# Number of finite simple groups of given order is at most $2$ - is a classification-free proof possible?

This Wikipedia article states that the isomorphism type of a finite simple group is determined by its order, except that:

• $L_4(2)$ and $L_3(4)$ both have order $20160$
• $O_{2n+1}(q)$ and $S_{2n}(q)$ have the same order for $q$ odd, $n > 2$

I think this means that for each integer $g$, there are $0$, $1$ or $2$ simple groups of order $g$.

Do we need the full strength of the Classification of Finite Simple Groups to prove this, or is there a simpler way of proving it?

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My guess is that something like this is part of the proof of the classification itself. – Qiaochu Yuan Aug 2 '10 at 21:19
I think this is an awesome question. If you get no satisfactory answer after a week or so, you should toss it up onMO. – BBischof Aug 2 '10 at 21:52
To be honest, I don't understand why the wait period :P – Mariano Suárez-Alvarez Aug 2 '10 at 22:58
I also think this would be an appropriate MO question. Reason: a generic mathematically literate person will guess "Yes, you probably need CFSG." However, to be satisfied with this answer, you want to hear it from an expert in group theory. – Pete L. Clark Aug 3 '10 at 1:20

There are many mathematicians outside finite group theory who asked whether important infinite fragments of the classification were possible without the entire classification. I believe the favorite questions has always been : Can you prove the finiteness of the sporadics without the full classification?

There is a good chance that third generation proof technology will reduce the entanglement between different portions of the classification because one knows the unipotent primes during earlier arguments where current methods only reveal semi-simple structure. There has been one remarkable success in this direction :

Theorem (Altinel, Borovik, Cherlin). A simple group of finite Morley rank containing an infinite elementary abelian 2-group is a Chevalley group over an algebraically closed field of characteristic 2.

There is no known proof that simply groups of finite Morley rank even have an involution, much less that groups with odd characteristic looking Sylow 2-subgroup are also algebraic. In consequence, there is a conjecture by Borovik that basically proposes one might classify finite simple groups who's 2-rank vastly exceeds the p-rank for any other p prime.

The final proof of [ABC] weighs in around 500 pages, but any finite analog would require many thousands of pages to deal with issues like twisted groups of Lie type and alternating groups, even assuming you find some trick for avoiding all the sporadics.

In short, there are an awful lot of interesting results that depend upon the full CFSG for the foreseeable future because only funky asymptotic fragments look even vaguely realistic as stand alone results and even those look extremely difficult.

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Although I am not (by any stretch) an expert on finite simple groups, let me flesh out my above comment.

Consider the following QCFSG (i.e., "qualitative" CFSG): with only finitely many exceptions, every finite simple group has prime order, is alternating, or is one of the finitely many known infinite families of Lie type. QCFSG must have been conjectured rather early on, whereas the exact statement of CFSG was much harder to come by, as much of the early work on the classification problem resulted in discovery of new sporadic groups.

I guess that early on someone must have looked at the nonsporadic finite simple groups and noticed that, except for the two exceptions listed above, they have distinct orders. [Assuming this is actually true, that is. I have no reason to doubt it, but I haven't checked it myself.] Once you notice that, if you believe QCFSG, then you certainly think that the order of a simple group determines the group up to finitely many exceptions. It is very hard for me to imagine how you could prove that the number of exceptions is precisely two without knowing the full CFSG.

I cannot resist conveying a story of Jim Milne, whose moral is that you shouldn't feel too bad when you say something absolutely stupid in public: better mathematicians than you or I have said stupider things.

Finally, a story to keep in mind the next time you ask a totally stupid question at a major lecture. During a Bourbaki seminar on the status of the classification problem for simple finite groups, the speaker mentioned that it was not known whether a simple group (the monster) existed of a certain order. "Could there be more than one simple group of that order?" asked Weil from the audience. "Yes, there could" replied the speaker. "Well, could there be infinitely many?" asked Weil.

For the source, and for some further fun stories, see

http://www.jmilne.org/math/apocrypha.html

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+1 for the great story. – Akhil Mathew Aug 3 '10 at 1:37
Weil reviewed Artin's collected works and noted the paper calculating the orders of the classical groups over finite fields, so he may have been aware of O(2n+1)=SO(2n) and trying to ask whether other series of cardinality coincidences could exist: "Well, (if one such hypothetical coincidence could exist), could there be infinitely many (additional coincidences)". – T.. Nov 29 '10 at 17:32