28
$\begingroup$

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

  • L4(2) and L3(4) both have order 20160
  • O2n+1(q) and S2n(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?

(Originally asked at math.stackexchange.com).

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ Here's a pedantic comment: my understanding (which might be wrong!) was that the final flourish of the classification was a computation which ruled out the existence of a 27th sporadic simple group; people knew what the order of this group would have to be, and enough of its character table (or perhaps the structure of its 2-Sylow or something) was constructed to get a contradiction and hence rule out its existence. Before this potential simple group had been ruled out one could almost certainly still prove the result you want. So strictly speaking the full strength isn't needed :-) $\endgroup$
    – Kevin Buzzard
    Commented Aug 3, 2010 at 18:40
  • 1
    $\begingroup$ I would be surprised if one could prove without using the classification that there is a universal constant $N$ such that for each integer $g$, there are at most $N$ simple groups of order $g$. However, I would be very happy to be proven wrong! $\endgroup$
    – Andy Putman
    Commented Aug 3, 2010 at 18:40
  • 1
    $\begingroup$ Chronologically speaking, there were several uniqueness proofs (of the form "there is exactly one simple group with order N, a centralizer of an involution of the form X, and possibly satisfying additional property P") that only appeared in the late 1980s, well after the proof of the classification was initially announced. These came at the end of chasing down lots of cases, so I would suspect the answer to your question is "essentially no" with anything resembling current technology. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 3, 2010 at 19:20
  • 4
    $\begingroup$ Sadly, I have nothing to say on the topic of the posting, but you may enjoy a related anecdote (involving A. Weil, but not a time machine) from J.S. Milne's webpage jmilne.org/math/apocrypha.html $\endgroup$
    – algori
    Commented Aug 3, 2010 at 23:58

2 Answers 2

Reset to default
53
$\begingroup$

It is usually extraordinarily difficult to prove uniqueness of a simple group given its order, or even given its order and complete character table. In particular one of the last and hardest steps in the classification of finite simple groups was proving uniqueness of the Ree groups of type $^2G_2$ of order $q^3(q^3+1)(q-1)$, (for $q$ of the form $3^{2n+1}$) which was finally solved in a series of notoriously difficult papers by Thompson and Bombieri. Although they were trying to prove the group was unique, proving that there were at most 2 would have been no easier.

Another example is given in the paper by Higman in the book "finite simple groups" where he tries to characterize Janko's first group given not just its order 175560, but its entire character table. Even this takes several pages of complicated arguments.

In other words, there is no easy way to bound the number of simple groups of given order, unless a lot of very smart people have overlooked something easy.

Improve this answer
$\endgroup$
2
  • 10
    $\begingroup$ Nice answer! A related comment: in Steinberg's 1967 Yale notes there is a discussion of known finite simple groups. "The group $H$ of D. Higman and Sims [...] Inspired by this construction, G. Higman then constructed his own group [$H'$] in terms of a very special geometry invented for the occasion. The two groups have the same order [44352000], and everyone seems to feel that they are isomorphic, but no one has yet proved this." $\endgroup$
    – fherzig
    Commented Aug 9, 2010 at 13:27
  • $\begingroup$ I don't know though how long it took for the two groups to be proven isomorphic, I see that $H$ was only found in 1967. $\endgroup$
    – fherzig
    Commented Aug 9, 2010 at 13:30
16
$\begingroup$

Emil Artin proved in 1955 in two papers that the above mentioned examples are the only instances of non-isomorphic finite simple groups having the same order. He proved the result only for the groups that were known till then. As new groups were being discovered Jacques Tits took the responsibility of checking that there were no such further cases. For an exposition of this, one may look in `Kimmerle and others, Proc. London Math. Soc. 60(3) (1990) 89–122'.

So, indeed the classification is used to some extent.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .