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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.2010-08-02
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    I think this is an awesome question. If you get no satisfactory answer after a week or so, you should toss it up onMO.2010-08-02
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    To be honest, I don't understand why the wait period :P2010-08-02
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    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.2010-08-03
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    Thanks for your answers and comments. I have re-asked it at MO (http://mathoverflow.net/questions/34424/number-of-finite-simple-groups-of-given-order-is-at-most-2-is-a-classification)2010-08-03

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