Hello, apparently finite groups which are n-transitive with n>5 are only the permutation groups Sn or the alternating groups An+2, see e.g. page 226 this book by Isaacs http://books.google.fr/books?id=pCLhYaMUg8IC&pg=PA226
Is this characterization useful at all? For instance, are there famous proofs (maybe in a geometric context), which use it to, say, show that a certain group is in fact some An ?
This fact is used in a nice way by Dunfield and Thurston to show that, for any finite simple group Q, the number of Q-covers of a "random" 3-manifold in their sense follows a Poisson distribution. (The multiple transitivity appears in Thm 7.4.)
Also: I don't have a reference for this in mind, but I've seen Nick Katz give talks where he uses a "linear" version of this, showing that an algebraic subgroup of GL(V) (typically a monodromy group) is "as big as you expect", using irreducibility of tensor powers of V.
EDIT: In response to Noah's query, I looked up a reference; the relevant theorem is due to Michael Larsen and is often called "Larsen's Alternative." You can read about it in section 1 of this paper of Katz.
For the specific problem of distinguishing a subgroup from a group by means of moments, the 2005 paper of Guralnick and Tiep is relevant.
[Edited to add the final field-theoretic paragraph]
Just noticed this in a list of "Related" problems. I've seen applications to the computation of Galois groups $G$ of explicit polynomials $P \in k[X]$. See for example Abhyankar's survey paper
[A] Abhyankar, Shreeram S., with an appendix by J.-P. Serre: Galois theory on the line in nonzero characteristic, *Bull. AMS (N.S.) 27 #1 (July 1992), 68–133.
(Note the dedication to "Walter Feit, J-P. Serre, and e-mail"!)
Let $N = \deg P$. If $G = S_N$ or $A_N$ then this can be proved by showing that $G$ is $n$-transitive for $n$ large enough (depending on $N$), at which point $G=A_N$ if ${\rm disc}(P\phantom.)$ is square in $F$, and $G=S_N$ if not. [This last assumes $2 \neq 0$ in $k$; there's a pseudo-discriminant criterion that works in characteristic 2.] As T. Sauvaget notes in his question, $n>5$ is always enough, but in fact $n=4$ suffices except for $N=11,12,23,24$ (Mathieu groups), and even $n=3$ brings it down to a usually-manageable list of possibilities (see [A, pages 86–87]).
This approach can be useful because showing (say) 4-transitivity amounts to proving that a few polynomials are irreducible. Indeed $P$ itself is irreducible iff $G$ is 1-transitive; in this case, we may adjoin to $k$ a root $X_0$ of $P$, and then the point stabilizer in $G$ is the Galois group of the degree-$(N-1)$ polynomial $P_1 := P(X)/(X-X_0)$ over $k_1 := k(X_0)$, so $G$ is 2-transitive iff $P_1$ is irreducible over $k_1$, in which case we can adjoin a second root, etc.; if $P_3$ is irreducible then $G$ is 4-transitive, and you're done (except in the four Mathieu cases where you must go one or two steps further). Again see [A], in particular Section 4 "Throwing away roots" (p.69).
This also has the following amusing consequence. Let $P\phantom.$ be an irreducible separable polynomial over $k$, and define $P_1, P_2, \ldots, P_n$ as before, as long as $P_m$ is irreducible for each $m<n$. Then:
i) If each of $P_1,P_2,P_3$ is irreducible but $P_4$ is reducible then $\deg P \phantom. \in \lbrace 6, 11, 23 \rbrace$.
ii) If each of $P_1,P_2,P_3,P_4$ is irreducible but $P_5$ is reducible then $\deg P \phantom. \in \lbrace 7, 12, 24 \rbrace$.
iii) Suppose $n \geq 5$. If $P_m$ is irreducible for each $m \leq n$, but $P_{n+1}$ is reducible, then $\deg P \phantom. = n + 3$.
Moreover, each of the allowed possibilities for $\deg P\phantom.$ in (i), (ii), and (iii) occurs for suitable $k$ and $P$.
This statement does not explicitly mention finite groups at all, but given Galois theory it is equivalent to the classification of 4-transitive permutation groups.
Half of my PhD was a proof that any smooth plane quartic can be recostructed by It's bitangents. This is classical if you know which bitangents corresponds to which odd theta characteristics. Here is the heart of the proof:
Say that A is moduli of sets of bitangents + level structure, B is the moduli of sets of bitangents, so A->B is Galois with Galois group at least SP(6,2) (the level-2 deck group), and at most S_28. I showed that this group is 2-transtivie, which meant that it is either SP(6,2), A28 or S28. Then I showed that it is not the last two, which brought it to the classical case.
