Let $G$ be a finite simple group such that $\pi (G)=\pi (A_{n})$, where $n\geq 23$ ($\pi (G)$ is the set of prime divisors of the order of $G$). Is it true that $G$ is isomorphic to an alternating group?
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$\begingroup$ In my question $A_{n}$ the alternating group on n letters. Thanks $\endgroup$– SaraCommented Apr 5, 2012 at 15:03
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1$\begingroup$ I guess you want to rephrase the question as "Is it true that $G$ is an alternating group?". If $n$ is not prime, then $\pi(A_n) = \pi(A_{n-1})$. $\endgroup$– Tom De MedtsCommented Apr 5, 2012 at 15:09
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3$\begingroup$ 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 $A_{19}$.) $\endgroup$– Tom De MedtsCommented Apr 5, 2012 at 15:32
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1$\begingroup$ 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? $\endgroup$– Will SawinCommented Apr 6, 2012 at 6:48
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2$\begingroup$ 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? $\endgroup$– Geoff RobinsonCommented Apr 6, 2012 at 10:01
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2 Answers
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I'm not sure what a definitive answer would look like, but maybe I can frame the question a little differently in a series of comments:
1) The underlying question really concerns an arbitrary pair of (not too small) simple groups $G, H$ for which $\pi(G) = \pi(H)$. As observed by others, the classification of finite (nonabelian) simple groups should be in hand already. Given only finitely many groups non-isomorphic to an alternating group or group of Lie type, the question basically involves $G, H$ of those two types. (But there is no obvious reason to require that one of the groups be alternating.)
2) There is a long tradition, going back at least to Emil Artin in the 1950s, of sorting out which of the known finite simple groups having different origins are actually isomorphic. So the question asked is not really about existence of isomorphisms: one knows already from the literature which groups of Lie type are isomorphic to others of Lie type or to alternating groups. In practice all such accidental isomorphisms involve relatively small orders. (On the other hand, simple groups coming from odd orthogonal groups and symplectic groups with the same rank and same finite field are well understood to be non-isomorphic in spite of having the same order.)
3) It's conceivable that some isolated group of Lie type will turn up whose order accidentally has the same prime divisors as some (not too small) alternating group. This would be a bit surprising, but probably not interesting unless something systematic is observed that involves more groups. The key word here is accidental. To prove rigorously that no accidents happen looks challenging, even if ultimately not too illuminating.
4) Why emphasize alternating groups? There is perhaps one good reason: here the primes dividing the group order form a list (from 2 to the largest prime) having no gaps. For groups of Lie type that are not very small, this behavior of $\pi(G)$ strikes me as purely accidental and somewhat unusual. If that's not the case, it would be interesting and need an explanation. But considering a family of groups such as $\rm{PSL}(2,p)$ (with $p>3$) suggests a skeptical attitude.
answered Apr 7, 2012 at 12:37
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A_8 has the same order as the non-isomorphic PSL(4,2).
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$\begingroup$ PSL(4,2) and A_8 are isomorphic; the other simple group of that order is PSL(3,4). $\endgroup$ Commented Apr 14, 2012 at 23:54