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Given a natural number k, are there only finitely many finite simple groups with the property that all elements have order at most k?

This holds if I only look at the finite simple groups I understand (e.g. alternating groups and SL(k,finite field)), but it's not clear to me whether this holds for all finite simple groups, even using their classification.

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3 Answers 3

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The classification of the finite simple groups implies that there are only finitely many finite simple groups of a given exponent $k$. To see this, first note that we can ignore the sporadic groups, as well as the cyclic groups of prime order. It is also also clear that there are only finitely many alternating groups of a given exponent. So we need only consider groups of (possibly twisted) Lie type over finite fields. Here we see that there are only finitely possibilities for the Lie type: otherwise, the Weyl groups would involve arbitrarily large alternating groups. Once the Lie type is fixed, there are only finitely many possibilities for the finite field: otherwise we would obtain semisimple/diagonal elements of arbitrarily large exponent.

There are currently no proofs of this result which do not use the classification of the finite simple groups.

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Well, every finite simple group is $2$ generated. The Restricted Burnside Problem implies that there are fintely many $2$-generated finite groups of exponent $k$. In particular, there are finitely many $2$-generated finite simple groups of exponent $k$.

This uses the classifications for the fact that every finite simple group is $2$ generated. And unfortunately it is also a circular logic because it uses the classification exactly for your own question since to solve the Restricted Burnside Problem you need to use Hall and Higman's paper that reduced the Restricted Burnside Problem to $p$-groups using the fact that there are only finite number of finite simple groups of exponent $k$.

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Indeed, a stronger claim is true: there are only finitely many finite simple groups $G$ such that the largest prime dividing $|G|$ (which is also the largest prime order of an element) is at most $k$. For a proof, see Section 5 of:

L. Babai, A. J. Goodman and L. Pyber. Groups without faithful transitive permutation representations of small degree. J. Algebra 195 (1997), no. 1, 1–29.

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thanks, on reflection, that's actually what I really needed. –  Stefan Friedl Jul 25 '10 at 21:45

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