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 auestion 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.)

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.)

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 $\ell = O((\log n)^2)$.

Can one do better in case (b)? In particular, does the bound $O(\log n)$ hold?

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 auestion 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.)

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 subgroup chain $$\{e\} = H_0 < H_1 < \dotsc < H_{\ell} = G$$ of length $\ell$ greater than $O(\log n)$.

In case (b), the trivial bound on the length $\ell$ of a subgroup chain is $\ell = O((\log n)^2)$.

Can one do better in case (b)? In particular, does the bound $O(\log n)$ hold?

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 $\ell = O((\log n)^2)$.

Can one do better in case (b)? In particular, does the bound $O(\log n)$ hold?