**Definition.** A finite group $G$ is *factorizable* if for any positive integer numbers $a,b$ with $ab=|G|$ there are subsets $A,B\subset G$ of cardinality $|A|=a$ and $|B|=b$ such that $AB=G$.

Problem 1.Is each finite group factorizable?

As I understood from these MO-posts (1, 2, 3), this problem is wide open and there is no intuition if it is true or not. So, we can ask a related

Problem 2.Which finite groups are factorizable?

The class of factorizable groups has a nice 3-space property that can be formulated in terms of bifactorizable subgroups.

A subgroup $H$ of a group $G$ is called *bifactorizable* if $H$ is factorizable and for any positive integer numbers $a,b$ with $ab=|G|$ there are sets $A,B\subset G$ such that $AHB=G$, $|A|\cdot |H|\cdot |B|=|G|$, $|A|$ divides $a$ and $|B|$ divides $b$.

Theorem.A finite group $G$ is factorizable if $G$ contains a bifactorizable subgroup $H$.

*Proof.* Given positive integers $a,b$ with $ab=|G|$ use the bifactorizability of $H$ to find subsets $A_1,B_1\subset G$ such that $AHB=G$, $|A_1|\cdot |H|\cdot |B_1|=|G|$, $|A_1|$ divides $a$, and $B_1$ divides $b$. The factorizability of $H$ yields two sets $A_2,B_2\subset H$ of cardinality $|A_2|=a/|A_1|$ and $|B_2|=b/|B_1|$ such that $A_2B_2=H$. Then the sets $A=A_1A_2$ and $B=B_2B_1$ have cardinality $|A|\le a$, $|B|\le b$ and
$AB=A_1A_2B_2B_1=A_1HB_1=G$. It follows from $ab=|G|=|A|\cdot|B|\le ab$ that $|A|=a$ and $|B|=b$. $\square$

It is easy to see that a subgroup $H$ of a group $G$ is bifactorizable if it is factorizable and has prime index in $G$. Moreover, as was observed by M.Farrokhi D.G. in his answer to this post, a subgroup $H$ of a group $G$ is bifactorizable if $H$ is factorizable and the index of $H$ in $G$ is a prime power $p^k$ such that $p^{2k-1}$ divides $|G|$.

A normal subgroup $H$ of a group $G$ is factorizable if both groups $H$ and $G/H$ are factorizable. This implies

Corollary.A finite group $G$ is factorizable if $G$ contains a bifactorizable subgroup $H$ with factorizable quotient $G/H$.

This corollary reduces Problems 1,2 to studying the factorizability of finite simple groups. According to the classification of finite simple groups, each finite simple group is either cyclic of prime order, or alternating, or belongs to 16 families of groups of Lie type or is one of 27 sporadic groups.

Among these families only the factorizability of finite cyclic groups is trivially true.

Problem 3.Is each alternating group $A_n$ factorizable?

It may happen that the argument of Ilya Bogdanov from his answer to this MO-problem can be helpful here. On the other hand, it can be shown that the subgroup $A_n$ is bifactorizable in $A_{n+1}$ if and only if $A_n$ is factorizable and $n\ne 3$ is a power of a prime. It follows from the answer to this MO-question that the subgroup $A_3$ is not bifactorizable in $A_4$.

Problem 4.Is any hope to prove that some infinite family of simple groups of Lie type consists of factorizable groups?