How many normal subgroups a primitive group can have?

Question

Let $G$ be a primitive permutation group of degree $n$, that is $G$ acts transitively and faithfully on a set consisting of $n$ elements and $G$ preserves no nontrivial partition of $X$. In a sense primitive groups are the 'simple' permutation groups.

For example one can show that a primitive group has at most 2 minimal normal subgroups. this can be regarded as a 'width' result.

My question is about the 'depth', namely:

Let $G$ be a primitive permutation group of degree $n$, Let $1\lneq G_m\lneq \cdots\lneq G_0=G$ be a strictly decreasing sequence of normal subgroups of $G$. Is there a good bound for $m$ in terms of $n$?

A stupid bound is $m\leq \log_2(n!)$. My guts feeling is that $m$ should be at most $n-1$, because it feels that every subgroup in the sequence should contribute at least one element to the orbit of a point.

Edit: I ran a computer check in the meanwhile up to degree $n=50$, and it turns out that $m$ is bounded by $n-1$ in this region. Moreover $m$ grows much slower than $n$, e.g., it never exceeded $7$.

Answer 1

I have been trying to find results in the literature on this topic, but finding it frustrating. The difficulty is that estimates on composition and chief series lengths of various types of groups are typically studied not as end in themselves, but as a means to proving other results, such as bounding the minimal generator numbers of groups.

But it is certainly possible to prove a bound that is linear in $n$ for general permutation groups of degree $n$ and logarithmic in $n$ for primitive permutation groups. It is harder to get sensible estimates of the constants involved.

It is proved in:

Cameron, P. J.; Solomon, R. G.; Turull, A. Chains of subgroups in symmetric groups. J. Algebra 127 (1989) 340--352.

that any chain of subgroups in $S_n$ is less than $3n/2$.

To study primitive permutation groups, you can use the O'Nan-Scott Theorem. For the affine case, the group $G$ is a semidirect product of elementary abelian $n=p^d$ with an irreducible subgroup of ${\rm GL}(d,p)$. In Theorem 2.3 of

Lucchini, A.; Menegazzo, F.; Morigi, M. On the number of generators and composition length of finite linear groups. J. Algebra 243 (2001) 427--447.

a bound is established on the composition length of a completely reducible subgroup of ${\rm GL}(d,p)$ that is linear in $d$ and logarithmic in $p$, so we get a bound on the composition length of $G$ that is logarithmic in $n$.

For the other cases of the O'Nan-Scott Theorem, the socle of $G$ is a direct product of isomorphic nonabelian simple groups, and the number of direct factors is logarithmic in $n$, and we can use the Cameron et al result to bound the composition length of the section of $G$ that permutes these factors. You still have to estimate the composition factors involving the automorphsim groups of these simple factors, but they cannot be very large compared with $n$, so it is possible to get the logarithmic bound.