Let $G$ be a group of order $n$ and $d$ a positive divisor of $n$. Is it true that there exists a subset $A$ of $G$ with $d$ elements and a subset $B$ such that $G=AB$ and $|AB|=|A||B|$ (equivalently, the product $AB$ is direct)?


Note: If $G$ has the property that there exists a subgroup of $G$ with order $d$ or $n/d$, then we can show that the answer is positive (see
[A question about finite groups (a weak version of the converse of Lagrange theorem)](https://mathoverflow.net/questions/143229/a-question-about-finite-groups-a-weak-version-of-the-converse-of-lagrange-theor)).


Now, is the answer positive for
${\rm PSL}(2,8)$, ${\rm PSL}(2,11)$, ${\rm PSL}(2,13)$, ${\rm SL}(2,11)$, ${\rm SL}(2,13)$,
${\rm PSL}(2,17)$, ${\rm A}\_7$, ${\rm PSL}(2,19)$, ${\rm A}\_5 \times {\rm A}\_5$?


About $PSL\_2(8)$ (in the last following comment, by Russ Woodroofe). If it has subgroups of indexes $6$, $14$ and $21$ with the property $P(\*,\*)$, then we can say that $PSL\_2(8)$ has $P(\*,\*)$. Note that $PSL\_2(8)$ has subgroups of indexes $6$ (that is solvable and so $P(\*,\*)$ is true for it), $14$ (that is also solvable) and $21$ (with order $84$). Now, is $P(\*,\*)$ valid for the last subgroup? (if yes, then $PSL\_2(8)$ has $P(\*,\*)$).



Factor subset of finite group
You can handle several of the groups on the list from your edit with a modification of the argument of Marty Isaacs. For the rest, you can find guidance as to what a counterexample would look like. 


Say that $G$ satisfies $P(a,b)$ if $ab = |G|$ and there are subsets $A$ and $B$ of cardinalities $a$ and $b$ such that $G = AB$. Write $P(\*,\*)$ when $G$ satisfies $P(a,b)$ for any $a,b$ with $ab = |G|$.


The following has exactly the same proof as the solvable argument.


**Lemma**: If $H$ is a subgroup of $G$ satisfying $P(a,b)$, and $[G:H] = t$, then $G$ satisfies $P(a,bt)$.


Then if $G$ has a maximal subgroup of prime index satisfying $P(\*,\*)$, then $G$ also satisfies $P(\*,\*)$. That handles $A\_5$, $PSL\_2(7)$, and $A\_5 \times A\_5$.


Moreover, if $G$ has a maximal subgroup of index $t$, then $G$ satisfies $P(a,b)$ for every pair $a,b$ with t dividing $b$ (and of course $ab=|G|$).


1. This lets us handle $A\_6$, which has maximal subgroups of index 6, 10, and 15. If $ab = 60$ then either $a$ or $b$ is divisible by one of these.

- $A\_7$ then follows, since $A\_6$ is a maximal subgroup of $A\_7$ having (prime) index 7.

- Similar (but slightly more involved) arguments to those for $A\_6$ show that $A\_8$ satisfies $P(\*,\*)$.

- The first open case is $PSL\_2(8)$, which has order $504 = 2^3 \cdot 3^2 \cdot 7$, and maximal subgroups with indices 9, 28, and 36. Since every subgroup satisfies $P(\*,\*)$, the possible places where $PSL\_2(8)$ could fail $P(\*,\*)$ are $P(12, 42)$ and $P(21, 24)$.


By the way, [GAP](http://www.gap-system.org/) will calculate the relevant data for these arguments for you pretty easily. The following commands will do it (replace $G$ with your favorite group):  

`G:=PSL(2,7);; Print(Size(G), " = ", FactorsInt(Size(G)), "\n"); Set(List(MaximalSubgroups(G), x->Index(G, x)));`


