Normal intermediate subgroup and normal core

Question

Let $G$ be a finite group and $H$ a subgroup.

The normal core of $H$ in $G$ is $core_G(H) := \bigcap_{g \in G}g^{-1}Hg$

Definition: $K$ is a normal intermediate subgroup of the inclusion $(H \subset G)$ if $H \subset K \subset G$, and $$\forall g \in G \text{ , } KgH=HgK$$ (This definition is motivated by the prop.3.3 p476 of this paper)

Examples : If $H=\{ e \}$ then $K$ is a normal intermediate subgroup iff $K$ is a normal subgroup of $G$.
$H_i$ and $G_i$ are obviously normal intermediate subgroups of the inclusion $(H_i \subset G_i)$, and
$H_1 \times G_2$ and $G_1 \times H_2$ are normal intermediate subgroups of $(H_1 \times H_2 \subset G_1 \times G_2)$.

Let $L$ and $K$ be normal intermediate subgroups of $(H \subset G)$, then $\langle K , L \rangle = KL=LK$.

Question : Is it true that $\forall k \in K$, $k.core_{KL}(K) \cap L \neq \emptyset$ ?

Remark: it's true for all the examples above: it's obvious if $H=\{ e \}$ , or if $\{ K , L \} \subset \{ H, G \} $,
and if $G = G_1 \times G_2$, $H = H_1 \times H_2$, $K=H_1 \times G_2$ and $L = G_1 \times H_2$, then $KL=G$ and $\{ e \} \times G_2 \subset core_{KL}(K)$, so if $k \in K$, $k=(h_1,g_2)$ and $(h_1,g_2).(e,g_2^{-1}) = (h_1,e) \in L$.

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Motivation: This question is (for me) the last step for getting a Jordan-Hölder theorem generalized to the inclusions of groups, as explained here in the context of group-subgroup subfactors.

Definition: An inclusion $(H \subset G)$ is simple if it admits no non-trivial normal intermediate subgroup.

Examples: The maximal inclusions are obviously simple, and if $H=\{ e \}$, it's simple iff $G$ is simple.
$(\mathbb{Z}_3 \subset A_5)$ is an example of simple inclusion which is neither maximal nor trivial.

Answer 1

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) and normal intermediate subgroups, thanks to a SAGE-GAP computation (see here for the generators):

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

$KL=A_6$ is simple, so $core_{KL}(K) = \{ e \}$, but $K \not\subset L$.
Then $\exists k \in K$ such that $k.core_{KL}(K) \cap L = \emptyset.$

Remark: Nevertheless, the inclusion $(D_{10} \subset A_6)$ checks Jordan-Hölder (see here).