The following GAP-output
gap> Display([G,H,K]);
[ Group( [ (3,4)(5,6), (1,2,3)(4,5,7) ] ), Group( [ (1,6,7)(2,4,3), (1,3)(4,7), (2,6)(4,7), (1,4)(3,7) ] ),
Group( [ (1,5,2,7,4,3,6) ] ) ]
gap> Display(G);Display(H);Display(K);List([G,H,K],IdGroup);List([G,H,K],StructureDescription);
Group( [ (3,4)(5,6), (1,2,3)(4,5,7) ] )
Group( [ (1,6,7)(2,4,3), (1,3)(4,7), (2,6)(4,7), (1,4)(3,7) ] )
Group( [ (1,5,2,7,4,3,6) ] )
[ [ 168, 42 ], [ 24, 12 ], [ 7, 1 ] ]
[ "PSL(3,2)", "S4", "C7" ]
gap> s:=List([]); for x in H do for y in K do s:=Union(s,List([x*y])); od;od;AsSortedList(s)=AsSortedList(G);
[ ]
true
gap>
shows that the groups $H$ and $K$ have the property that every element in $G$ can be uniquely written as a product $hk$ with $h\in H,k\in K$.
And we also have $H\cap K={e}$.
So, in some sense the simple group $G=PSL(3,2)$ should be a product of $H$ and $K$, but it cannot be a semidirect product because $H$ and $K$ no normal subgroups ($G$ is even simple).
I guess it is a non-split extension (correct me if I am wrong).
My questions :
Under which considitions can a finite group $G$, not being a semidirect product of non-trivial groups, be decomposed in this way ? Is it enough that subgroups with coprime order $m$ and $n$ and $mn=|G|$ exist ?
Can the group $G$ in the given example be written as $H.K$ ?