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Apr 7, 2012 at 12:37 answer added Jim Humphreys timeline score: 6
Apr 6, 2012 at 16:32 answer added John McKay timeline score: 1
Apr 6, 2012 at 10:36 history edited Tom De Medts CC BY-SA 3.0
added tag "finite-groups", minor grammar fixes
Apr 6, 2012 at 10:01 comment added Geoff Robinson That sort of number theory might be do-able, but it can't really be classification-free, because you surely need to know there aer only finitely many sporadic simple groups to touch a problem like this, don't you?
Apr 6, 2012 at 8:11 comment added Tom De Medts I would be surprised if there were a classification-free proof. Just think of how hard it already is to prove that $2 \in \pi(G)$! But even with the classification in hand, I have the impression that the problem involves some hard number-theoretical questions for the groups of Lie type, having to do with getting hold on the prime divisors of expressions of the form $\prod (q^k -1)$.
Apr 6, 2012 at 6:48 comment added Will Sawin Given the existence of a nontrivial counterexample at $19$, it seems likely that the easiest proofs is just by checking the classification of finite simple groups. Is there any evidence that there might exist a nicer proof?
Apr 5, 2012 at 15:32 comment added Tom De Medts I've checked all the examples from madore.org/~david/math/simplegroups.html, and it's true for those at least. (The condition $n \geq 23$ is strictly necessary, since the group $^2E_6(4)$ has the same prime spectrum as $A_{19}$.)
Apr 5, 2012 at 15:20 history edited Sara CC BY-SA 3.0
added 13 characters in body
Apr 5, 2012 at 15:09 comment added Tom De Medts I guess you want to rephrase the question as "Is it true that $G$ is an alternating group?". If $n$ is not prime, then $\pi(A_n) = \pi(A_{n-1})$.
Apr 5, 2012 at 15:03 comment added Sara In my question $A_{n}$ the alternating group on n letters. Thanks
Apr 5, 2012 at 13:28 history asked Sara CC BY-SA 3.0