Symmetric groups which are not quotients of Z/2Z*Z/3Z

Somehow this question made me think of instances of small exceptions in general, and I remembered the statement I heard once that $S_5,A_6,S_6,A_7,A_8,S_8$ are the only instances of symmetric/alternating groups that are not quotients of $Z/2Z*Z/3Z=PSL_2(Z)$ (see this MathSciNet entry, which I just found). Does anyone have an idea of a conceptual explanation for this fact?

Edit: I also find this article which mentions the same result. It's quite interesting that the positive part (for all $n>8$ these groups are quotients) is proved using Bertrand's postulate. I think it's cool that Bertrand's postulate can be used for group theory.

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clearly, a group is a quotient of Z/2Z*Z/3Z if and only if it is generated by one element of order 2 and one element of order 3. Of course one knows precisely, how elements of order 2,3 look like. It would be nice if the upcoming answer also includes these generators of S_n (for n>8). – HenrikRüping Feb 18 2010 at 13:46
@Henrik: "order 2"-->"order dividing 2" and the same for 3 ;-) – Kevin Buzzard Feb 18 2010 at 14:30
@Vladimir: "Does anyone have an idea of a conceptual explanation for this fact?". It's surely a "low-dimensional phenomenon", whereby sometimes some arguments need a certain amount of room before they can work (in this case the construction of the large cycles you need to prove the result). Sort of the same reason why pi_1(X) might not be abelian but pi_n(X) is abelian for all n>=2. – Kevin Buzzard Feb 18 2010 at 14:32
Regarding the coolness of Bertrand's postulate in group theory: Betrand originally used it to prove (in modern terminology) that the index of a proper subgroup of $S_n$ is either 2 or is $\geq n$. – Konrad Swanepoel Feb 18 2010 at 18:26
During the summer of 2008, I supervised three undergraduates at Northwestern who came up with their own proof this this theorem. It is possible to write down explicit generators, and to avoid an appeal to Bertrand's postulate. The real content is Jordan's 1872 theorem on highly transitive subgroups of S_n. – Lavender Honey Feb 19 2010 at 6:03

Yes, this is well known. In fact, there is a name for such groups - this is a (2,3)-generation property. And yes, by now there is a conceptual understanding why all sufficiently large finite simple groups have this property - the basic ideas are outlined in this helpful MathSciNet review gently explaining the major breakthrough by Liebeck and Shalev (1996). There are more recent developments in the field, both in the asymptotic direction and in the explicit construction, such as figuring out which $PSL(n,q)$ are (2,3)-generated - see papers by Tamburini, etc. - the literature is too big to be reviewed here.

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@Igor: could you give a MR review number (and perhaps the title and authors of the reviewed paper) in addition to the link? – Victor Miller Feb 18 2010 at 17:57
Yes. MR1405944 on a paper by M. Liebeck and A. Shalev: Classical groups, probabilistic methods, and the $(2,3)$-generation problem, Ann. of Math. 144 (1996), 77-125. The reviewer is Alex Lubotzky. – Igor Pak Feb 18 2010 at 18:27
@Igor: That's very interesting. Thanks a lot! I hope you don't mind me not accepting this answer for a while hoping that someone will also explain what's so special in numbers 5,6,8 for $S_n$... – Vladimir Dotsenko Feb 19 2010 at 9:43
@Vladimir: Sure. As for $S_5$ et al. this is basically "the small values effect", kind of like in the Frobenius problem. Say, in $S_5$ you need to have a 3-cycle for one generator, and (12) or (12)(34) for the involution. In the former case, they are either disjoint or don't have full support (thus, don't generate $S_5$), while in the second case both generators are even. A similar analysis deals with the other exceptions. But as soon as n is large enough, you can fit them both so every 3-cycle intersects some 2-cycle and vice versa. – Igor Pak Feb 19 2010 at 20:01
@Vladimir: Oh, $S_7$ is still like in the Frobenoius problem. E.g. both 10 and 12 can be changed into {5,7}, but not 11 - this "non-monotonicity" is inherent for small values. Now, $S_7$ can be generated by (123)(456) and (12)(34)(67). Note that for larger values, Jordan's theorem (on primitive transitive permutation groups) allows you to avoid excessive checking and reduces the problem to checking the combinatorics of cycles. – Igor Pak Feb 22 2010 at 17:30
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