Let $p$ be a prime. Let $G$ be a solvable, non-regular, transitive permutation group such that some element fixes no point, and each element fixing some point fixes exactly $p$ points. Suppose that for $g \notin N_G(G_{\alpha})$ we have $$ G_{\alpha}^g \cap G_{\alpha} = 1 $$ and that $p$ does not divide the order of $G_{\alpha}$. Then $G$ has a regular normal subgroup $R$ or $G$ has an intransitive normal subgroup $F$ of index $p$ which acts as a Frobenius group on its $p$ orbits.

Proof: Let $G$ be a counter-example of minimal order. Let $N$ be an elementary abelian normal subgroup of $G$. As $N_{\alpha} \le G_{\alpha}$, but $p$ does not divide the order of $G_{\alpha}$ we have $N_{\alpha} = 1$, i.e. $N$ acts semi-regularly on $\Omega$. Since $G$ is a counter-example, $N$ is not transitive. Let $\Sigma$ be the orbits of $N$, and let $\Delta$ be the orbit containing $\alpha$. Then $G/N$ acts on the set $\Sigma$ and $G_{\alpha}N / N$ is the stabilizer of $\Delta$. Let $\overline \alpha$ be the set of fixed points of $G_{\alpha}$. Since $G$ acts transitive, the size of each block divides the order of $G$, and hence $\overline \alpha$ is a minimal $G$-block. Also since the orbits of $N$ are $G$-blocks, we have that either $\Delta \cap \overline \alpha = \overline \alpha$ or $\Delta \cap \overline \alpha = \{\alpha\}$.

Let $g \in G_{\alpha}$ and assume that $g$ fixes an orbits $\Delta'$ of $N$, i.e. $\langle g \rangle$ opertes on $\Delta'$. If $g$ would have no fixed point, then we would have by an orbit decomposition that $|\Delta'| = k\cdot o(g)$ for some $k$. But $|\Delta'| = |N|$, which is a $p$-group, and $o(g)$ does not divide $p$, so $g$ must have a fixed point in $\Delta'$ (which must come from $\overline a$). Hence if $\overline \alpha \subseteq \Delta$, $G / N$ acts as a Frobenious group on $\Delta$ and the action of $G/N$ on $\Sigma$ is faithful.

I do not understand why $G/N$ acts as a Frobenious group, i.e. that it is non-regular and each $gN$ fixes at most one point. If $gN$ fixes some point, then this is equivalent to the statement that $g$ fixes some orbit $\Delta'$ of $N$, but in general we do not have $g \in G_{\alpha}$, so I do not see how the previous comments apply?

Let $R / N$ be the Frobenious kernel. Then $R$ acts regularly on $\Omega$, which is impossible.

If $r \in R, r \ne 1$ would fix some $\beta^r = \beta$ and $\Delta'$ denotes the orbit of $N$ containing $\beta$, then as this is a block we would have $(\Delta')^r = \Delta'$, i.e. $rN$ fixes a point, which is not possible by definition of the kernel. So this is clear to me.

Therefore each of the orbits of $N$ contains at most one of the points of $\overline \alpha$ and $G / N$ acts in such a way that some element fixes no point, and each element fixing some point fixes exactly $p$ points.

Here I do not understand why $G/N$ acts in that way?

If $|\Sigma| = p$ then $G$ would not be a counter-example.

What condition is precisely violated so that this is not a counter-example?

Hence $|\Sigma| > p$ and $G/N$ acts faithfully on $\Sigma$. Therefore the assertion holds for $G/N$.

I do not see why the point stabilizers are not divisible by $p$ and the intersection of different conjugates of the point stabilizers is trivial?

Again, if $R/N$ acts regularly on $\Sigma$ then $R$ acts regularly on $\Sigma$, which is impossible. Thus $G/N$ has an intransitive normal subgroup $F/N$ of index $p$ acting as a Frobenious group on its $p$ orbits. Each of the $p$ orbits of $F$ on $\Sigma$ contains exactly one of the points of $\overline \alpha$. Hence $F$ acts as a Frobenious group on $\Omega$, and we have a contradiction, proving the assertion. $\square$

I have some difficulty following the arguments in the proof. I have marked the parts I am unsure about. Maybe some additional remark, by saying $\overline \alpha$ is the set of fixed points of $G_{\alpha}$ we used the fact that each element from the point stabilizer has the same set of $p$ fixed points, i.e. the set $\overline \alpha$. This uses the assumption that different point stabilizers intersect trivial, I can supply a proof of this fact if wanted!

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    $\begingroup$ Just a minor comment: in several of your questions you have a transitive permutation group and you have assumed that some element fixes no point, But that is true of any transitive permutation group of degree greater than $1$, so there is no need to assume that! $\endgroup$ – Derek Holt Nov 1 '15 at 17:07
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    $\begingroup$ I started looking through your proof, but I could not follow the third sentence in the proof. You don't know at this stage that $N$ is a $p$-group for your given prime $p$. $\endgroup$ – Derek Holt Nov 1 '15 at 17:11
  • $\begingroup$ Regarding your first comment: Yes, that each transitive permutation group has a fixed point free element is a classical result by Jordan which I forgot by the moment, thanks for pointing out! For your second question, really that I asked myself too! This is the single other unclear point for me, but I choosed not to ask this to focus on the other points (I hoped that this point becomes clearer later as I started reading and learning about solvability, and I guessed it has something to do with solvability, as solvable groups contain elementary abelian normal subgroups for some $q$ (continue). $\endgroup$ – StefanH Nov 1 '15 at 17:23
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    $\begingroup$ OK, I understand that now. The orbits of $N$ all have the same length and, since $N$ is abelian, $N$ acts regularly on each of its orbits. Now if $N_\alpha \ne 1$, then $N$ fixes exacly $p$ points (i.e. $\overline{\alpha}$), but then $\overline{\alpha}$ is a union of orbits of $N$, so $N$ is indeed a $p$-group, and their argument works. $\endgroup$ – Derek Holt Nov 1 '15 at 17:59
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    $\begingroup$ Suppose $H$ is a transitive non-regular permutation group on a set $X$, and there exists $x \in X$ with the property that any element of $H$ that fixes $x$ fixes no other point of $X$. Then $H$ is a Frobenius group. Just apply this with $H=G/N$, $X=\Sigma$, and $x = \Delta$. $\endgroup$ – Derek Holt Nov 1 '15 at 20:41

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