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.?



Factorization of a finite group by two subsets
There are subsets $A$ and $B$ of orders $21$ and $24$ of $G=PSL(2,8)$ with $G=AB$. A naive search of course does not work. However, in trying to find first $A$ with $A^{-1}A$ being small, a good candidate to work with is $A=G\_7G\_3$, where $G\_3$ and $G\_7$ are subgroups of order $3$ and $7$. The smallest possible size of such a set $A^{-1}A$ is $48$. Finding $B$ with $AB=G$ is an exact cover problem: For each $g\in G$ set $S\_g=\{ag\;|\;a\in A\}$. Then a set $B$ has the required property if and only if the sets $S\_b$, $b\in B$, yield a disjoint union of $G$.


I can't offer a GAP code, rather a quick and naive Sage code which finds such good sets $A$ and $B$:



```

g = PSL(2,8)
n = g.order()
l = {i:x for i,x in enumerate(g)}
li = {y:x for x,y in l.items()}
g7 = g.sylow_subgroup(7)
s3 = g.subgroup([[z for z in g.conjugacy_classes_representatives()\
                  if z.order() == 3][0]])
n3 = g.normalizer(s3)
tv = [z[0] for z in g.cosets(n3)]
for g3 in [s3.conjugate(z) for z in tv]:
    A = [g(zz)*g(z) for zz in g7 for z in g3]
    AA = set([z1^(-1)*z2 for z1 in A for z2 in A])
    print len(AA)
    if len(AA) > 48:
        continue
    M = matrix(n,n,0)
    for i, x in l.items():
        for j in [li[a*x] for a in A]:
            M[i,j] = 1
    c = sage.combinat.matrices.dlxcpp.OneExactCover(M)
    if c == None:
        continue
    B = [[l[i] for i in range(n) if M[i] == row][0] for row in c]
    AB = set([a*b for a in A for b in B])
    print len(AB) == n # Check if AB = G
    break

```

One solution we get is $A=G\_7G\_3$ with $G\_7$ and $G\_3$ generated by $(3,8,6,4,9,7,5)$ and $(1,3,4)(2,7,8)(5,9,6)$, and $B=\{(),
 (1,2,4,3,5,6,7,9,8),
 (2,5,8,3,7,6,4),
 (1,2,5,7,3,6,8),
 (2,6,9,4,8,7,5),
 (1,2,7,8,6,4,5),
 (2,8)(3,7)(4,9)(5,6),
 (1,6,5,2,4,8,9,3,7),
 (1,4,6)(2,5,7)(3,8,9),
 (1,5,7,4,6,2,3,9,8),
 (2,7,6,3,4,9,5),
 (2,4)(3,6)(5,7)(8,9),
 (1,5)(3,6)(4,9)(7,8),
 (1,4,8,5,2,3,9,7,6),
 (2,9,5,7,4,3,8),
 (1,7)(2,5)(3,8)(4,9),
 (1,5,8,2,4,6,3,7,9),
 (2,3)(4,6)(5,9)(7,8),
 (1,2,6,8,4,7,9),
 (1,4,5,6,2,9,7,3,8),
 (1,4,8,6,9,3,5),
 (1,3,7,8,5,2,6,4,9),
 (1,2)(3,5)(6,9)(7,8),
 (1,3,9,8,2,4,7)\}$.


**Added:** (in view of the modification of the question) The groups of orders $24^2$ and $28^2$ are solvable by Burnside's $p^aq^b$-Theorem. Hence they don't give counterexamples, as pointed out in the answers to [this related question by the OP](https://mathoverflow.net/questions/155986/factor-subset-of-finite-group). These answers, by the way, give more promising indications where possible counterexamples may reside. (As to $PSL(2,8)$, the two potential candidates $(a,b)=(21,24)$ and $(a,b)=(12,42)$ are given there. The first doesn't exist by the result from above, and a slight modification of the above code gives $G=AB$ with $|A|=12$, $|B|=42$, so the other one doesn't exist either.)


