Cyclic Sylow $p$-subgroup in finite simple groups

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I came accross a beautiful and "simple" (no pun intended) theorem, mentioned here in slide 14 by Jack Schmidt.

All finite simple groups have a cyclic Sylow $p$-subgroup for some $p$

I found references to proofs that involve the classification of finite simple groups, for instance in Composition factors from the group ring and Artin's theorem on orders of simple groups by Wolfgang Kimmerle, Richard Lyons, Robert Sandling, David N. Teague, MR1023806 (Theorem 4.9).

Is there a known proof that does not involve the classification of finite simple groups? That would be really nice.

Thank you.

Frieder Ladisch

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asked Oct 14, 2012 at 13:40

Portland

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$\begingroup$ I do not think there is a simpler proof. There are other "easy" facts that cannot be proved without the Classification. For example, every finite simple group is generated by (at most) two elements, has very little automorphism group, etc. $\endgroup$
– user6976
Oct 14, 2012 at 14:55
2

$\begingroup$ It's hard to state categorically that an elementary proof is impossible, but it's hard to envisage one. How would you prove that a simple group of odd order had a cyclic Sylow $p$-subgroup for some prime $p$ without using Feit Thompson, for example? $\endgroup$
– Geoff Robinson
Oct 14, 2012 at 15:47
1

$\begingroup$ I'd add to the comments by Mark and Geoff that there is definitely no known proof without the classification. And such a proof seems impossible for mere mortals to achieve. $\endgroup$
– Jim Humphreys
Oct 14, 2012 at 19:57
$\begingroup$ Thank you very much Mark, Geoff and Jim. I won;t spend anymore time finding a simpler proof. $\endgroup$
– Portland
Oct 15, 2012 at 11:44

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