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.


1 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
sage: P3==P4

$KL=A_6$ 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.