A second isomorphism theorem for action on cosets II Let $G$ be a finite group, and $K$, $L$ subgroups of $G$ and $H=K \cap L$ such that:  


*

*$G = \langle K,L \rangle$.  

*$\forall g \in G$ : $HgK=KgH$ and $HgL=LgH$


Remark: These assumptions imply that $G=KL=LK$, but the converse is false,
 because the example of Jack Schmidt here doesn't check the second assumption.
Let $\Omega = G/K$ and $\pi: G \to S_{\Omega}$ the canonical action on cosets.  

Question: Is it true that $\forall g \in G$ $\exists l \in L $ such that $\pi(g)=\pi(l)$ ?   

If yes, then $\forall k \in K$ $\exists s \in K \cap L $ such that $\pi(k)=\pi(s)$, because $ker(\pi) \subset K$.
 A: No, $(D_{10} \subset A_6)$ gives a counterexample.   
It has exactly two non-trivial intermediate subgroups $K$ and $L$, each isomorphic to $A_5$ (see here). They  check the assumptions, thanks to a SAGE-GAP computation (see generators here):  
sage: G=AlternatingGroup(6)
sage: H=G.subgroup([(1,2,3,4,5),G("(2,5)(3,4)")])
sage: K=G.subgroup([(1,2,3,4,5),(1,2,3)])
sage: L=G.subgroup([(1,2,3,4,5),G("(1,4) (5,6)")])
sage: P1=[Set([G(i)*k*G(j) for i in H for j in K]) for k in G]
sage: P2=[Set([G(j)*k*G(i) for i in H for j in K]) for k in G]
sage: P3=[Set([G(i)*k*G(j) for i in H for j in L]) for k in G]
sage: P4=[Set([G(j)*k*G(i) for i in H for j in L]) for k in G]
sage: P1==P2
True
sage: P3==P4
True

Now $KL=A_6$ is simple, so $ker(\pi) = \{ e \}$, but $L \subsetneq G $.
Then $\exists g \in G$ such that $\forall l \in L$ then  $\pi(g) \neq \pi(l)$   
Remark: The assumptions are not stronger enough for being able to generalize the second isomorphism theorem to  actions on cosets.   We have to find a natural additional assumption.  
