Is the solvability of finite groups of order coprime to 15 essentially easier to prove than the entire Classification of Finite Simple Groups?

1$\begingroup$ Can you use Thompson's Ngroups en.wikipedia.org/wiki/Ngroup_(finite_group_theory) ? $\endgroup$ – Mark Sapir Jun 21 '13 at 10:29

7$\begingroup$ The answer is yes, because Thompson proved that the only finite simple groups of order coprime to 3 are the Suzuki groups, and Glauberman later extended this to a classification of simple groups that do not have $S_3$ as a subgroup. Bothe of thsoe results are preclassification. But I have a suspicion that neither of them was ever published! $\endgroup$ – Derek Holt Jun 21 '13 at 10:35

$\begingroup$ @Derek, Mark: Thank you very much! (The result would follow also from the classification of minimal simple groups (Thompson, 1968) if one knows that every nonabelian finite simple group contains a minimal simple group, but is there a proof of the latter which is not based on CFSG?  Barry and Ward (dmle.cindoc.csic.es/pdf/…) write "It is a consequence of the classification of finite simple groups that every nonabelian simple group contains a subgroup which is a minimal simple group".) $\endgroup$ – Stefan Kohl Jun 21 '13 at 13:37

$\begingroup$ But what is the meaning of "without CFSG"? For instance does it allow making use of FeitThompson? Does Thompson's result about groups of order prime to 3 use FeitThompson? $\endgroup$ – YCor Jun 21 '13 at 15:15

1$\begingroup$ @Yves: The proof of FeitThompson's Odd Order Theorem has something like 200 pages, while even the secondgeneration classification proof will have about 5000 pages. So I'd say definitely something can still be essentially easier to prove than CFSG if the proof needs FeitThompson. $\endgroup$ – Stefan Kohl Jun 21 '13 at 16:07
As Derek Holt has pointed out, the answer to the question is yes.  Thompson proved that the only finite simple groups of order coprime to 3 are the Suzuki groups, and Glauberman later extended this to a classification of simple groups that do not have ${\rm S}_3$ as a subgroup. Both of these results are preclassification, though they might not have been published.

1$\begingroup$ Taking into account the suggestions on Meta (meta.mathoverflow.net/questions/493/…), I have turned Derek's comment into an answer. This is mainly to make clear that the question has already been answered. $\endgroup$ – Stefan Kohl Jul 16 '13 at 23:27

1$\begingroup$ Glauberman's classification of S_4free groups was definitely published. Also, a simple group of minimal order subject to having order prime to 3 has every proper subgroup solvable, so is an $N$group, and would appear in Thompson's (published) list of $N$groups, so the answer is yes on the basis of published results. $\endgroup$ – Geoff Robinson Oct 6 '13 at 21:16

$\begingroup$ @Geoff: Thanks for your remark!  By the way, is it possible that you rather mean ${\rm S}_3$free groups than ${\rm S}_4$free ones? $\endgroup$ – Stefan Kohl Oct 6 '13 at 21:21

2$\begingroup$ No, I meant what I sai. I think Glauberman published an AMS Monograph on S_4free group. Clearly, an S_3free group is S_4 free, so Glauberman's S_4free classification includes the classification of S_3free groups. $\endgroup$ – Geoff Robinson Oct 6 '13 at 22:16