A normal intermediate subgroup in $B_3$ lattice with an additional index condition?

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This post is a sequel of: A normal intermediate subgroup in $B_3$ lattice?

Let $G$ be a finite group and $H$ a subgroup.
Let $\mathcal{L}(H \subset G )$ be the lattice of intermediate subgroups between $H$ and $G$.
An intermediate subgroup $H \subset K \subset G$ is a normal intermediate subgroup if $HgK = KgH$, $\forall g \in G$

Suppose that $\mathcal{L}(H \subset G ) \sim B_3$, as follows:

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Let $K_i$ ($i=1,2,3$) be the minimal overgroups of $H$ such that:
$$ [G:K1] + [G:K2] + [G:K3] > [G:H]$$

Question: Is there $i$ such that $K_i$ is a normal intermediate subgroup of $(H \subset G )$ ?

Remark: I've checked it's true by GAP, for $\vert G \vert <512$.

Remark: The counter-example given in the previous post does not check the inequality.

Motivation: I try to prove a result in the theory of subfactors with a distributive lattice of intermediates. If the lattice has $<8$ vertices then I have a proof. But now for the lattice $B_3$ I've not (yet) a proof, so that I'm looking for counterexamples in the group theory. If the last inequality is not checked then the subfactor result is automatically true. Now the condition asked in the question is a sufficient condition for my result, so any counter-example could be useful.

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No, the first counter-example is $(D_8 \subset S_6)$:

gap> G:=SymmetricGroup(6);;
gap> H:=Group( [ (1,3,2,4)(5,6), (1,3)(2,4), (1,2)(3,4) ] );;
gap> IsomorphismGroups(H,DihedralGroup(8));
[ (1,3,2,4)(5,6), (1,3)(2,4), (1,2)(3,4) ] -> [ f2, f1*f2, f3 ]
gap> int:=IntermediateSubgroups(G,H);
gap> int.inclusions;
[ [ 0, 1 ], [ 0, 2 ], [ 0, 3 ], [ 1, 4 ], [ 1, 5 ], [ 2, 4 ], [ 2, 6 ], 
  [ 3, 5 ], [ 3, 6 ], [ 4, 7 ], [ 5, 7 ], [ 6, 7 ] ]
gap> sub:=int.subgroups;;    
gap> List(sub,K->Size(K));
[ 16, 24, 24, 48, 48, 360 ]
gap> [Size(H),Size(G)];
[8,720]
gap>  List(sub,K->IsNormalIntermediate(G,H,K));
[ false, false, false, false, false, true ]

Remark: It's the only counter-example for $\vert G \vert < 768$ (except $512$).