# Question about doubly transitive groups with an n-cycle

Let $G$ be a doubly transitive subgroup of $S_n$ which contains an $n$-cycle, and let $G_{12}$ be the subgroup of $G$ consisting of all elements $g\in G$ for which $g(1)=1$ and $g(2)=2$. Then $G_{12}$ has an orbit on $\{3,4,...,n\}$ which is preserved by $N_{12}$, where $N$ is the normalizer of $G$ in $S_n$.

I can prove this by using the classification of doubly transitive groups with an $n$-cycle, which depends on the classification of finite simple groups. But I think there should be a 5-line proof which uses nothing more than the definition of double transitivity. Can anyone find such a proof?

The reason I believe that such a proof should exist is as follows. The main result of a 1954 paper by Uchiyama asserts that if $f(x)$ is a degree-$n$ polynomial in $\mathbf{F}_q[x]$ for which $\displaystyle\frac{f(x)-f(y)}{x-y}$ is absolutely irreducible, and if in addition the characteristic of $\mathbf{F}_q$ is sufficiently large compared to $n$, then the image set $f(\mathbf{F}_q)$ contains at least $q/2$ elements. In fact Uchiyama gives only part of the proof, and says he will give the full proof elsewhere; but his published papers do not contain the full proof. I can deduce Uchiyama's result from the group-theoretic assertion in a few lines, using tools Uchiyama knew. So I think this might be the proof he had in mind. Since he did not indicate that he used any nontrivial group theory, I suspect that if he had a proof of the group-theoretic result then it must have been very simple. The group-theoretic result is stronger than what Uchiyama needed, but still it seems like the most natural statement he might have tried to prove. I note that nowadays one can prove much more precise results than Uchiyama's by using tools developed in the past 60 years; but still it would be interesting to know whether there is indeed a simple proof along the lines I suggested, in case such a proof might provide further insights.

• Start aiming for a (to give room) 15 line proof that begins "By double transitivity of G, the orbit / action of G12 on orbit is ...", and see where you get stuck. Then ask about that point on MO. – The Masked Avenger May 5 '14 at 15:39

This is not an answer, rather a comment about this appealing question: I believe that an elementary argument would be surprising, because the presence of an $n$-cycle is a condition which is hard to use, and the assertion doesn't hold anymore without this assumption.

Here is one potential approach: If $G=\text{AGL}_1(\mathbb F_p)=C_p\rtimes C_{p-1}$ for a prime $p$, then $G=N$ and there is nothing to prove. Also, the case that $G$ is triply transitive is trivial. So assume that neither holds. Then, again by the classification of the simple groups, $G_{12}$ has just $2$ orbits on $\{3,4,\dots,n\}$. So it would be interesting to have an easy proof for that independent of the finite simple groups classification! Anyway, to finish things, one would have to show that $N_{12}$ can't switch these two orbits. This is easy if $G_{1}$ is imprimitive on $\{2,3,4,\dots,n\}$. If $G_{1}$ is primitive though, maybe Higman's theory helps.

• How can $G_{12}$ be primitive on $\{3,4,\ldots,n\}$? It is not even transitive. – Derek Holt May 5 '14 at 16:25
• Perhaps representation theory can be used? The number of orbits of $G_{12}$ on $\{3,4,\ldots,n\}$ is the number of non-trivial irreducible constituents of the permutation module for $G_1$ acting on $\{2,\ldots, n\}$. On the other hand, 2-transitivity means that the permutation module for $G$ has only one non-trivial irreducible constituent. $N_1$ must permute the non-trivial irreducible constituents of the permutation module for $G_1$ on $\{2,\ldots, n\}$ and this is tied to the action of $N_{12}$ on the irreducible constituents by the connections between orbitals, suborbits & centralizers. – Benjamin Steinberg May 5 '14 at 19:13
• @Ben: one difficulty is that the conclusion is not true for arbitrary 2-transtive groups; somehow the $n$-cycle hypothesis must be used, but it isn't clear how to make use of it. – Michael Zieve May 6 '14 at 1:39
• Agreed. That is the tricky business. – Benjamin Steinberg May 6 '14 at 1:49