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](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-50/issue-2/Normal-intermediate-subfactors/10.2969/jmsj/05020469.full))


**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](https://mathoverflow.net/questions/156311/jordan-h%C3%B6lder-theorem-for-subfactors/158685#158685) 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.



Normal intermediate subgroup and normal core
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](http://homepages.ulb.ac.be/~tconnor/atlaslat/alt6.pdf)) and normal intermediate subgroups, thanks to a SAGE-GAP computation (see [here](http://groupprops.subwiki.org/wiki/Subgroup_structure_of_alternating_group:A6) 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](https://mathoverflow.net/questions/156311/jordan-holder-theorem-for-subfactors/159112#159112)).


