My question is in the title. Some context: there are two versions of the O'Nan-Scott theorem. The first, weaker version, is due to O'Nan and Scott (independently) and gives the structure of the maximal subgroups of a finite symmetric group:

A maximal subgroup of $S_n$ is one of:

  • $S_k\times S_{n-k}$, the stabiliser of a $k$-set (that is, intransitive),

  • $S_a \mathrm{wr} S_b$ with $n=ab$, the stabiliser of a partition into b parts of size a (that is, imprimitive), or

  • primitive (that is, preserves no nontrivial partition) and of one of the following types:

    1. $\mathrm{AGL}(d,p)$ with $n=p^d$,

    2. $S_\ell \mathrm{wr} S_k$ with $n=\ell^k$, the stabiliser of a product structure,,

    3. a group of diagonal type, or

    4. an almost simple group.

Question 1: Am I right in thinking that this theorem does not depend on the Classification of Finite Simple Groups (CSFG)?

(Note, of course, that you need CFSG if you want to explicitly list the almost simple groups at (4). But such an explicit list is not a part of the stated theorem.)

The second, stronger, version of the O'Nan-Scott theorem is due to Aschbacher, O'Nan and Scott, and gives the structure of all finite primitive permutation groups. I'll state an abbreviated version.

A primitive subgroup of $S_n$ is of affine type, of wreath type in a product action, of diagonal type, of twisted wreath type, or almost simple.

All the proofs of this result that I have seen depend on the Schreier Conjecture, which asserts that the outer automorphism group of a finite simple group is solvable. This Conjecture is known to be true but only as a consequence of CFSG. So....

Question 2: Is it true that all known proofs of the Aschbacher-O'Nan-Scott theorem depend on CFSG?


According to the book "Permutation groups" by Peter Cameron, your first version, concerning maximal subgroups of $S_n$ was announced in 1979 before the announcement of the completion of CFSG, so it did not depend on CFSG at that point.

There is a complication however, because the earliest versions of O'Nan-Scott erroneously omitted the twisted wreath product case, which was pointed out later by Aschbacher. But the twisted wreath products are not maximal in $S_n$ - they are contained in the product type $S_l \wr S_k$ maximals, so I don't think that affects the maximality theorem.

According to Chapter 4 of Dixon and Mortimer's book on Permutation Groups, whihch deals with the O'Nan-Scott Theorem, the only place where the Schreier conjecture is used (Theorem 4.7B of Dixon and Mortimer) is in the analysis of the primitive groups with a regular nonabelian normal subgroup (which is exactly the twisted wreath product case), and I believe that no proof of that result is known that does not use the Schreier Conjecture. So if you want to avoid it, then you just have to leave that case without further information.

I heard Chris Parker from Birmingham give a talk for students on the O'Nan-Scott Theorem a few years ago, and I remember he did address the question of use of CFSG, and I believe it was exactly as I have described here, but you could write to him to check!

Addendum: Just to emphasize the dependence on the Schreier Conjecture, if that conjecture had been false, then it is conceivable that there could be two simple groups $S,T$ with $T \le {\rm Aut}(G)$ but $T \not\le {\rm Inn}(G)$, in which $T$ leaves no nontrivial proper subgroup of $S$ invariant, in which case the semidirect product $S \rtimes T$ acting on the cosets of $T$ would be a primitive permutation group having $S$ as regular normal subgroup.

  • $\begingroup$ Hi Derek, thanks for your answer. Your summary is exactly in line with how I understand the situation.... It'd be great if someone could give a reference for a full proof of the weak version that didn't use Schreier. (It should be extractable from Dixon & Mortimer, but I haven't the wherewithal to do that just now.) I'm also intrigued as to how seriously people have tried to prove the strong version without using Schreier? If one didn't use Schreier, I wonder what could be said? $\endgroup$ – Nick Gill Jan 23 '14 at 14:57
  • $\begingroup$ The account in Cameron's book is more concise, but he refers to Dixon and Mortimer for more details. The Schreier Conjecture is used in the proof of Thm 4.7 B (ii) of D&M, and without that you could have groups with a nonabelian regular normal subgroup $N$ in which the point stabilizer had a nonabelian simple subgroup that fixed, and induced outer automorphisms of, all of the simple direct factors of $N$. So you would need to rule that out somehow. $\endgroup$ – Derek Holt Jan 23 '14 at 15:50
  • $\begingroup$ @Derek: I've inserted the LaTeX command for the wreath symbol, which is just \wr $\endgroup$ – Jim Humphreys Jan 23 '14 at 17:35
  • $\begingroup$ It is rarely the case in applications that you need the structure of twisted wreath product groups. Usually it is just enough to know that there is a nonabelian regular minimal normal subgroup which is unique. Have you looked at the paper by Liebeck, Praeger and Saxl? $\endgroup$ – Michael Giudici Jan 28 '14 at 9:06
  • $\begingroup$ From memory there is a full proof of what you are calling the weak version in the paper by Scott, but I don't have it to hand at the moment. $\endgroup$ – Michael Giudici Jan 28 '14 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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