Unique intermediate subgroup and double coset relation II

Question

Let $G$ be a group and $H$ a subgroup such that there is a unique (non-trivial) intermediate subgroup $K$ (i.e. $H < S < G$ implies $S=K$).

Question: Is there $\alpha \ge 1$ such that if $1/\alpha \le \frac{[G:K]}{[K:H]} \le \alpha$ then $HgK=KgH$, $\forall g \in G$ ?

Experiment : If $[G:H] \le 30$ then it's true for all $\alpha \ge 1$ (see this comment of Derek Holt).

If it's true in general, let $\alpha_M$ be the maximum of the allowed $\alpha$.
Thanks to this answer we know that $\alpha_M \le 55/2$.

What do we know about $\alpha_M$ ? (a better upper-bound? integer, rational or irrational? its value?)

Answer 1

No, see this comment of Derek Holt announcing a counterexample for $\alpha=1$

Counter-example: << $A_{10}$ with a subgroup of index $120$ (the setwise stabilizer of three points), which itself has a maximal subgroup of index $120$ >>.

Some details:

gap> G:=AlternatingGroup(10);   
Alt( [ 1 .. 10 ] )
gap> H:=Group([ (8,9,10), (1,2,3,4,5,6,7), (2,3,5)(4,7,6), (2,7)(3,6)(4,5)(9,10) ]);
Group([ (8,9,10), (1,2,3,4,5,6,7), (2,3,5)(4,7,6), (2,7)(3,6)(4,5)(9,10) ])
gap> IntermediateSubgroups(G,H);
rec( subgroups := [ Group([ (2,7)(3,6)(4,5)(9,10), (2,3,5)(4,7,6), (1,2,3,4,5,6,7), (8,9,10), (6,7)(9,10) ]) ], 
  inclusions := [ [ 0, 1 ], [ 1, 2 ] ] )
gap> K:=Group([ (2,7)(3,6)(4,5)(9,10), (2,3,5)(4,7,6), (1,2,3,4,5,6,7), (8,9,10), (6,7)(9,10) ]);
Group([ (2,7)(3,6)(4,5)(9,10), (2,3,5)(4,7,6), (1,2,3,4,5,6,7), (8,9,10), (6,7)(9,10) ])
gap> Size(G); Size(K); Size(H);
1814400
15120
126
gap> IsNormalIntermediate(G,H,K);
false

Using the following function:

IsNormalIntermediate:=function(G,H,K)
    local D1,D2,s,i,j,E1,E2,c;
    D1:=DoubleCosets(G,H,K);
    s:=Size(DoubleCosets(G,H,K));
    c:=0;
    if s=Size(DoubleCosets(G,K,K)) then
        return true;
    else
        D2:=DoubleCosets(G,K,H);
        for i in [1..s] do 
            for j in [1..s] do
                E1:=Elements(D1[i]);
                E2:=Elements(D2[j]);
                if Size(E1)=Size(E2) then
                    if E1=E2 then
                        c:=c+1;
                    fi;
                fi;
            od;
        od; 
    fi;
    return c=s;
end;;