Factor subset of finite group 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)).
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 $P(*,*)$).
 A: 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 $ab=|G|$).


*

* 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 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)));
A: This is true for solvable groups, at least. We are given a group $G$ of order $ab$, where $a$ and $b$ are integers, and we want subsets $A$ and $B$ of $G$ such that $|A| = a$,
$|B| = b$ and $G = AB$. We work by induction on $|G|$ to prove that $A$ and $B$ exist if $G$ is solvable. We can certainly assume that $|G| > 1$. Since $G$ is solvable, it has a subgroup $H$ of prime index $p$, and we can assume without loss that $p$ divides $b$. Then $|H| = (a)(b/p)$, so by the inductive hypothesis, we can write $H = AX$, where $|A| = a$ and $|X| = b/p$. Now let $T$ be a set of representatives for the right cosets of $H$ in $G$. Then $|T| = p$ and
$G = HT = AXT$. Let $B = XT$, so $AB = G$. Also, $|B| = |XT| \le |X||T| = (b/p)p = b$. Equality must hold here since $AB = G$.
