I heard that there are no 3 nonisomorphic simple groups of the same order.

Question: Is there an elementary proof of this?

In case this is not the case, here a modified question:

Question: Is there an elementary proof that there are not $m$ nonisomorphic simple groups of the same order with $m \geq 4$ as small as possible?

  • A related question: How spread out are the (non-abelian) simple groups in terms of their orders? What are some "near misses"? For example, 168-60 is already quite small. Are there smaller differences, perhaps scaling to account for size? – David Richter Dec 7 at 17:17
up vote 13 down vote accepted

No, there are no known proofs of any results of this type that do not rely on the complete classification of finite simple groups

In particular, the result of Pyber (1993) giving an upper bound on the number of isomorphism classes of finite groups of order $n$ (see Jack Schmidt's answer to this question for details) could only be incorrect if there were vast numbers of isomorphism classes of simple groups of the same order, but it still relies on the classification.

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