*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 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, 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.)

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.)

isprimitive: If $\Delta$ is a block, then $\Delta \cap \operatorname{Mov} \sigma$ is empty or a block for $\langle \sigma \rangle$, and the same holds for $\tau$. By going through the different possibilities, I get that $\Delta$ must be trivial (or did I forget a case?). Of course, degrees p, p+1, p+2 remain. (I would be interested in a hint or sketch of your proof, by the way.) $\endgroup$ – Frieder Ladisch Jun 19 '14 at 15:20