Yes if the lattice $[H,G]$ has less than $64$ elements, or if the index $\vert G : H \vert$ is less than $648$.
Let $[H,G]$ be a distributive interval of finite groups and $K_1, \dots , K_n$ its atoms.
We will first prove that the answer is yes for $n<6$ (the case $n \ge 6$ is still open).
Lemma: There is at most one atom $K_i$ with $|K_i:H| = 2$.
Proof: First of all if $|K_i:H| = 2$ then $H \triangleleft K_i$. If there is an other atom $K_j$ with $|K_j:H| = 2$, then $H \triangleleft (K_i \vee K_j)$. But the interval $[H, K_i \vee K_j]$ is a sublattice of the distributive interval $[H,G]$, so it is also distributive. But by normality, $[H, K_i \vee K_j] \simeq [1, (K_i \vee K_j)/H]$, so by distributivity and Ore's theorem, $(K_i \vee K_j)/H$ is a cyclic group, moreover it would have exactly two minimal subgroups, $K_i/H$ and $K_j/H$. But if a cyclic group has exactly two minimal subgroups, they must be of different order, contradiction with $|K_i:H| = |K_j:H| = 2$. $\square$
Now we will use the following theorem (proved in this paper):
Theorem: If $[H,G]$ is distributive with $\sum_{i=1}^n \vert K_i : H \vert^{-1} \le 2$, then it is linearly primitive.
Corollary: If the distributive interval $[H,G]$ has $< 6$ atoms, then it is linearly primitive.
Proof: by the lemma $\sum_i \vert K_i : H \vert^{-1} \le 1/2 + (n-1)/3 $.
But $1/2 + (n-1)/3 \le 2$ iff $n \le 11/2$. $\square$
Corollary: It is also true if $|[H,G]| < 2^6 = 64$ or $\vert G : H \vert < 2 \cdot 3^5 = 486$.
Proof: A distributive lattice with $6$ atoms has at least height $6$ and $2^6$ elements (see here). Then a distributive interval $[H,G]$ with $6$ atoms, has an index $|G:H| \ge 2^6$, but if $|G:H| = 2^6$ then any atom $K_i$ satisfies $|K_i:H| = 2$, which contradicts the lemma; so there exists an atom $K_i$ with $|K_i:H| \neq 2$, i.e. $|K_i:H| \ge 3$. Next we apply an induction on the interval $[K_i,G]$ and we get $|G:K_i| \ge 2 \cdot 3^4$. $\square$
Now we will go up to index $< 648$. We will use the following theorem (see this paper).
Theorem: If $[H,G]$ is a rank $n$ boolean interval with a nonzero dual Euler totient $$\hat{\varphi}(H,G):=\sum_{K \in [H,G]} (-1)^{\ell(H,K)}|G:K|,$$ then $[H,G]$ is linearly primitive.
Proposition: A rank $n$ boolean interval $[H,G]$ of index $|G:H| = 2 \cdot 3^{n-1}$, is linearly primitive.
Proof: Under this assumption, we deduce that:
- For every maximal chain of $[H,G]$, exactly one edge has index $2$.
- For every element $K \in [H,G]$, $K \neq G$, then at most one edge $[K,X]$ has index $2$.
- For every element $K \in [H,G]$, $K \neq H$, then at most one edge $[X,K]$ has index $2$.
The point 1. is immediate. The point 2. and 3. use arguments similar to the previous lemma's proof.
By this answer, there exists an atom $A$ of $[H,G]$ such that an edge $[X,Y]$ has index $2$ iff $Y = X \vee A$. But then $\hat{\varphi}(H,G) = \hat{\varphi}(H,A) \cdot \hat{\varphi}(A,G) = (2-1) \cdot (3-1)^{n-1} > 0$.
The result follows by the previous theorem. $\square$
Corollary: A distributive interval $[H,G]$ of index $< 2 \cdot 3^4 \cdot 4 = 648$, is linearly primitive.
Proof: By this answer and the first corollary, we can assume the lattice to be boolean of rank $6$. By the lemma, there exists a maximal chain with at most one edge of index $2$. So $|G:H|$ equals a product of six integers $>1$ with at most one equal $2$. Now the next product like that after $2 \cdot 3^5$ is $2 \cdot 3^4 \cdot 4$. So the result follows by the previous corollary and proposition. $\square$
Remark: By some similar arguments, we can probably upgrade this bound.
