Length of composition series in a primitive group

Question

Let $G$ be a primitive group acting on a set $\Omega$ with $n$ elements. By Cameron/Liebeck (essentially a consequence of the Classification + O'Nan-Scott), there are two possibilities:

(a) $G$ has a subgroup of index $\leq n$ isomorphic to an alternating group,

(b) $G$ is of size $\leq n^{O(\log n)}$.

In case (a), $G$ cannot have a composition series $$\{e\} = H_0 \triangleleft H_1 \triangleleft \dotsc \triangleleft H_{\ell} = G$$ of length $\ell$ greater than $O(\log n)$.

In case (b), the trivial bound on the length $\ell$ of a composition series is $\ell = O((\log n)^2)$. In fact, the trivial bound on the length of a chain of subgroups $\{e\} = H_0 \lneq H_1 \lneq \dotsc \lneq H_{\ell'} = G$ is also $O((\log n)^2)$.

Questions:

1. (The question I asked at first.) In case (b), can one give a better bound on $\ell'$ than $O((\log n)^2)$? (Answer: no; see below.)
2. (The question I meant to ask.) In case (b), can one give a better on bound on $\ell$ than $O((\log n)^2)$? Perhaps $O(\log n)$? (Answer: yes; see below.)

Answer 1

To answer your new question, Theorem 1.3 of this arXiv paper proves that the composition length $c(G)$ of primitive $G \le S_n$ satisfies $$c(G) \le \frac{8}{3} \log_2 n - \frac{4}{3}$$ and also identifies the examples in which equality holds.

Answer 2

Note: This is an answer to an earlier version of the question about whether there is an $O(\log n)$ bound on the length of an arbitrary subgroup chain of a group in case (b). The example below shows that the answer to that question is no, and we cannot do better than $O((\log n)^2)$.

${\rm GL}(n,p)$ acts primitively (in fact $2$-transitively) on $(p^n-1)/(p-1)$ points, and it has (for $n$ even) an elementary abelian $p$-subgroup of order $p^{n^2/4}$ (think of upper unitriangular matrices with an $n/2 \times n/2$ block in the top right corner), and hence a subgroup chain of length $n^2/4$.