# Normal intermediate subgroup and normal core

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 $$(h_1,g_2).(e,g_2^{-1}) = (h_1,e) \in L$$.

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.

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).