.** A finite group $G$ is called *multifactorizable* if for any positive integer numbers $a\_1,\dots,a\_n$ with $a\_1\cdots a\_n=|G|$ there are subsets $A\_1,\dots,A\_n\subset G$ such that $A\_1\cdots A\_n=G$ and $|A\_i|=a\_i$ for all $i\le n$.
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> In this case we shall write that the group $G$ is *$a\_1{\times}\cdots{\times}a\_n$-factorizable.*
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It can be shown that each finite Abelian group is multifactorizable.



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> **Problem 1.** Is each finite (simple) group multifactorizable?
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As was observed by Geoff Robinson in his answer to this question, each finite nilpotent group is multifactorizable.



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> **Problem 2.** Is each finite solvable group multifactorizable?
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**Added in Edit.** It turns out that the alternating (solvable) group $A\_4$ is not multifactorizable, more precisely, $A\_4$ is not $2{\times}3{\times}2$-factorizable.


Now it remains to find an example of a finite simple group which is not multifactorizable.



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> **Problem 3.** Is the alternating group $A\_5$ multifactorizable? In particular, is $A\_5$ $2{\times}15{\times}2$-factorizable?
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**Added in Edit 2.** By computer calculations, @Fracois Brunault proved that the alternating group $A\_5$ is not multifactorizable. More precisely, the group $A\_5$ is not $2{\times}3{\times}5{\times}2$-factorizable.


**Added in Edit 3.**
On the other hand, as was remarked by @Gro-Tsen, it is an open problem (of minimal logarithmic signature) if any finite (simple) group $G$ can be written as the product $G=A\_1\cdots A\_n$ of subsets $A\_i\subset G$ whose cardinality $|A\_i|$ is a prime number or 4 such that $|G|=|A\_1|\cdots|A\_n|$. This problem is resolved for some classes of finite simple groups, see [this paper](http://mcccc.sites.unlv.edu/22/Abstract-Singhi1.pdf) for more information.



Is each finite group multifactorizable?
I wrote a Magma procedure to test whether $A\_5$ is multifactorizable. A brute force search does not seem feasible, so I used the ideas in Taras Banakh's answer and in the comments.


Let $A\_5 = ABCD$ with $|A|=2$, $|B|=3$, $|C|=5$, $|D|=2$. We may assume $A,B,C,D$ all contain the identity element. Then $A=\{e,a\}$ and $D=\{e,d\}$ where $a$ and $d$ distinct elements of order 2. Applying an automorphism of $A\_5$ and using Taras's argument, we may reduce to the case $a=(1 2)(3 4)$ and $d=(2 3)(4 5)$.


I loop over all possible subsets $B$ of $A\_5$ of size 3 containing $e$ such that $\# ABD = 12$ (there are 1248 such subsets). For each such $B$, I compute the list of all subsets of the form $ABcD$ with $c \in A\_5$ which are disjoint from $ABD$ and are of size 12 (typically, there are about ten such subsets) and I test whether any of them add up to $A\_5$. It turns out that there is no solution, so the group $A\_5$ is not multifactorizable.


