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 $P(*,*)$).