abelian quotients of permutation groups

Question

Let $G$ be a subgroup of the permutation group $S_n$, and let $H$ be a normal subgroup of $G$ such that the quotient group $G/H$ is abelian. What is the best known upper estimate for the cardinality $|G/H|$?

If $p_1, p_2, \ldots, p_m$ are the first $m$ prime numbers and $n=p_1+\cdots+p_m$, let $G$ be the group generated by $m$ independent cycles of orders $p_1, \ldots, p_m$. Then $G$ itself is abelian, and $|G|=p_1\cdots p_m$. The Prime Number Theorem implies that $\log|G|\sim(n\log n)^{1/2}$, so in general one cannot expect anything better than $\log|G/H|\le (n\log n)^{1/2}(1+o(1))$. Is this true? If this is not true, or unknown to be true, what is the best known estimate?

In the special case of transitive $G$, what is the best known estimate? Here one cannot expect anything better than $|G/H|\le n$.

Answer 1

In Finite permutation groups with large abelian quotients Kovacs and Praeger show (among other things) the following: If $G$ is a (not necessarily transitive) permutation group of degree $n$ and if the derived subgroup $G'$ of $G$ is a proper subgroup of $G$, then $\lvert G/G'\rvert\le p^{n/p}$ for some prime divisor $p$ of $\lvert G/G'\rvert$. In particular, we always have $\lvert G/H\rvert\le 3^{n/3}$.

The proof does not require the classification of the finite simple groups.

Remark: In On abelian quotients of primitive groups, Aschbacher and Guralnick prove the sharper bound $\lvert G/H\rvert\le n$ if $G$ is primitive (and also reprove the Kovacs-Prager result from above).