# abelian quotients of permutation groups

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 $$|G/H|\le n$$.

• Oh, I see. So my construction with consecutive primes is no longer relevant. Still, the question persists: can one show anything better than the trivial estimate $|G/H|\le n!$ ? Jun 5 at 8:37
• Are you asking for an upper estimate not depending on $G$ (in which case, in other words, you're asking about the maximum cardinal of abelian subquotients of $S_n$).
– YCor
Jun 5 at 13:22
• Peter, many thanks! This is exactly what I am looking for! Jun 6 at 14:09

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 $$\lvert G/H\rvert\le 3^{n/3}$$.
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).