[This Wikipedia article](http://en.wikipedia.org/wiki/List_of_finite_simple_groups) 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](http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups) to prove this, or is there a simpler way of proving it?


*(Originally asked at [math.stackexchange.com](https://math.stackexchange.com/questions/1423/number-of-finite-simple-groups-of-given-order-is-at-most-2-is-a-classification)).*



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


