Is each finite group multifactorizable?

Further to Gerhard Paseman's comment, it is true that finite $p$-groups are multifactorizable when $p$ is prime (and hence finite nilpotent groups are multifactorizable). Let $G$ be a $p$-group, and let each $a_{i}$ be a power of $p$ with $\prod_{i=1}^{n}a_{i} = |G|$. We show by double induction, first on $|G|$, then on $n,$ that there are subsets $A_{1},A_{2}, \ldots , A_{n}$ such that $A_{1}A_{2} \ldots A_{n} = G$ and that $A_{1}$ may be chosen to be a subgroup of $G$ of order $a_{1}.$

If $n = 1,$ we may take $A_{1} = G,$ so suppose that $n > 1.$ If $n =2,$ take $A_{1}$ to be a subgroup of $G$ of order $a_{1}$ and take $A_{2}$ to be a right transversal to $A_{1}$ in $G.$

Suppose then that $n > 2$ and the result is established for $n-1$ (and for $p$-groups of order less than $|G|)$.

Then $G$ has a subgroup $H$ of order $a_{1}a_{2},$ and there are subsets $A_{3},A_{4}, \ldots, A_{n}$ (with each $A_{i}$ of size $a_{i})$ such that $G = HA_{3} \ldots A_{n}.$

Since $n >2$, we have $a_{1}a_{2} < |G|$ and $|H| < |G|$. By induction ( or the case $n = 2$), $H$ has a subgroup $A_{1}$ of order $a_{1}$ and there is a subset $A_{2}$ of size $a_{2}$ of $H$ such that $A_{1}A_{2} = H.$

Then $G = A_{1} \ldots A_{n}$ with $A_{1}$ a subgroup of $G$ of order $a_{1}$ and each $A_{i}$ subset of $G$ of size $a_{i}$ for $i > 1.$

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.

I've just discovered that the alternating group $A_4$ is not multifactorizable. Namely, it can not be written as the product $A_4=ABC$ of subsets $A,B,C\subset A_4$ of cardinality $|A|=2$, $|B|=3$, and $|C|=2$.

The argument is as follows. To derive a contradiction, assume that $A_4=ABC$ for some subsets $A,B,C\subset A_4$ of cardinality $|A|=2$, $|B|=3$, $|C|=2$. Observe that for any $a\in A$, $b\in B$ and $c\in C$ the equality $ABC=A_4$ implies that $$(b^{-1}a^{-1}Ab)(b^{-1}B)(Cc^{-1})=b^{-1}a^{-1}ABCc^{-1}=A_4,$$ so we can replace the sets $A,B,C$ by the sets $b^{-1}a^{-1}Ab$, $b^{-1}B$, $Cc^{-1}$ and assume that the sets $A,B,C$ contain the neutral element $e$ of $A_4$.

Then $A=\{e,a\}$ and $C=\{e,c\}$ for some $a,c\in A_4$.

We claim that $a^2=e$. Assuming that $a^2\ne e$, we would conclude that $a^3=e$. Since the sets $BC$ and $aBC\ni a$ are disjoint, $a\notin BC$. Then $a^2\in ABC=BC\cup aBC$ implies $a^2\in BC$ and hence $e=a^3\in aBC$, which is not possible as the sets $aBC$ and $eBC\ni e$ are disjoint. This contradiction shows that $a^2=e$. By analogy we can prove that $c^2=e$.

Since the sets $B\ni e$ and $aBc\ni ac$ are disjoint, $a\ne c$. Then $a,c,ac$ are the unique elements of order 2 in $A_4$.

Since the sets $B$, $aB\ni a$, $Bc\ni c$ and $aBc\ni ac$ are pairwise disjoint, the set $B$ contains no elements of order 2.

Fix any element $b\in B$ (or order 3). The group $A_4$ can be thought as the group of orientation-preserving isometries of a regular tetrahedron. Selecting a suitable enumeration of the vertices of the tetrahedron, we can assume that $a$ is the permutation $(12)(34)$ and $b$ is the cycle $(123)$. It is easy to check that $b^{-1}ab^{-1}=aba$.

One can check that $\{c,ac\}=\{bab^{-1},b^{-1}ab\}$ and $A_4\setminus\{e,a,c,ac\}=\{b,b^{-1},ab,ba,aba,ab^{-1},bab,b^{-1}a\}$.

It follows that $c=bab^{-1}$. Otherwise $c=b^{-1}ab$ and then the set $aBc\ni abb^{-1}ab=b$ is not disjoint with $B$.

Let $b'$ be the unique point of the set $B\setminus\{e,b\}$. Taking into account that the set $B$ is disjoint with $aB\cup Bc\cup aBc$, we conclude that $b'\notin\{b,ab,bc,abc\}=\{b,ab,b^{-1}ab^{-1},ab^{-1}ab^{-1}\}=\{b,ab,aba,ba\}$ and hence $b'\in\{b^{-1},ab^{-1},bab,b^{-1}a\}$.

If $b'=b^{-1}$, then $ab'c=ab^{-1}bab^{-1}=b^{-1}=b'$.

If $b'=ab^{-1}$, then $ab'c=ab^{-1}bab^{-1}=ab^{-1}=b'$.

If $b'=bab$, then $ab'c=ababbab^{-1}=abab^{-1}ab^{-1}=abaaba=ab^{-1}a=ab^{-1}ab^{-1}b=aabab=bab=b'$.

If $b'=b^{-1}a$, then $ab'c=ab^{-1}abab^{-1}=ab^{-1}ab^{-1}b^2ab^{-1}=aabab^{-1}ab^{-1}=baaba=b^{-1}a=b'$.

In all cases, we get $ab'c=b'$, which is not possible as the sets $aBc$ and $B$ are disjoint. This contradiction completes the proof.

Conclusion. Since the group $A_4$ is solvable, it answers negatively Problem 2 and (partly) Problem 1.

It remains find a finite simple group which is not multifactorizable.