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 $|G/H|\le n$.

  • $\begingroup$ 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 $|G/H|\le n!$ ? $\endgroup$
    – Yuri Bilu
    Jun 5 at 8:37
  • 2
    $\begingroup$ 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 $S_n$). $\endgroup$
    – YCor
    Jun 5 at 13:22
  • $\begingroup$ Peter, many thanks! This is exactly what I am looking for! $\endgroup$
    – Yuri Bilu
    Jun 6 at 14:09

1 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.