I want to write a GAP program for checking the following question.

Let $G$ be a given finite group with order $n$. Is it true that for every factorization $n=ab$ there exist subsets $A$ and $B$ such that $|A|=a$, $|B|=b$ and $G=AB$?

We have many candidates for a counterexample such as $PSL(2,8)$, $PSL(2,11)$, $PSL(2,13)$, $SL(2,11)$, $SL(2,13)$, $PSL(2,17)$ and $PSL(2,19)$, and need to check them.

About $PSL_2(8)$, if it has a subgroup of index $21$ with the property (in the above question), then we can say that the answer for $PSL_2(8)$ is positive. Can some body check it by GAP?

Any help would be highly appreciated.

Now, can any one write a GAP code for $G=PSL_2(8)$, $a=21$ and $b=24$? (i.e. Are there subsets $A$ and $B$ such that $|A|=21$, $|B|=24$ and $PSL_2(8)=AB$? or equivalently, are there subsets $A$, $B$ of $PSL_2(8)$ such that $|A|=21$, $|B|=24$ and $|A^{-1}A\cap BB^{-1}|=1$?)

Considering the answer from Peter Mueller, we need to **try some other cases** for getting a counterexample. With do attention to https://math.stackexchange.com/questions/885778/subgroups-of-groups-of-order-36 and https://math.stackexchange.com/questions/882859/groups-of-order-n2-that-have-no-subgroup-of-order-n?lq=1 , I propose the following groups:

Groups of order $n=m^2$ with no subgroup of order $m$ (so $|G|=mm, a=b=m$). As one can see in the second link $n\geq 24^2$ ($m\geq 24$).

Now, is it true for every group $G$ of order $n=24^2$ and $a=b=24$? What about $n=28^2$, $n=30^2$, etc.?

subgroupsof $G$. $\endgroup$ – Chip Eastham Jul 24 '14 at 10:20