Let $G$ be a finite permutation group on $\Omega$ acting nonregular and transitive such that each nontrivial element fixes at most two points of $\Omega$.

Suppose that for $\alpha \in \Omega$ the stabilizer $G_{\alpha}$ has even order and $|\Omega|$ is even too. Let $S \in \mbox{Syl}_2(G)$ and suppose $S_{\alpha} \ne 1$, but $S \nleq G_{\alpha}$ and that there exists $\beta \in \Omega$ distinct form $\alpha$ such that $S_{\alpha} = S_{\beta}$ with $|S : S_{\alpha}| = 2$ and some element of $S$ interchanges $\alpha$ and $\beta$ (note that by normality of $S_{\alpha}$ this holds for each $S \setminus S_{\alpha}$).

As $S_{\alpha}$ already has two fixed points, this subgroup must have regular orbits (i.e. acts semiregular) on the remaining points of $\Omega$. It follows that $\{\alpha,\beta\}$ is the unique $S$-orbit of length $2$ and all other orbits have length $|S|$.

I do not see why all other $S$-orbits have length $|S|$? As $S_{\alpha}$ is normal in $S$ the $S_{\alpha}$ orbits form a system of blocks for each $S$-orbit and hence have all equal size on each $S$-orbit, so if $\gamma \notin \{\alpha,\beta\}$ then $$ |\gamma^S| = |S : S_{\gamma}| = k \cdot |S_{\alpha} : S_{\alpha}\cap S_{\gamma}| = k\cdot |S_{\alpha}| $$ for some $k$. Now $|S| = 2|S_{\alpha}|$, so that $k \in \{1,2\}$. What I want is $k = 2$ (which is equivalent to $S_{\gamma} = 1$, i.e. $S$ itself acts regular on the orbit $\gamma^S$). If $k = 1$ we must have $|S_{\gamma}| = 2$ and as $S_{\alpha} \cap S_{\gamma} = 1$ we have $S = S_{\gamma}S_{\alpha}$. Further $\gamma^S = \gamma^{S_{\alpha}}$, so that $S_{\alpha}$ already acts transitive on the orbit. But I do not see where this all might lead to a contradiction, thereby excluding the case $k = 1$?

Do you see why $|\gamma^S| = |S|$? Can you help me seeing it!?

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    $\begingroup$ What about $G \cong A_4$ with $|\Omega|=6$, $|G_\alpha|=2$. Unless I have overlooked something, that satisfies the hypotheses. Then $S$ has order $4$ and has three orbits each of length $2$. $\endgroup$ – Derek Holt Nov 10 '15 at 22:34
  • $\begingroup$ Why $|G_{\alpha}| = 2$? A point stabilizer of $A_4$ has order three, for example for $\alpha = 4$ we have $G_4 = \langle (1 ~ 2 ~ 3) \rangle$, or not? $\endgroup$ – StefanH Nov 10 '15 at 22:51
  • $\begingroup$ Derek is considering a different action of $A_4$, on the cosets of the group generated by a double transposition. (Equivalently, acting on the 6 unordered pairs of the natural $4$-set.) This group is imprimitive with three blocks of size 2, and the Sylow $2$-subgroup is normal, with these blocks as orbits. It does seem like a counter-example. $\endgroup$ – verret Nov 11 '15 at 6:07
  • $\begingroup$ $S_4$ acting on the $6$ cosets of $\langle (1,2),(3,4) \rangle$ is also a counterexample. Here $|S|=8$, and $S$ has orbits of lengths $2$ and $4$. $\endgroup$ – Derek Holt Nov 11 '15 at 9:28
  • $\begingroup$ Okay, this is truly an error in the paper. Thank you all! And Derek, I would prefer if you put your counterexample in an answer so I can award the bounty... $\endgroup$ – StefanH Nov 11 '15 at 10:39

The transitive action of $A_4$ of degree $6$ on the cosets of the subgroup $\langle (1,2)(3,4) \rangle$ is a counterexample. Here we have $|S_\alpha|=2$, $|S|=4$, and $S$ has three orbits of zise $2$. You could take $$G = \langle (1, 2)(5, 6), (1, 3, 5)(2, 4, 6) \rangle$$ with $$S = \{ 1,\,(1,2)(3,4),\,(1,2)(5,6),\,(3,4)(5,6) \}.$$

Another counterexample, which contains this one, is $S_4$ acting on the $6$ cosets of $\langle (1,2),(3,4) \rangle$. Here $|S|=8$, and $S$ has orbits of sizes $2$ and $4$. This is the intersection of $S_2 \wr S_3$ with $A_6$.

I don't know whether there are any larger counterexamples - possibly not.

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