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 $7$.

transitivepermutation groups (not necessarily primitive) for which the number of chief factors exceeds the degree... $\endgroup$ – Tom De Medts Jun 23 '11 at 9:33