TITLE: Number of finite simple groups of given order is at most 2 - is a classification-free proof possible?
QUESTION [27 upvotes]: 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).

REPLY [16 votes]: 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.