# Is applying Feit–Thompson’s theorem for the nonexistence of a simple group of order $1004913$ really a circular argument?

In p.212 of Dummit–Foote’s Abstract Algebra, 3rd Edition, an analysis of a hypothetical simple group $$G$$ of order $$1004913 = 3^3 \cdot 7 \cdot 13 \cdot 409$$ is carried out. The authors write:

We might then guess that there is a simple group of this order, but the Feit–Thompson Theorem asserts that there are no simple groups of odd composite order. (Note, however, that the configuration for a possible simple group of order $$3^3 \cdot 7 \cdot 13 \cdot 409$$ is among the cases that must be dealt with in the proof of the Feit–Thompson Theorem, so quoting this result in this instance is actually circular. We prove no simple group of this order exists in Section 19.3; see also Exercise 29.)

However, as far as I searched, neither $$409$$ nor $$1004913$$ appears in the original paper Solvability of groups of odd order of Feit–Thompson. Therefore, I am not convinced that applying Feit–Thompson’s theorem for the nonexistence of $$G$$ is really a circular argument. I would like to ask you to judge whether this is circular or not.

For example, if Feit–Thompson’s paper relies on another previous work and it deals with the nonexistence of $$G$$, it can be described that this argument is actually circular. (Of course, even if so, there may be another proof without dealing with $$G$$. But in this case I can understand that the description of Dummit–Foote is not wrong.)

As a side note, ”Exercise 29” asks us to prove nonexistence of a simple group of order $$3^3 \cdot 7 \cdot 13 \cdot 409$$ by the knowledge of Chapters 1–6, which actually I am trying to solve. I am not going to verify the details of the proof of Feit–Thompson’s theorem by myself, which is clearly beyond my ability to learn. The reason I want to know whether it is circular or not is that just my curiosity: Is this exercise just a trivial exercise to learn elementary group theory or an exercise that has some meaning for general group theory?

P.S. After I thought twice again, I obtained my solutions to the exercises in two or three ways. I think that I have given it enough effort. So you do not have to avoid spoilers any more. Thanks for your consideration.

• It is unlikely that the specific primes appear in the paper; rather, it would likely be a class of groups (of potential minimal simple groups of odd order) whose order has prime factorization that satisfies certain properties, and which this particular order satisfies. Sep 28, 2023 at 20:17
• I've never fully understood what people mean by "circular" in this sort of situation. Suppose, for example, that Feit and Thompson explicitly analyzed that group in their proof. Then citing Feit-Thompson would be using a sledgehammer to kill a fly, but there would be nothing logically suspect about it. It would just be a "corollary of the proof" (or a "porism" as some people call it). "Circular" has strong connotations of logical fishiness, which don't seem to be present here, unless this result is supposed to be a step in a new proof of Feit-Thompson. Sep 28, 2023 at 23:55
• @TimothyChow: In the literature as a whole, Feit–Thompson is proved and can be certainly used without circularity. But a textbook also aims to avoid circularity in both its own internal exposition, and the student-reader’s mental development of the topic. The potential circularity from things like this is: student X learns this and similar lemmas by appeal to Feit–Thompson; X later learns a proof of Feit–Thompson, which relies on some such lemmas; X remembers they’ve already seen these lemmas proven, so doesn’t follow up references to find elementary pre-Feit–Thompson proofs of them. [cont’d] Sep 30, 2023 at 10:40
• [cont’d] In the worst case, this can become near-circularity in the literature too. If all modern textbooks appeal to F–T for some difficult lemma that’s needed in F–T, then a couple of generations later, most mathematicians have only learned the “circular” version, and the original pre-F–T proofs of the lemma may be very hard to understand — their techniques largely forgotten. As I understand, guarding against this and similar issues was a big motivation for the “second-generation” proof of the CFSG. Minimising such near-circularity keeps the literature more robust long-term. Sep 30, 2023 at 10:53
• @PeterLeFanuLumsdaine I agree that the scenario you describe is undesirable. But I still wouldn't describe the problem as circularity. The problem is that the proof is not self-contained. By the way, I would say that the big motivations for the 2nd-gen CFSG were (1) to make sure there weren't yawning gaps that you could drive a truck through, and (2) to write down a proof that was readable. What you're calling "circularity" was undoubtedly on the radar screen as well, but there were much bigger flaws that needed fixing. Sep 30, 2023 at 13:10

I do doubt that Feit–Thompson explicitly dealt with that particular group in their Pacific Journal of Mathematics paper. But from fairly early in their paper, one can see that in a group $$G$$ of the given order, the Sylow $$3$$-subgroup of the supposedly simple $$G$$ is elementary Abelian of order $$27$$ (because $$3$$ is the smallest prime divisor of $$|G|$$).

Then by transfer/Frobenius normal $$p$$-complement theorem, you can see that $$[N_{G}(S):C_{G}(S)]$$ is divisible by $$13$$ for a Sylow $$3$$-subgroup $$S$$ of $$G$$ which means that there is only one conjugacy class of subgroups of order $$3$$ in $$G$$ and so on…, you see that you can begin to get a grip on the local structure of $$G$$. Also, by the rather deeper Feit–Thompson uniqueness theorem, a Sylow $$3$$-subgroup of $$G$$ is contained in a unique maximal subgroup of $$G$$….

I think perhaps what Dummitt and Foote might mean is that just saying "by Feit–Thompson, $$G$$ is solvable" gives no understanding of why $$G$$ is solvable unless you understand how you can deduce the solvability of $$G$$ by specializing the relevant results of Feit–Thompson you need to exhibit the solvability of the particular group $$G$$.

Later edit: I had some further thoughts about this. When I looked further into it (I don't want to interfere too much with the OP's attempts at "Exercise 29" of Dummitt and Foote), I realised (using Sylow type congruences) that the supposedly simple group $$G$$ would have to be a CA-group (a group in which the centralizer of every non-identity element is Abelian). While, as indicated above, the quickest way for me to see that involved the use of some of Feit and Thompson's general theory, I found this interesting, and possibly relevant to the question.

For it is part of the history of the proof of the odd order theorem that M. Suzuki proved some years before Feit and Thompson's general proof, that there is no finite simple CA-group of odd order. After Suzuki's proof appeared, Feit, M. Hall and Thompson proved that there is no finite simple CN-group of odd order (a CN-group is one in which the centralizer of every non-identity element is nilpotent), and later Feit and Thompson proved the full odd order theorem.

While I suspect that the Feit–Thompson proof does not ultimately rely on the CA and CN results — ( later edit— but see @David A. Craven's remark below), but would re-prove them as a corollary of the more general result, this does point out a hypothetical scenario in which circularity might arise.

If Feit–Thompson had taken Suzuki's result as an assumed prerequisite of their proof, then there would be some circular reasoning in quoting Feit–Thompson to prove that a given CA-group of odd order is solvable.

It is worth pointing out that the end-point of the CA-group, the CN-group, and the general Feit–Thompson proof are all rather similar. The set $$\pi$$ of prime divisors of the order a minimal counterexample $$G$$ is partitioned into subsets $$\{ \pi_{i} : 1 \leq i \leq t \}$$ such that for each $$i$$, $$G$$ has a nilpotent Hall $$\pi_{i}$$-subgroup $$H_{i}$$, and $$M_{i} = N_{G}(H_{i})$$ is a maximal subgroup of $$G$$ for each $$i$$. In the CA and CN group papers, the $$M_{i}$$ are all Frobenius groups. In the Feit–Thompson paper, the structure of the $$M_{i}$$ may be somewhat more complicated. In each paper, character theory plays a large role in the proof.

Much later edit: In view of some of the the comments below, let me be more specific about what I meant by "a hypothetical scenario in which circularity might arise".

If Feit and Thompson assumed the odd CA-theorem of M. Suzuki as a "given", then to use the Feit–Thompson of odd order theorem to justify the fact that a given odd order CA-group (such as the one in the question) is solvable, while certainly logically consistent and correct, has what I would consider a degree of “circularity” about it, albeit in a colloquial sense. In any case, there would be some redundancy involved, since the fact that the group is solvable is a consequence of Suzuki's theorem, which pre-dates the Feit–Thompson theorem.

As I said above, the point is moot, in that I am pretty sure that the Feit–Thompson proof gives an alternative proof of Suzuki's theorem, since the case that all maximal subgroups are Frobenius groups (which happens in a minimal counterexample to the CA-theorem) is covered by the analysis of a minimal counterexample to Feit–Thompson's odd order theorem.

• There would be no circular reasoning no matter what. The Feit–Thompson theorem is proved, and it is perfectly valid to quote it to infer any consequence the statement implies, irrespective of whether that consequence is related to something that was used in the Feit–Thompson proof. It would only be circular reasoning if you quoted the Feit–Thompson theorem to prove something with the intention to use it in a new proof of the Feit–Thompson theorem, as pointed out in Timothy Chow’s comment above. Sep 30, 2023 at 10:26
• @EmilJeřábek : I really don't want to get into semantics. I think there is some circular reasoning that has the potential to lead to a contradiction, and some circular reasoning which is logically harmless, but is effectively tautological. I agree that there is no possibility circular reasoning of the former kind in this situation, Sep 30, 2023 at 11:14
• Thank you very much for your kind answer from the viewpoint of a professional. Though I do not know higher theory, I notice that your beginning arguments are similar to my solution to the exercise, such as determining that $P \in Syl_{3}(G)$ is elementally Abelian or considering N/C of $P$. Hence, by reading your answer, my understanding is that solving this exercise gives learners an opportunity to touch an elementary prototype of a Feit-Thompson-like argument. Oct 2, 2023 at 16:28
• Just as a comment, FT assume the CN-theorem. See p.983 of their paper, the final line of Section 33. Nov 22, 2023 at 22:23
• @David A.Craven : Ah, thanks. I wasn't sure about that. Nov 22, 2023 at 22:46