I noticed that the following is true, and that there is a reasonably elementary proof of it (in particular, the classification of finite simple groups is not needed). Let $G$ be a finite permutation group which contains two $p$-cycles which do not commute (where $p$ is any odd prime other than a Mersenne prime). Then $G$ is not solvable ( more precisely, $G$ has a non-Abelian composition factor of order divisible by $p$). Since the methods are reasonably elementary, I wonder if anyone has come across this or similar results (possibly in a Galois Theory context) and can point me to a reference? 


For every Mersenne prime $p,$ there is a solvable permutation group $G$ of degree $p+1$ and order $p(p+1)$ which contains two non-commuting $p$-cycles.


(Later note: The analogous result is not true for $p^{2}$-cycles ($p$ prime). For $p=2$ take $G = S\_{4}$ and for $p>2$ take $G$ to be a Sylow $p$-subgroup of $S\_{p^{2}}.$ In each case, $G$ is a solvable (even nilpotent when $p$ is odd) permutation group containing two $p^{2}$-cycles which do not commute).



Permutation Groups Containing non-commuting $p$-cycles
*Fixed several inaccuracies, many thanks to Frieder Ladisch for spotting them:*


Let $x$ and $y$ be two non-commuting $p$-cycles, $G=\langle x,y\rangle$, and $G$ be considered as a transitive permutation group on the support $\Omega$ of $G$.


We show that either $\text{AGL}\_1(\mathbb F\_q)\le G$ for a Mersenne prime $q$, or $G$ is simple non-abelian. 


*Proof.* $G$ is primitive (as pointed out by Frieder Ladisch already). This can be seen as follows: Let $\Delta$ be a block of a non-trivial block system. The action of the $p$-cycle $x$ on the block system is trivial, for otherwise $x$ would move $p\lvert\Delta\rvert>p$ points. On the other hand $G$ transitively moves the blocks, a contradiction.


Next we show that $G$ is doubly transitive. This follows from Burnside's classical theorem if $\lvert\Omega\rvert=p$. If $\lvert\Omega\rvert>p$, then the pointwise stabilizer in $G$ of the points fixed by $x$ is transitive (via $x$) on the remaining points. By [Jordan's Theorem](http://books.google.de/books?id=4bNj8K1omGAC&pg=PA179&lpg=PA179&dq=%22jordan%20set%22%20doubly%20transitive&source=bl&ots=q4i_hG76Rd&sig=R-OGPQCPeMKXGvgBcoWjoBFPz6c&hl=de&sa=X&ei=klmpU4isLMj44QTrlICoCg&ved=0CDkQ6AEwBA#v=onepage&q=%22jordan%20set%22%20doubly%20transitive&f=false) on primitive groups with Jordan sets we again see that $G$ is doubly transitive.


Let $N$ be a minimal normal subgroup of $G$. By Burnside, $N$ is either (a) elementary abelian and regular, or $N$ is (b) simple, primitive, and not regular. The Mersenne exceptions follow from looking at $p$-cycles in $\text{GL}\_m(q)$ where $n=q^m$ for a prime $q$: Such a $p$-cycle fixes $q^r<q^m$ points, hence $p=q^r(q^{m-r}-1)$. We obtain $r=0$ and $q=2$. By Schur's Lemma, we can identify $\langle x\rangle$ with the multiplicative group of $\mathbb F\_q$. This yields $\text{AGL}\_1(\mathbb F\_q)\le G$ as claimed. (If one uses [Kantor's paper on Singer cycles](http://dx.doi.org/10.1016/0021-8693%2880%2990214-8), then one gets more precisely the possibilities $G=\text{AGL}\_1(\mathbb F\_q)$, $\text{A$\Gamma$L}\_1(\mathbb F\_q)$, and $\text{AGL}\_m(\mathbb F\_2)$. However, that paper relies on a wrong paper of Cameron/Kantor, see [here](http://www.maths.qmul.ac.uk/~pjc/odds/antiflag.pdf).)


Now assume case (b). We show that $G=N$, so $G$ is actually simple. In order to do so, we show that $p$ divides the order of $N$. Note that $\lvert\Omega\rvert<2p$, so $p^2$ does not divide $\lvert N\rvert$, hence $x,y\in N$ in this case.


The case $p=\lvert\Omega\rvert$ is clear.


So $p<\lvert\Omega\rvert$ from now on. We let $\omega$ be a fixed point of $x$, and set $\Omega'=\Omega\setminus\{\omega\}$.


First suppose $p=\lvert\Omega\rvert-1$. The point stabilizer $N\_\omega$ is a normal subgroup of $G\_\omega$. As $G\_\omega$ is transitive on $\Omega'$, all orbits of $N\_\omega$ on $\Omega'$ have equal length. But $\lvert\Omega'\rvert=p$, so these orbit lengths are either $1$ or $p$. The former cannot hold, because then $N$ were regular on $\Omega$. Thus $N$ is doubly transitive on $\Omega$, so $p=\lvert\Omega\rvert-1$ divides $\lvert N\rvert$.


If $p<\lvert\Omega\rvert-2$, then $N=\text{Alt}(\Omega)$ by Jordan, so $G=N$ again.


If $p=\lvert\Omega\rvert-2$, then $G\_\omega$ contains the cycle $x$ of length $p=\lvert\Omega'\rvert-1$ on $\Omega'$, so $G\_\omega$ it is doubly transitive on $\Omega'$. The argument in the case $p=\lvert\Omega\rvert-1$ shows that $N\_\omega$ is either doubly transitive on $\Omega'$, or regular. In the former case $p=\lvert\Omega'\rvert-1$ divides $\lvert N\_\omega\rvert$, hence $G=N$ again. In the latter case, $N$ is sharply doubly transitive on $\Omega$, which implies that $N$ has a regular normal subgroup, contrary to $N$ being simple. (For this last step, simple counting suffices. One does not need Frobenius' theorem about the existence of Frobenius kernels.) 


