TITLE: Is each finite group multifactorizable?
QUESTION [10 upvotes]: 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 $i\le n$.
In this case we shall write that the group $G$ is $a_1{\times}\cdots{\times}a_n$-factorizable.

It can be shown that each finite Abelian group is multifactorizable.

Problem 1. Is each finite (simple) group multifactorizable?

As was observed by Geoff Robinson in his answer to this question, each finite nilpotent group is multifactorizable.

Problem 2. Is each finite solvable group multifactorizable?

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.

Problem 3. Is the alternating group $A_5$ multifactorizable? In particular, is $A_5$ $2{\times}15{\times}2$-factorizable?


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 for more information.

REPLY [6 votes]: 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.