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

Question

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?

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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.

Answer 1

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.